Class 8 Maths

Dividing A Line Segment Into Three Or Five Equal Parts

Question 1. I with the help of pencil, scale and compass, trisect a straight line 9 cm long and will measure the length of each part by scale.
Solution:

A 9 cm long straight line AB is drawn. A On point A of any measure ∠BAC is drawn. At AB on point B equal to ∠BAC, ∠ABX is drawn.

From straight line AC taking same radius two equal parts AD and DE are cut.

From straight line BD taking same radius two equal parts BF and FG cut. E, F and D, G are joined. EF and DG both straight lines cut AB straight line respectively points P and Q points.

AB straight line is divided into three equal parts on points P and Q.

“WBBSE Class 8 Maths Chapter 23 solutions, Dividing a Line Segment into Three or Five Equal Parts”

i.e., AP = PQ = QB = \(\frac{1}{3}\) AB = \(\frac{1}{3}\) × 9 cm = 3 cm

Length of each part is 3 cm.

Read and Learn More WBBSE Solutions For Class 8 Maths

Dividing A Line Segment Exercise 23.1

Question 1. Rihana draw a straight line PQ of length 10 cm. I divided the segment PQ into five equal parts by scale and compass. Let’s verify by scale that each part of the segment is 2 cm or not.
Solution:

Given

Rihana has drawn in her copy a 10 cm long straight line PQ. I divided the segment PQ into five equal parts by scale and compass.

At PQ on point P an acute angle ∠RPQ is drawn and on PQ straight line on the opposite side of ∠RPQ equal to that angle ∠PQS is drawn.

Now from straight line PR taking the same radius four equal parts PA, AB, BC and CD are cut.

Similarly the straight line QS, taking same radius, is cut into four equal parts QE, EF, FG and GH. A,H; B,G; C,H and D,E are joined which cut PQ straight line at points I, J, K and L respectively.

PQ straight line is divided by I, J, K and L points into five equal parts and length of each part 2 is cm,

i.e., PI = IJ=JK=KL=LQ= \(\frac{1}{5}\) PQ = \(\frac{1}{5}\)×10 cm

= 2 cm.

“WBBSE Class 8 Maths Chapter 23 solutions, Dividing a Line Segment into Three or Five Equal Parts PDF”

Each part of the segment is 2 cm.

Question 2. Ajij divided a line segment XY of length 12 cm in some parts in such a way that each part is of length 2.4 cm. First, decide how many parts are there and then divide the line XY into that number of equal parts.
Solution:

Given

Ajij divided a line segment XY of length 12 cm in some parts in such a way that each part is of length 2.4 cm.

To divide 12 long straight line into parts of length 2.4, it should be cut into 5 equal parts.

∴ Ajij will cut XY straight line into 5 equal parts.

At point X on XY an acute angle ∠ZXY is drawn and on XY straight line on the opposite side of ∠ZXY, equal to that angle ∠XYP is drawn.

Now from straight line XY taking same radius, it is cut into 4 equal parts XA, AB, BC and CD.

Similarly taking the same radius YP straight line is cut into 4 equal parts YE, EF, FG and GH. A, H; B,G; C,H and D,E are joined by which XY straight line is divided into five equal parts at points l, J, K and L; and length of each part is 2.4 cm, i.e.,

XI = IJ = JK= KL= LY = \(\frac{1}{5}\) XY = \(\frac{1}{5}\)×12 cm

=2.4 cm.

“Class 8 WBBSE Maths Chapter 23 solutions, Dividing a Line Segment study material”

Question 3. Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then draw the median AD. I trisect AD in AE, EF and FD. Now we join B and F by scale and extend it to X such that it intersects AC at X.
Solution:

Given

Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then draw the median AD.

I trisect AD in AE, EF and FD. Now we join B and F by scale and extend it to X such that it intersects AC at X.

Measuring by scale we see,

AX =  CX [Let’s put the number]

In ΔABC side BC is bisected with the help of a pencil and compass and the AD median is drawn. Median AD is divided into three parts AE, EF and FD with help of scale and compass.

Now, with help of scale, I join two points B and F, and extend them which cut a straight line AC at point X.

Measuring by scale we see,

AX = CX, i.e., AX= 1 CX

Question 4. Let’s divide a line segment of length 12.6 cm into 7 equal parts, by scale and compass. Using the above construction, let’s draw an equilateral triangle of length 7.2 cm.
Solution:

A 12.5 cm long straight line AB is drawn. At point A on AB an acute angle ∠XAB is drawn and on AB on the opposite side of ∠XAB, equal to that angle, ∠ABY is drawn.

Now from AX straight line taking same radius six equal parts AC, CD, DE, EF, FG and GH are cut off.

Similarly by taking same radius BY straight line is cut into six equal parts Bl, IJ, JK, KL, LM and MO. C,0 ; D,M; E,L; F,K; G,J and H,l are joined by which AB which straight line is divided into 7 equal parts respectively P, Q, R, S, T and U points and length of each part is 1.8 cm. i.e.,

AP = PQ = QR = RS = ST = TU = UB = 1.8 cm

“Class 8 WBBSE Maths Chapter 23, Dividing a Line Segment into Equal Parts easy explanation”

AS = 7.2 cm. Now taking radius equal to AS and taking centre A and S two arcs are drawn which cut each other at W point. A, W and S, W are joined.

ASW equilateral triangle is formed whose length of each side is 7.2 cm.

Question 5. Rama pradhan drew a parallelogram ABCD where AB = 6 cm, BC = 9 cm and ABC = 60°. Let’s select two points P and Q on the diagonal BD such that BP = PQ = QD. Joining the points A, P; P, C; C, Q and Q, A, let’s write the nature of quadrilateral APCQ.
Solution:

Given

Rama pradhan drew a parallelogram ABCD where AB = 6 cm, BC = 9 cm and ABC = 60°.

Let’s select two points P and Q on the diagonal BD such that BP = PQ = QD. Joining the points A, P; P, C; C, Q and Q, A,

ABCD is a parallelogram whose sides AB = CD = 6 cm and BC = AD = 9 cm and ABC = 60°. I choose with help of pencil, scale and compass, on BD diagonal of the parallelogram ABCD two points at P and Q such that BP = PQ = QD.

Now joining A,P; P,C; C,Q and Q, A, APCQ quadrilateral is formed. In it ∠AQC = 130°, ∠PAQ = ∠PCQ = 50° and AP = CQ and AQ = CP.

∴ It is a parallelogram.

“WBBSE Class 8 Maths Chapter 23, Dividing a Line Segment into Three or Five Equal Parts solved examples”

Question 6. Sujata drew three line segments of length 4 cm, 6 cm and 10 cm. By scale and compass, Rahul bisects the 1st line segment, trisects the 2nd line segment and divides the 3rd line segment into five equal parts. Sabnarh drew a triangle PQR taking \(\frac{1}{3}\) th of 1 1 the first line segment, \(\frac{1}{5}\) th of the 2nd line segment and \(\frac{1}{2}\) th of the 3rd line segment. Let’s classify the triangles on the basis of their sides.
Solution:

Given

Sujata drew three line segments of length 4 cm, 6 cm and 10 cm.

By scale and compass, Rahul bisects the 1st line segment, trisects the 2nd line segment and divides the 3rd line segment into five equal parts.

Sabnarh drew a triangle PQR taking \(\frac{1}{3}\) th of 1 1 the first line segment, \(\frac{1}{5}\) th of the 2nd line segment and \(\frac{1}{2}\) th of the 3rd line segment.

Half of the length of the first straight line = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) × 4 cm = 2 cm

“WBBSE Class 8 Dividing a Line Segment solutions, Maths Chapter 23”

= \(\frac{1}{3}\) rd of the length of the 2nd straight line = \(\frac{1}{3}\) PQ = \(\frac{1}{3}\) × 6 cm = 2 cm

= \(\frac{1}{5}\) th of length of the third straight line = \(\frac{1}{5}\) XY = \(\frac{1}{5}\) × 10 cm = 2 cm

∵ Half of the length of the first straight line, \(\frac{1}{3}\) rd of that of the second and \(\frac{1}{5}\) th of that of the third, are equal.

∴ PQR is an equilateral triangle whose length of each side is 2 cm.

∴ Shabnam has drawn equilateral triangles.

A Few Interesting Problems

Question 1. Interesting problems with matchsticks.

1. I constructed an equilateral triangle with three matchsticks.
Solution:

2. My brother constructed 6 equilateral triangles with 12 matchsticks.
Solution:

Replacing only 4 sticks from the 12 sticks I will construct 3 equilateral triangles where their areas are not equal.

The first triangle is A, the Second triangle is (A + B) and 3rd triangle is (A + B + C). All three are equilateral triangles whose measurements are not equal.

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. Megha arranged 26 matchsticks as shown in the figure beside.
Solution:

“WBBSE Class 8 Maths Chapter 24 solutions, A Few Interesting Problems”

Replacing only 14 matchsticks in this arrangement I constructed 3 squares where their areas are not same.

A, B, and C are three squares whose areas are not the same.

Question 3. Rokeya had 20 matchsticks. She made a square-type house with 4 matchsticks as the figure beside and she also made a square-type fence around the garden with the remaining 16 sticks.
Solution:

Given

Rokeya had 20 matchsticks. She made a square-type house with 4 matchsticks as the figure beside and she also made a square-type fence around the garden with the remaining 16 sticks.

Let’s divide the garden into five equal shapes and sizes with 10 other matchsticks.

I inserted Rokeya’s arrangement with 10 new sticks 1, 2, 3, 4, 5, 6, 7,8, 9, and 10; and divided this garden into five equal parts A, B, C, D, and E.

“Class 8 WBBSE Maths Chapter 24 solutions, Interesting Problems study material”

Question 4. Let’s put numbers from 1 to 19 in the circles of the following wheel in such a way that the sum of the numbers in 3 circles along a straight line segment is 30.
Solution:

Question 5. 

1. 

Solution:

= \(\frac{3}{2}-1=\frac{1}{2}\)

= \(\frac{8}{3}-2=\frac{2}{3}\)

= \(\frac{19}{3}-3=\frac{10}{3}\)

2. 

Solution:

(4+9) × 2 = 26

(9 + 16) × 2 = 50

(16 + 4) × 2 = 40

Question 6. 

1.

Solution:

\((2)^2+(4)^2=20\),

\((1)^2+(5)^2=26\),

\((3)^2+(9)^2=90\)

\((1)^3+(2)^2=9\),

\((4)^3+(5)^3=189\),

\((2)^3+(3)^3=35\)

Question 7.

1. 

Solution:

1. 7×3 + 8 = 29

2. 4×3 + 7 = 19

3. 5×? + 6 = 31

∴ ? = (31 – 6)÷5 = 5

2.

Solution:

1. 4×2 – 1 =7

2. 5×3-3 = 12

3. 6×7 – ? = 39

∴ ? = 42 – 39 = 3

Solution:

1. (10-9)+(15-12)

Solution:

(10-9)+(15-12) = 1+3

(10-9)+(15-12) = 4

(10-9)+(15-12) = 4

“WBBSE Class 8 Maths Chapter 24, A Few Interesting Problems solved examples”

2. (16-20)+(28-12)

Solution:

(16-20)+(28-12)  =-4+16

(16-20)+(28-12)  =12

(16-20)+(28-12) =12

3. (15-16)+(23-11)

Solution:

(15-16)+(23-11) =-1+12

(15-16)+(23-11) =11

(15-16)+(23-11) =11

Let’s see the rules of the game and find out the correct numbers.

Question 1. If the sign ‘÷’ stands for the sign ‘×’ and the sign ‘+’ stands for the sign ‘÷’ and the sign ‘#’ stands for the sign ‘+’, then let’s write which of the following numbers will be the value of 2÷5+5#100.

1. 100
2. 102
3. 108
4. 105

Solution:

2 × 5 ÷ 5 + 100

= 2 × 1 + 100

= 102

Question 2. If 7 * 1= 64 and 3*9=144, then let’s write which one of the following numbers will be the value of 5*6 :

1. 2
2. 45
3. 101
4. 121

Solution:

7*1 = 64

= \((7+1)^2\) = 64 :

= \((3+9)^2\) = 144

∴  5 * 6

= \((5+6)^2\)

= 121

Question: 3. If 84 ‘+’ 72 = 45 and 73 ‘×’ 41=43, then find the value of 94 ‘×’ 72 from the following:

1. 55
2. 59
3. 56
4. 66

1. 84 = 8-4 = 4

Solution:

72 = 7-2 = 5

∴ Number = 45

2. 73 = 7-3= 4

Solution:

41=4-1=3

∴ Number = 43

3. 94 = 9-4 = 5

Solution:

72 = 7-2 = 5

∴ Number = 55

“WBBSE Class 8 Interesting Problems solutions, Maths Chapter 24”

Question 4. If the sign ‘÷’ and the sign ‘+’ and the numbers ‘6’ and ‘3’ interchange their positions, then find which one of the following relations is true-

1. 3+6-2=5

Solution:

=6÷3 + 2

= 2 + 2

= 4

2. 6÷3+2=8

Solution:

= 3 + 6 ÷ 2

= 3 + 3

= 6

3. 3+6÷5=7

Solution:

6÷3 + 5

=2+5

=7

4. 3÷6+1 =6

Solution:

= 6 + 3÷1

= 6 + 3

= 9

∴ 3. Relation is true

Question 5. If the sign ‘+’ and the sign ‘-‘ and the numbers ‘4’ and ‘8’ interchange their positions, then find which of the following relations is true

1. 4+8-12=16

Solution:

8-4+12

=20-4

= 16

2. 4-8+12=6

Solution:

8 + 4-12

= 12-12

= 0

3. 8+4-12=24

Solution:

4 – 8 + 12

= -4+12

= 8.

4. 8-4+12=8

Solution:

4 + 8-12

= 12-12

= 0

∴ 1. Relation is true

Question 6. Let’s find out a few more interesting numbers –

Solution:

Let’s see why 142857 is a (Revolving Number) –

142857×1 = 142857

142857×2 = 285714

142857×3 = 428571

142857×4 = 571428

142857×5= 714285

142857×6= 857142

Question 7. Let’s find 24 using a one-digit number thrice: 3³-3 = 24 Let’s find 24 using another one-digit number thrice except 3.

Solution:

8 + 8 + 8 = 24

8. Let’s find 30 using a one-digit number thrice: 3³+3 = 30. Let’s find 30 using another one-digit number thrice except 3.

Solution:

6 x 6 – 6 = 30

5 x 5 + 5 = 30

9. Imran arranged 8 pieces of papers in two columns writing the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9.

Solution:

Let’s try to get the same sum from the columns replacing only 2 pieces of paper.

Question 10. Maria’s father kept an envelope with some money on the table for 10.buy books and go to the office. Maria returned home and saw that 98 was written on the envelope. So she went to the bookshop and bought a book priced at Rs. 92. However, at the time of paying the money, she saw that there were less than 92 rupees in the envelope. How did this happen? Let’s think and write.
Solution:

Given

Maria’s father kept an envelope with some money on the table for 10.buy books and go to the office.

Maria returned home and saw that 98 is written on the envelope. So she went to the bookshop and bought a book priced at Rs. 92.

“Class 8 WBBSE Maths Chapter 24, A Few Interesting Problems easy explanation”

However, at the time of paying the money, she saw that there were less than 92 rupees in the envelope.

Maria’s father had written 86 on the envelope but Maria had seen the envelope in othe pposite direction.

Construction Of Parallel Lines

Question 1. With a scale and pencil let’s join two points C and Q and produce the line obtained on both sides. Thus, I got a straight-line PR.
Solution:

Now I prove logically step-by-step that PR||AB.

I joined D,Q and O,Q.

In Δ s CDQ and DOQ.

DC = OQ, CQ= DO and DQ is their common side.

∴ ΔCDQ = ΔDOQ (by s-s-s criteria of congruency)

∴ ∠CQD = ∠QDO, but they are alternate angles.

∴ CQ || DO

Therefore, PR || AB

Read and Learn More WBBSE Solutions For Class 8 Maths

∴ I got straight line PR through the point C, which is parallel to the straight line AB, i.e., I draw a line PR through the point C which was not on the given line AB and parallel to the line PR AB.

Parallel Lines Exercise  22.1

Question 1. Let’s see how many line segment parallel to the line XY are possible to draw through the point Z which is not on XY.
Solution:

On straight line XY a point Q is taken. Points Z and Q are joined, Consequently ∠ZQY is formed. Now on point Z on the opposite side of ∠ZQY ∠QZP is drawn equal to angle ∠ZQY.

Joining P and Z with scale and extending on both sides AB straight line is got.

∵ ∠PZQ = ∠ZQY, are alternate angles, AB // XY

Now outside the line XY from a point Z, no other parallel line can be drawn because the alternate angle at Z will not be equal in measurement.

∴ From point Z parallel to XY only one straight line can be drawn.

“WBBSE Class 8 Maths Chapter 22 solutions, Construction of Parallel Lines”

Question 2. Habib has drawn a line segment PQ in his exercise book and he has also considered a point R outside the line segment PQ. Let’s draw a line parallel to the line segment PQ that passes through R by a scale and a compass.
Solution:

Given

Habib has drawn a line segment PQ in his exercise book and he has also considered a point R outside the line segment PQ.

On straight line PQ a point S is taken Both points. R and S are joined with scale, consequently, ∠RSQ is drawn.

Now on point R on opposite side of ∠RSQ ∠SRT is drawn equal to ∠RSQ.

∵ ∠SRT = ∠RSQ, alternate angles

∴ PQ and TU straight lines are parallel, i.e., PQ // TU

“Class 8 WBBSE Maths Chapter 22 solutions, Construction of Parallel Lines study material”

Question 3. By a scale and a compass, Megha draws an angle ∠ABC = 60° between rays BA and BC respectively. Let’s take two points P and Q. Let’s draw a straight line through point P parallel to the ray BC and also draw a straight line through the point Q parallel to the ray BA. Let D be the intersecting point. Let’s write the type, of which quadrilateral PBQD is.
Solution:

Given

By a scale and a compass, Megha draws an angle ∠ABC = 60° between rays BA and BC respectively. Let’s take two points P and Q.

Let’s draw a straight line through point P parallel to the ray BC and also draw a straight line through the point Q parallel to the ray BA. Let D be the intersecting point.

∠ABC = 60°

“WBBSE Class 8 Maths Chapter 22, Construction of Parallel Lines solved examples”

At AB on point P ∠APE is drawn equal to ∠ABC. At BC on point Q ∠FQC is drawn equal to ∠ABC. PE and QF are joined which cut each other at D.

∠ABC = ∠APE (Corresponding angles)

∴ PE//BC

∠ABC = ∠FQC (Corresponding angles)

∴ AB // FQ

Now in PBQD,

BP // QD and BQ // PD

∴ PBQD is a parallelogram.

Formation Of An Equation And Its Solution

Today we will play an interesting game. Shibani has collected some marbles in a bag made of cloth. Dhruba, Mahua, Ashoke, Murad and I will play with marbles in the room on the roof of Sibani’s house.

The game is like this : first Shibani will distribute some marbles to Dhruba and Mahua in a certain way and then I will tell the number of marbles each of them has got.

To Mahua, Shibani gives 18 marbles more than twice the number of marbles she gives to Dhruba. Let’s find the number of marbles Sibani might give to Mahua.

Let Shibani gives × marbles to Dhruba.

∴ Shibani gives – (2 × x + 18) = 12x + 18I marbles to Mahua.

Number of marbles 1 2 3 4 5 6 7 ……………… x given to Dhruba

Number of marbles 20 22 24 26 28 30 32 12x + 18 marbles to Mahua. given to Mahua.

∴ Shibani can give 20, 22, marbles to Mahua.

I come to know that Shibani has given 108 marbles to Mahua.

2x + 18 = 108

or, 2x = 108 – 18

or, 2x = 90

or, x= 90/2

∴ x = 45

So, Shibani has given 45 marbles to Dhruba.

∴ x = 45 is the root of the equation : 2x + 18 = 108.

Read and Learn More WBBSE Solutions For Class 8 Maths

Now, Shibani gives Mahua 4 less marbles than half of marbles she gives to Murad. Let’s find out the number of marbles which Shibani can give to Mahua.

Let’s think that Shibani gives x marbles to Murad.

∴ Shibani gives \(\frac{x}{4}\)-4  marbles to Mahua.

Number of marbles given 8 10 12 20 28 ……… n to Ashoke

Number of marbles given 0 1 2 6 10 \(\frac{n}{2}\)-4 to Mahua

After counting, we see that Shibani has given 86 marbles to Mahua.

= \(\frac{x}{2}-4=86\)

∴ Shibani has given 180 to Murad.

∴ x = 180

We get that the root of the equation.

“WBBSE Class 8 Maths Chapter 19 solutions, Formation of an Equation and Its Solution”

= \(\frac{x}{2}-4=86\) is 180

Again, Shibani gives to Mahua 3 less marbles than \(\frac{5}{2}\) part of marbles she gives to Ashoke. Let’s find the number of marbles which Shibani can give to Mahua.

Let’s think that Shibani gives x marbles to Ashoke.

∴ Shibani gives \(\frac{5 x}{2}-3\) marbles to Mahua.

Number of marbles 2 4 8 10 20 ………………. n given to Ashoke

Number of marbles 2 7 17 22 47 ……………… \(\frac{5 x}{2}-3\) given to Mahua

If Shibani gives 127 marbles to Mahua, let’s find the number of marbles Shibani has given to Ashoke.

Let’s find the root of the equation

= \(\frac{5 x}{2}-3=127\)

Formation Of An Equation Exercise 19.1

Question 1. Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than \(\frac{x}{2}\) part of the marbles mine, then let’s calculate the possible number of marbles of Murad.
Solution:

Given

Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than \(\frac{x}{2}\) part of the marbles of mine

Let Shibani has given me x marbles.

∴ Murad has been given \(\frac{7 x}{3}-2\) marbles.

Number of marbles 3 6 9 15 18 21

Number of Murad’s marbles 5 12 19 33 40 47

∴ Murad can be given 5,12,19,33,40,47……………marbles.

B.T.P.,

= \(\frac{7 x}{3}-2=40\)

or, \(\frac{7 x}{3}=40+2\)

or, x = \(\frac{42 \times 3}{7}\)

or, x = 18

If Murad has 40 marbles, then Shibani has given me 18 marbles.

This type of n number of designs contain 4n number of sticks.

If we use 80 sticks then the number of squares is 20

Therefore, 4n = 80. The root of the equation is 20

Again, Rokeya made another design –

To make such n number of ‘H’ the number of sticks will be needed is (3 x n + 2).

Let’s form an equation and find the root of the equation to find the number of ‘IT if we form H,s by 35 sticks. Rokeya made another design

This type of n number of designs contain 4n number of sticks. Now I will make another design with matchsticks and find the number of sticks required to from n number of such designs.

Make equation and write in language.

x = -19

2x + 38 = 0

t = 21

5t-105 = 0

All of us will contribute some money for organising a feast. I have 2, 100 rupee notes. I want to change these two notes. I want to change these two notes in the form of 5 and 10 rupee notes.

Shibani’s father changed my money and gave me 32 notes in total.

Let’s form an equation and solve it to get how many notes of each type I have.

Let, I have x number of 5 rupee notes.

∴ I have (32 – x) number of 10 rupee notes.

Value of 1 number of 5 rupee notes is Rs. 5.

∴ Value of 1 number of 10 rupee notes is Rs. 10.

Value of x number of 5 rupee notes is Rs 5x.

∴ Value of (32 – x) number of 10 rupee notes is Rs. 10(32 – x).

So I have Rs. 5x and Rs. 10(32 – x) and my total money is Rs. 200.

’ We have, 5x + 10 (32-x) = 200

or, 5x + 320 – 10x = 200

or, 5x – 10x = 200-320

or, -5x = -120 -120

or, x = \(\frac{-120}{-5}\)

∴  x = 24

So I have 24 number of 5 rupee notes and 32-24 = 8 number of 10 rupee notes.

“Class 8 WBBSE Maths Chapter 19 solutions, Formation of Equation study material”

Question 2. If I change 3 hundred rupee notes in the form of 5 and 10 rupee notes then I get 48 notes in total. Let’s find out how many notes of each type I would have.
Solution:

Given

If I change 3 hundred rupee notes in the form of 5 and 10 rupee notes then I get 48 notes in total.

Let I have x notes of Rs. 5.

∴ I have 10 number of (48-x) notes.

Value of 5 notes of Rs. x is Rs. 5x

Value of 10 notes of Rs. (48-x) is Rs. 10 (48-x)

B.T.P.,

5x + 10 (48 – x) = 300

or, 5x + 480 – 10x = 300

or, -5x = 300 – 480

or, -5x = -180 .

or, x = \(\frac{-180}{-5}\)

or, x = 36

∴ I have 36 notes of Rs. 5 and (48-36) = 12 notes of 10.

Question 3. There are 35 students in Dhruba’s class. Their average age is 14 years. New 7 more students got admission and then the average age becomes is 13.9 years. Let’s find the equation and calculate the average age of the seven new students of Dhruba’s class.
Solution:

Given

There are 35 students in Dhruba’s class. Their average age is 14 years. New 7 more students got admission and then the average age becomes is 13.9 years.

Let the average age of the 7 new students is x years.

∴ Total age of the 7 new students is (7xx) years = 7x years.

Total age of the previous 35 students was 35×14 years = 490 years.

Now the total number of students is Dhruba’s class is (35+7)

= 42

∴ Total age of these 42 students is = (7x + 490) years.

Average age of these 42 students is = 13.9 years.

∴ Total age of these 42 students is = (13.9 x 42) years.

B.T.P., 7x +490 = 13.9×42

or, 7x = 13.9 × 42 – 490

or, x = \(\frac{13.9 \times 42-490}{7}\)

∴ x = 13.4

∴The average age of the 7 new students is h 3.4l years.

Question 4. Manash has written a fraction whose denominator is 1 more than twice of the numerator. If we add 4 to the numerator and the denominator, then the fraction will be \(\frac{7}{11}\) Let’s form an equation and find out the fraction written by Manash.

Solution: 

Manash has written a fraction whose denominator is 1 more than twice of the numerator. If we add 4 to the numerator and the denominator, then the fraction will be \(\frac{7}{11}\)

Let the numerator of the fraction = x

Hence the denominator = 2x + 1

∴ The fraction = \(\frac{x}{2 x+1}\)

Let’s add 4 to both the numerator and the denominator of the fraction and let’s see what we get,

= \(\frac{x+4}{2 x+1+4}=\frac{x+4}{2 x+1+4}=\frac{x+4}{2 x+5}\)

B.T.P.,

= \(\frac{x+4}{2 x+5}=\frac{7}{11}\)

or, 14x+35 = 11x+44

or, 14x-11x = 44-35

or, 3x = 9

∴ x = 3

The required fraction  = \(\frac{x}{2 x+1}=\frac{3}{2 x 3+1}=\frac{3}{7}\)

Question 5. Ashoke has written a fraction whose numerator is 2 less than the denominator. If we add 1 to the numerator and to the denominator then the fraction will be \(\frac{4}{5}\); let’s form an equation and hence find the fraction written by Ashoke.

Solution:

Given

Ashoke has written a fraction whose numerator is 2 less than the denominator. If we add 1 to the numerator and to the denominator then the fraction will be \(\frac{4}{5}\);

Let denominator is x.

∴ Number of fraction = x-2

∴ Fractions = \(\frac{x-2}{x}\)

B.T.P.,

= \(\frac{x-2+1}{x+1}=\frac{4}{5}\)

or,  \(\frac{x-1}{x+1}=\frac{4}{5}\)

or, 5x – 5 = 4x + 4

or, 5x – 4x = 4 + 5

or, x = 9

∴ Denominator = 9

∴ Numerator = 9-2 = 7

∴ Fractions = \(\frac{7}{9}\)

Question 6. Murad has written a two digit number where the sum of the two digits is 11; if we add 63 to the number then the digits change their positions. Let’s form an equation and hence try to find the two digit number written by Murad.

Given

Murad has written a two digit number where the sum of the two digits is 11; if we add 63 to the number then the digits change their positions.

Let the digit in the unit’s position of the two digit number is x.

∴ Digit in tenth place = (11 — x)

∴ Two digit number = 10x number in tenth place + number in unit place

= 10 × (11-x) + x

= 110 – 10x + x

Two digit number = 110- 9x

If we interchange the digits of the two digit number, then the number becomes

= 10 × x+11-x

= 10x +11-x

= 9x + 11

B.T.P.,

= 110-9x+63 = 9x+11

or, -9x-9x = 11-110-63

or, -18x = -162

or, x = \(\frac{-162}{-18}\) = 9

∴ x = 9

∴ Digit in unit place = 9

∴ Digit in tenth place = 11 -9 = 2

∴ Two digit number = 10×2 + 9

= 20 + 9

Two digit number = 29

Question 7. Half of a number is 6 more than \(\frac{1}{5}\) part of that number. Let’s form an equation and hence find the number.
Solution:

Given

Half of a number is 6 more than \(\frac{1}{5}\) part of that number.

Let the number be x.

∴ Half of the number = \(\frac{x}{2}\)

Half of the number is 6 more than of that \(\frac{1}{5}\) number.

Therefore, \(\frac{x}{2}-\frac{x}{5}\) = 6

or, \(\frac{5 x-2 x}{10}c\) = 6

or, \(\frac{3 x}{10}\) = 6

∴ x = 20

The required number is 20

Formation Of An Equation Exercise 19.1

Let’s form equation and work out.

Question 1. 2 more than twice the number that Sima has written is equal to 5 less than thrice the number she has written. Find the number she has written.

Solution:

Given

2 more than twice the number that Sima has written is equal to 5 less than thrice the number she has written.

Let Sima has written number x.

∴ Twice of the numbers = 2x

∴ Thrice of the number = 3x

B.T.P., required equation –

2x + 2 = 3x – 5

or, 2x – 3x = – 5 – 2

or, -x = -7

or, x = 7

∴ Number written by Sima = 7

Equation = 2x + 2 = 3x – 5, Number = 7

Question 2. Let’s write down three consecutive integers such that 5 less than the sum of the numbers is equal to 11 more than twice the second number. Let’s find the three consecutive integers.

Solution:

Let three consecutive numbers be x, x + 1 and x + 2.

Twice of the second number = 2 (x + 1).

B.T.P., required equation –

x + (x + 1) + (x + 2)-5 = 2 (x + 1)+11

or, x + x+1+x + 2- 5 = 2x + 2+11

or, 3x – 2x = 2 + 11 + 5 – 1-2

or, x = 15

∴ First number = 15

Second number = 15 + 1 = 16

Third number = 15 + 2 = 17

Required equation : x + x+1 + x + 2- 5 = 2 (x+1) +11

Three consecutive numbers are 15, 16 and 17.

Question 3. Let’s find a number such that one-fourth of the number is 1 less than one-third of the number.

Solution:

Let the number is = x

= \(\frac{1}{3}\) rd of the number = \(\frac{x}{3}\)

= \(\frac{1}{4}\) th of the number =\(\frac{x}{4}\)

B.T.P.,

= \(\frac{x}{3}-\frac{x}{4}=1\)

or, \(\frac{4 x-3 x}{12}=1\)

or, x = 12

Required equation : \(\frac{x}{3}-\frac{x}{4}=1\)

Required number = 12

Question 4. Let’s find a fraction whose denominator is 2 more than the numerator; and when 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes equal to \(\frac{7}{3}\)

Solution.

Let the numerator of the fraction is x.

∴ The denominator of fraction = x + 2.

∴ Fraction = \(\frac{x}{x+2}\)

B.T.P., required equation-

= \(\frac{x+3}{x+2-3}=\frac{7}{3}\)

= \(\frac{x+3}{x-1}=\frac{7}{3}\)

or, 7x-7 = 3x+9

or, 7x-3x = 9+7

or, 4x = 16

or, x = \(\frac{16}{4}\) = 4

Required equation = \(\frac{x+3}{x+2-3}=\frac{7}{3}\) and fraction = \(\frac{4}{6}\)

Question 5. Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again, she added 2 to the numerator and subtracted 1 from the denominator and also subtracted 1 from the numerator and added 2 to the denominator to get 2 fractions whose product is \(\frac{2}{5}\). Let’s write the fraction written by Sucheta.

Solution:

Given

Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again, she added 2 to the numerator and subtracted 1 from the denominator and also subtracted 1 from the numerator and added 2 to the denominator to get 2 fractions whose product is \(\frac{2}{5}\).

Let the numerator of fraction is x.

∴ Denominator = x + 3

∴ Fraction = \(\)

B.T.P., required equation

= \(\frac{x+2}{x+3-1} \times \frac{x-1}{x+3+2}=\frac{2}{5}\)

= \(\frac{x+2}{x+2} \times \frac{x-1}{x+5}=\frac{2}{5}\)

= \(\frac{x-1}{x+5}=\frac{2}{5}\)

or, 5x – 5 = 2x + 10

or, 5x – 2x = 10 + 5

or, 3x = 15

or X= \(\frac{15}{3}\) = 5

∴ Numerator of the fraction = 5 and denominator =5 + 3 = 8

∴ Required fraction = \(\frac{5}{8}\)

Required equation :  \(\frac{x+2}{x+3-1} \times \frac{x-1}{x+3+2}=\frac{2}{5}\)

Fraction written by Sucheta = \(\frac{5}{8}\)

Question 6. Raju wrote a number having two digits where the digit in the tenth position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number. Let’s find the number written by Raju.

Solution:

Given

Raju wrote a number having two digits where the digit in the tenth position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number.

Let the digit in the unit’s place of the two digit number is x.

∴ Digit in the tenth’s place = 3x

∴ Two digit number = 10x digit in the tenth’s place + digit in the unit’s place.

= 10 × 3x + x

= 30x + x

Number got by interchanging the digits of the two digit number-

= 10 × x + 3x

= 10x + 3x

B.T.P., required equation

30x + x – 36 = 10x + 3x

or, 31 x – 13x = 36

or, 18x = 36

or, x = \(\frac{36}{18}\) = 2

∴ Digit in the units place of the two digit number = 2 and digit in the tenths place = 3×2 = 6

Required number = 10×6 + 2 = 62

Required equation : 30x + x – 36 = 10x +3x

Required number = 62

The number written by Raju = 62

“WBBSE Class 8 Maths Chapter 19, Formation of an Equation and Its Solution solved examples”

Question 7. If the sum of two numbers is 89 and their difference is 15, let’s find the two numbers.

Solution:

Given

If the sum of two numbers is 89 and their difference is 15,

Let the greater number is x.

∴ The smaller number = 89 – x

B.T.P., required equation –

= x – (89 – x) = 15

or, x – 89 + x = 15

or, 2x = 15 + 89

or, 2x = 104

or, x  = \(\frac{104}{2}\)= 52

∴ Greater number = 52, smaller number = 89 – 52 = 37

Required equation : x – (89-x) = 15

Required number : 52 and 37.

Question 8. Divide 830 into two parts in such a way that 30% of one part will be 4 more than 40% of the other.

Solution:

Let one part is x.

∴ Second part = (830 – x)

30% of one part = 30% of x

= \(\frac{x \times 30}{100}=\frac{3 x}{10}\)

40% of second part = 40% of (830 – x)

= \(\frac{(830-x) \times 40}{100}=\frac{2(830-x)}{5}\)

B.T.P., required equation-

= \(\frac{x \times 30}{100}=\frac{(830-x) \times 40}{100}+4\)

= \(\frac{3 x}{10}=\frac{2(830-x)}{5}+4\)

= \( \frac{3 x}{10}=\frac{1680-2 x+20}{5}\)

= \(\frac{3 x}{2}=\frac{1660-2 x}{1}\)

= \(\frac{3 x}{2}=\frac{1660-2 x}{1}\)

or, 3x = 3360-4x

or, 3x+4x = 3360

or, x = \(\frac{3360}{7}\) = 480

∴ One part = 480

Second part = 830 – 480 = 350

Required equation: = \(\frac{x \times 30}{100}=\frac{(830-x) \times 40}{100}+4\)

Two parts are respectively 480 and 350.

Question 9. Divide 56 into two parts so that thrice of the first part becomes 48 more than one third of the second part.

Solution:

Let the first part is x.

∴ Second part = 56 – x

Thrice of the first part = 3x

= \(\frac{1}{3}\) rd of the second part = (56 – x)

∴ B.T.P., required equation –

3x = \(\frac{56-x}{3}\) + 48

or, 9x = 56 – x + 48 x 3

or, 9x + x = 56 + 144

or, 10x = 200

or x= \(\frac{200}{20}\) =20

First part is 20 and the second part is 56-20 = 36 ,

Required equation : 3x = \(\frac{56-x}{3}\)+ 48

Required numbers are 20 and 36.

Question 10. \(\frac{1}{5}\) part of a pole lies in mud, \(\frac{3}{5}\) part lies in water and the remaining 5 metres lies above the water. Let’s find the length of the pole and write it.

Solution:

Given

\(\frac{1}{5}\) part of a pole lies in mud, \(\frac{3}{5}\) part lies in water and the remaining 5 metres lies above the water.

Let the length of the pole is x meters.

∴ Part of pole in mud = \(\frac{1}{5}\) part x m of = \(\frac{x}{5}\)

∴ Part of pole in water = \(\frac{3}{5}\) part x m of = \(\frac{3x}{5}\)m

B.T.P., required equation-

= \(x-\left(\frac{x}{5}+\frac{3 x}{5}\right)=5\) = 5

= \(x-\frac{4 x}{5}\) = 5

= \(\frac{5 x-4 x}{5}\) = 5

or, x = 25

∴ length of the pole = 25 meters

B.T.P., required equation : \(x-\frac{4 x}{5}\) = 5

Length of the pole = 25 m

“WBBSE Class 8 Formation of an Equation and Its Solution solutions, Maths Chapter 19”

Question 11. At present, my father’s age is 7 times of my age. After 10 years my father’s age will be 3 times of my age. Let’s find and write the present age of my father and me.

Solution:

Given

At present, my father’s age is 7 times of my age. After 10 years my father’s age will be 3 times of my age.

Let my present age is x years.

∴ Present age of my father = 7x years.

10 years later my age – (x + 10) years.

10 years later father’s age = (7x + 10) years.

B.T.P., required equation-

7x + 10 = 3 (x + 10)

or, 7x + 10 = 3x + 30

or, 7x – 3x = 30 – 10

or, 4x = 20

or, x = \(\frac{20}{4}\) = 5

My present age = 5 years

Present age of my father = 7 × 5 years = 35 years

Required equation : 7x + 10 = 3 (x + 10)

My present age is 5 years and the present age of my father is 35 years.

Question 12. My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee notes, if he received 137 notes in total, then find the number of 5 rupee notes that he got.

Solution:

Given

My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee notes, if he received 137 notes in total

Let uncle got 10 number of Rs. x notes.

∴ (137 – x) number of Rs. 5 notes got.

B.T.P., required equation –

10x + 5 (137-x) = 1000

or, 10x + 685 – 5x = 1000

or, 5x = 1000 – 685

or, 5x = 315

or, x = \(\frac{315}{5}\) = 63

∴ Number of 5 rupee notes = (137-x) = 7 (137 – 63) = 74

Required equation : 10x + 5 (137 – x) = 100

Number of Rs. 5 notes = 74

“Class 8 WBBSE Maths Chapter 19, Formation of an Equation easy explanation”

Question 13. Salem uncle of our village used half of his savings to buy a house after his retirement from a government job. One day after falling into trouble, he sold his house and got 5% more than its cost price. If he would have taken 3450 rupees more than he would get 8% more than his cost price. Let’s find the amount of money Salem uncle used to buy his house, and his entire savings.

Solution:

Given

Salem uncle of our village used half of his savings to buy a house after his retirement from a government job. One day after falling into trouble, he sold his house and got 5% more than its cost price. If he would have taken 3450 rupees more than he would get 8% more than his cost price.

Let the total savings of Salem uncle be Rs. x

Expenditure on buying house = \(\frac{1}{2}\) part of Rs. x Rs. = Rs. \(\frac{x}{2}\)

5% of the cost price of house = 5% of Rs. \(\frac{x}{2}\) = Rs. \(\frac{x}{2} \times \frac{5}{100}\) x

8% of the cost price of house = 8% of Rs. \(\frac{x}{2}\) = Rs. \(\frac{x}{2} \times \frac{8}{100}\)

B.T.P., required equation –

= \(\frac{x}{2} \times \frac{5}{100}+3450=\frac{x}{2} \times \frac{8}{100}\)

or, \(\frac{x}{40}+3450=\frac{x}{25}\)

or, \(\frac{x}{40}-\frac{x}{25}=-3450\)

or, \(\frac{5 x-8 x}{200}=-3450\)

or, \(\frac{-3 x}{200}=-3450\)

or, \(x=\frac{3450 \times 200}{2}\)

or, x = 230000

Cost price of the house = = \(\frac{230000}{2}\) = Rs.115000

Amount of the savings of Salem Uncle = Rs. 230000

Required equation : – = \(\frac{x}{2} \times \frac{5}{100}+3450=\frac{x}{2} \times \frac{8}{100}\)

Salem uncle bought house of Rs. 115000 and his total savings was Rs. 230000.

Question 14. There was food provision for twenty days in a refugee camp of the village Gopalpur. After 7 days, 100 more refugees joined the refugee camp and they consumed the food grains within 11 days. Let’s write the number of refugees that were present initially.

Solution.

Given

There was food provision for twenty days in a refugee camp of the village Gopalpur. After 7 days, 100 more refugees joined the refugee camp and they consumed the food grains within 11 days.

Let there were x number of people in the refugee camp.

∴ For x people 20 days’ food was there. After 7 days for x people (20-7) or 13 days’ food was there.

7 days later number of people = (x + 100)

∴ (x + 100) people will eat x people’s food for 13 days.

B.T.P., required equation

= \(\frac{(20-7) x}{x+100}=11\)

or, \(\frac{13 x}{x+100}=11\)

or, 13x = 11x+1100

or, 13x-11x=1100

or, 2x = 1100

or, x = \(\frac{1100}{2}\) = 550

∴ Earlier there were 550 number of people in the refugee camp.

Requied equation \(\frac{(20-7) x}{x+100}=11\) and 550 number of people.

Question 15. Let’s find the roots of the equations :

1. \(\frac{5}{3 x+4}=\frac{4}{5(x-3)}\)

Solution:

or, 25x-75 = 12x+16

or, 25x-12x = 16+75

or, 13x = 91

or, x = \(\frac{91}{13}\) = 7

or, x = 7

2. 14(x-2)+3(x+5) = (x+8)+5

Solution:

or, 25x-75 = 12x+16

or, 25x-12 = 16+75

or, 13x = 91

or, x = \(\frac{42}{14}\) = 3

∴ x = 3

3. \(\frac{x}{2}+5=\frac{x}{3}+7\)

Solution:

= \(\frac{x}{2}+5=\frac{x}{3}+7\)

= \(\frac{x}{2}-\frac{x}{3}=7-5\)

or, \(\frac{3 x-2 x}{6}=2\)

or, x = 12

4. \(\frac{x+1}{8}+\frac{x-2}{5}=\frac{x+3}{10}+\frac{3 x-1}{20}\)

Solution:

or, \(\frac{5(x+1)+8(x-2)}{40}=\frac{2(x+3)+3 x-1}{-1^{20}}\)

or, \(\frac{5 x+5+8 x-16}{2}=\frac{2 x+6+3 x-1}{1}\)

or, \(\frac{13 x-11}{2}=\frac{5 x+5}{1}\)

or, 13x-11 = 10x+10

or, 13x-10x = 10+11

or, 3x = 21

or, x = \(\frac{21}{3}\) = 7

∴ x = 7

5. \(\frac{x+1}{4}+3=\frac{2 x+4}{5}+2\)

Solution:

or, \(\frac{x+1+12}{4}=\frac{2 x+4+10}{5}\)

or, \(\frac{x+13}{4}=\frac{2 x+14}{5}\)

or, 16x+2 = 3x+80

or, 16x-3x = 80-2

or, 13x = 78

or, x = \(\frac{78}{13}\) = 6

∴ x = 6

6. \(\frac{3}{5}(x-4)-\frac{1}{3}(2 x-9)=\frac{1}{4}(x-1)-2\)

Solution:

or, \(\frac{3(x-4)}{5}-\frac{(2 x-9)}{3}=\frac{(x-1)}{4}-2\)

or, \(\frac{9 x-36-10 x+45}{15}=\frac{x-1-8}{4}\)

or, \(\frac{9-x}{15}=\frac{x-9}{4}\)

or, 15x-135 = 36-4x

or, 15x+4x = 36+135

or, 19x = 171

or, x = \(\frac{171}{19}\) = 9

∴ x = 9

7. \(\frac{x+5}{3}+\frac{2 x-1}{7}=4\)

Solution:

or, \(\frac{7 x+35+6 x-3}{21}=4\)

or, 13x+32 = 84

or, 13x = 84-32

or, 13x = 52

or, x = \(\frac{52}{13}\) = 4

∴ x = 4

“WBBSE Class 8 Maths Chapter 19 solutions, Formation of Equation and Its Solution PDF”

8. 25 + 3(4x-5) +8(x+2) = x+3

Solution:

or, 25 + 12x-15 + 8x+16 = x + 3

or, 12x + 8x-x = 3 + 15-25-16

or, 20x =18-41

or, 19x = -23

or, \(x=\frac{-23}{19}=-1 \frac{4}{19}\)

∴  x = \(-1 \frac{4}{19}\)

9. \(\frac{x-8}{3}+\frac{2 x+2}{12}+\frac{2 x-1}{18}=3\)

Solution:

or, \(\frac{12(x-8)+3(2 x+2)+2(2 x-1)}{36}=3\)

or, 12x – 96 + 6x + 6 + 4x – 2 = 108

or, 22x = 108+ 96+ 2-6

or, 22x = 206 – 6

or, 22x = 200

or, \(x=\frac{200}{22}=\frac{100}{11}\)

∴ \(x=9 \frac{1}{11}\)

10. \(\frac{t+12}{6}-t=6 \frac{1}{2}-\frac{1}{12}\)

Solution:

or, \(\frac{t+12-6 t}{6}=\frac{13}{2}-\frac{1}{12}\)

or, \(\frac{12-5 t}{6_1}=\frac{78-1}{72_2}\)

or, \(\frac{12-5 t}{1}=\frac{77}{2}\)

or, 24 -10t = 77

or, -10t = 77 – 24

or, -10t = 53

or, t = \(-\frac{53}{10}\)

or, t = \(-5\frac{3}{10}\)

11. \(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)

Solution:

or, \(\frac{14 x+14-5 x-9}{28}=\frac{x+6+105-7 x+84}{21}\)

or, \(\frac{14 x+14-5 x-9}{-28}=\frac{x+6+105-7 x+84}{-243}\)

or, \(\frac{9 x+5}{4}=\frac{195-6 x}{3}\)

or, 27x + 15 = 780 – 24x

or, 27x + 24x = 780 – 15

or, 51x = 765

or, x = \(\frac{765}{51}\) = 15

∴ x = 15

12. \(\frac{9 x+5}{14}+\frac{8 x-7}{7}=\frac{18 x+11}{28}+\frac{5}{4}\)

Solution:

or, \(\frac{9 x+5+16 x-14}{14}=\frac{18 x+11+35}{28}\)

or, \(\frac{25 x-9}{1}=\frac{18 x+46}{2}\)

or, 50x-18 = 18x+46

or, 50x-18x = 46+18

or, 32x = 64

or, x = \(\frac{64}{32}\) = 2

or, x = 2

13. \(\frac{3 y+1}{16}+\frac{2 x-3}{7}=\frac{y+3}{8}+\frac{3 y-1}{14}\)

Solution:

or, \(\frac{3 y+1}{16}=\frac{y+3}{8}=\frac{y+3}{8}+\frac{3 y-1}{14}\)

or, \(\frac{3 y+1-2 y-6}{16}=\frac{3 y-1-4 y+6}{14}\)

or, \(\frac{y-5}{8}=\frac{5-y}{7}\)

or, 7y – 35 = 40 + 8y

or, 7y + 8y = 40 + 35

or, 15y = 75

or, y = \(\frac{75}{15}\) = 5

∴ y = 5

14. 5x -(4x-7)(3x-5) = 6-3(4x-9)(x-1)

Solution:

or, 5x-12 x²+ 20x + 21 x – 35 = 6-3 (4x²-4x – 9x + 9)

or, -12x² + 46x – 35 = 6 – 12x² + 12x + 27x – 27

or, -12x² + 12x² + 46x – 12x – 27x = 6 + 35-27

or, 46x – 39x = 41-27

or, 7x = 14

or, x =\(\frac{14}{7}\)= 2

∴ x = 2

15. 3(x-4)² + 5(x-3)²=(2x-5) (4x-1)-40

Solution:

or, 3 (x² – 8x + 16) +5 (x² – 6x + 9) = 8x² – 2x – 20x + 5-40

or, 3x² – 24x + 48 + 5x² – 30x + 45 = 8x² – 22x – 35

or, 8x² – 8x² – 24x – 30x + 22x =-35 – 48 – 45

or, -32x = – 128

or x = \(\frac{-128}{-32}\) =4

∴ x = 4

16. 3(y-5)²+5y = (2y-3)²-(y+1)² + 1

Solution:

or, 3 (y-5)² + 5y = (2y-3)²– (y+1 )²+1

or, 3(y²-10y+25) + 5y = 4y²– 12y + 9 – (y² + 2y+ 1) + 1

or, 3y² – 30y + 75 + 5y = 4y² – 12y+ 9- y²– 2y – 1 + 1

or, 3y²+y²-4y²-30y + 12y + 2y = 9 – 75

or, -11 y = -66

or, y = \(\frac{-66}{-11}\) = 6

“WBBSE Class 8 Maths Chapter 19, Formation of an Equation and Its Solution important questions”

Question 17. Let’s write mathematical stories and form equations.

Solution:

x = 5 → 2x-10 = 0

y = -11 → 5y+55 = 0

t = \(\frac{7}{8}\) → 8t-7 = 0

x = 24 → 3x-72 = 0

x = 7 → 7x-49 = 0

Time And Work

There are 18 looms in Mansur’s factory in Shantipur. But last week 3 looms were closed. So Last week only 165 dhotis and sarees were weaved. This week all the looms are running.

Question 1. Let’s calculate in a proportional method how many dhotis and sarees will be weaved this week in Mansur’s factory.

Solution:

Given

There are 18 looms in Mansur’s factory in Shantipur. But last week 3 looms were closed. So Last week only 165 dhotis and sarees were weaved. This week all the looms are running.

The problem in mathematical language is

If the time is constant and the number of looms increases or decrease, then the number of dhotis and sarees will Increase or Decrease (increase/decrease).

∴ There is a direct proportion between the number of looms and the number of dhotis and sarees.

∴ The direct proportion is –

15 : 18:: 165:? (Required number of dhotis and sarees)

∴ The required number of dhotis and sarees is = \(\frac{165 \times 18}{15}\) = 198.

So in this week, all looms being run, 198 dhotis and sarees will be made.

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. If all the looms run, calculate in how many days 594 dhotis and sarees will be made.

Solution:

In mathematical language, the problem is –

The number of looms being constant, as the number of dhoti-sarees increases/decreases, the required time will Increase or Decrease (increase / decrease)

∴ The direct proportion is –

198 : 594:: 7:?

∴ Required time = \(\frac{594 \times 18}{198}\) days = 21 days

“WBBSE Class 8 Maths Chapter 17 solutions, Time and Work”

Question 3. If all the looms, i.e., 18 looms would run, 594 dhotis and sarees can be woven in 21 days. Let’s calculate how many looms will run to weave 594 dhotis and sarees in 14 days.

Solution:

Given

If all the looms, i.e., 18 looms would run, 594 dhotis and sarees can be woven in 21 days.

In mathematical language the problem is

If the number of dhotis and sarees is constant then the number of looms will Increase (increase/decrease) with the increase of time and the number of looms will Decrease (increase / decrease) with the decrease of time.

The number of looms and time are inversely proportional.

The inverse proportion is –

21 : 14:: ? (required number of looms) : 18 ∴ 14 : 21:: 18 : ? (required number of looms)

∴ Required number of looms = \(\frac{21\times18}{14}\) = 27

∴ In 21 days, to weave 594 dhotis and sarees 27 looms, i.e., (27 – 18) = 9 more looms will run.

So we see that there is a relation between time and work in Monsur’s loom. Let’s write the relation between the required time, the amount of work, and the number of workers is the chart given below.

Question 4. In a village of Niamatpur 15 men can cultivate 10 bighas of land in a week. Let’s calculate how many men can cultivate 18 bighas of land in a week.

Solution:

Given

In a village of Niamatpur 15 men can cultivate 10 bighas of land in a week.

In mathematical language the problem is –

As work increases or decreases, the number of workers will, time being constant, Increase or Decrease (increase/decrease).

The amount of work and the number of workers are in Direct (direct / inverse) relation.

Let’s calculate in the proportional method.

So, 10 : 18 :: 15 : ? (required number of workers)

∴ Required number of people = \(\frac{15 \times 18}{10}\) = 27 people

∴ 27 workers can cultivate 18 bighas of land in a week.

Let’s do the above problem by unitary method.

“Class 8 WBBSE Maths Chapter 17 solutions, Time and Work study material”

Question 5. 250 volunteers of Bakultala gram panchayat have constructed half of a dam in 24 days. Let’s calculate how many men will be needed to construct the rest half part in 20 days.

Solution:

Given

250 volunteers of Bakultala gram panchayat have constructed half of a dam in 24 days.

The problem in mathematical language is –

Keeping work constant, with an increase or decrease of time, the number of workers will respectively d] or CH (increase/decrease).

∴ The time and the number of workers are in (direct/inverse) relation.

∴ The inverse proportion is

24 : 20 :: ? (the required number of workers) : 250

∴  20 : 24 :: 250 : ? (the required number of workers) .

∴ The required number of workers = \(\frac{250 \times 24}{20}\)

So, additional workers 300-250 = 50 need to be absorbed in the team of workers. Let’s calculate this problem by unitary method.

Time And Work Exercise 17.1

Question 1. In Aman’s factory 216 parts of machine are made in 3 days. Let’s calculate how many parts of machine will be made in 7 days.

Solution:

Given

In Aman’s factory 216 parts of machine are made in 3 days.

In mathematical language the problem is –

If number of days increases or decreases, number of machine parts will increase or decrease.

∴ Days and number of machine parts are in direct proportion.

∴ 3 : 7 :: 216 : ?

or, \(\frac{3}{7}=\frac{216}{?}\)

or, Required number of parts = \(\frac{7 \times 216}{3}\) = 504

Question 2. In loom factory at Atpur 12 looms can weave 380 sarees. During the puja season for doing more work 3 looms have been established. Let’s make a proportion and calculate how many sarees can be woven in this month.

Solution:

Given

In loom factory at Atpur 12 looms can weave 380 sarees. During the puja season for doing more work 3 looms have been established.

In mathematical language the problem is –

If number of looms increases, number of sarees will increase.

∴ Number of looms and number of sarees are in direct proportion.

∴ 12 : 15 :: 380 : ?

or, Required number of sarees = \(\frac{15 \times 380}{12}\)

Question 3. Let’s make the story of mathematics and find out the relation seeing the above chart.

Solution:

In a factory a given number of looms can weave 45 m cloth in 25 days, in 15 days the same number of looms will weave how much cloth?

Relation : If number of days increases, length of cloth will increase.

∴ Days and length of cloth are in direct proportion.

∴ 25 : 15 :: 45 : ?

or, Length of cloth woven in 15 days =\(\frac{15 \times 45}{25}\) = 27 m

Question 4. We can see after 15 days of digging a canal of 1200 metre length, 3/4 parts of the canal have been dug. Let’s calculate how many days are needed to dig the rest of the part.

Solution:

Given

We can see after 15 days of digging a canal of 1200 metre length, 3/4 parts of the canal have been dug.

Length of the canal dug in 15 days

Length of remaining part of canal = (1200 – 900) = 300 m

In mathematical language the problem is –

Relation : If length of canal increases, days will increase.

∴ Length of canal and days are in direct relation.

∴ 900 : 300 :: 15 : ?

or, \(\frac{900}{300}=\frac{15}{?}\)

∴ Required days = \(\frac{300 \times 15}{900}\)

Question 5. 3 tractors can plough 18 bighas in a day. By rule of three method let’s calculate how many bighas can be ploughed in a day.

Solution:

Given

3 tractors can plough 18 bighas in a day.

In mathemetical language the problem is –

Relation : If number of tractors increases, quantity of land will increase.

∴ Number of tractors and quantity of land are in direct proportion.

or, Required quantity of land = g Bighas = 42 Bighas

Question 6. In Kusum’s factory 35 men can cast 10 ton iron parts in a week. The owner has got an order of casting 14 ton iron parts in a week. Let’s find out the proportion and calculate how many new men will be needed.

Solution.

Given

In Kusum’s factory 35 men can cast 10 ton iron parts in a week. The owner has got an order of casting 14 ton iron parts in a week.

In mathematical language the problem is –

Relation : If quantity of iron increases, number of people will increase.

∴ Quantity of land and number of people are in direct proportion.

∴ \(\frac{10}{14}=\frac{35}{?}\)

or, Required number of people = \(\frac{14 \times 35}{10}\) = 49

Number of extra people = (49 – 35) = 14

So 14 more men will be appointed.

Question 7. Let’s make the story in mathematical language and find out the relation seeing the above chart.

Solution:

If 9 people make 6 cycles every day then 72 people will make how many cycles?

Relation : If the number of people increases, amount of work will increase.

∴ Number of people and number of cycles are in direct proportion.

∴ 9 : 72 :: 6 : ?

or, \(\frac{9}{72}=\frac{6}{?}\)

or, Required number of cycles = \(\frac{72 \times 6}{9}\) = 48

Question 8. A pond will be dug in our locality. 24 men need 12 days to cut that pond. Let’s find out the proportion and find relation how many men will be needed to dig that pond in 8 days.

Solution:

Given

A pond will be dug in our locality. 24 men need 12 days to cut that pond.

In mathematical language the problem is –

Relation : If days decrease, number of people will increase.

∴ Days and number of people are in inverse proportion.

∴ \(\frac{12}{8}=\frac{?}{24}\)

∴ Required number of men = \(\frac{12 \times 24}{8}\) = 36

To dig that pond in 8 days, more men will be needed.

Question 9. 5 members in a bulb manufacturing co-operative factory can make 10000 bulbs in 12 days. Based on a sudden order, 10000 bulbs have to be made in 9 days. Let’s calculate how many extra members will be appointed to supply the bulbs according to the agreement.

Solution:

Given

5 members in a bulb manufacturing co-operative factory can make 10000 bulbs in 12 days. Based on a sudden order, 10000 bulbs have to be made in 9 days.

In mathematical language the problem is –

Relation : If days decrease, number of people will increase.

∴ Days and number of people are in inverse proportion.

∴ \(\frac{12}{9}=\frac{?}{45}\)

∴ Required number of people = \(\frac{12 \times 45}{9}\)   = 60

More number of members = (60-45) = 15. To complete the work in time, 15 more members will be appointed.

“WBBSE Class 8 Maths Chapter 17, Time and Work solved examples”

Question 10. 250 men need 18 days to dig a pond of 50 metre length and 35 metre breath. Let’s calculate how many days will be needed by 300 men to dig a pond of 70 metre length and 40 metre breadth of the same depth.

Solution:

Given

250 men need 18 days to dig a pond of 50 metre length and 35 metre breath.

Relation : If days decrease, number of people will increase.

Relation : If number of men increases, days will decrease and if area of pond increases, days will increase. So number of people and days are in inverse proportion and area of pond and days are in direct proportion.

∴ Required number of days To dig this kind of pond of length 70 m and breadth 40 m it will take 300 people days.

Question 11. Anawarabibi and Mihir uncle started to plaster three rooms of same size. But Harun. chacha, Anawarabibi and Mihir uncle finished the work in 10 days, 12 days and 15 days respectively. If the three members worked together in one room, the work would finish in less time. How will we calculate the number of days for plastering 1 room where three men work together?

Solution:

Given

Anawarabibi and Mihir uncle started to plaster three rooms of same size. But Harun. chacha, Anawarabibi and Mihir uncle finished the work in 10 days, 12 days and 15 days respectively. If the three members worked together in one room, the work would finish in less time.

Let’s see how many parts of the work was done by every one in a day.

Haran chacha plasters 1 room in 10 days.

Haran chacha plasters \(\frac{1}{10}\) part of the total work in 1 day.

Anawarabibi plasters 1 room in 12 days.

Anawarabibi plasters \(\frac{1}{12}\) part of the total work in 1 day.

Mihir uncle plasters 1 room in 15 days.

Mihir uncle plasters \(\frac{1}{15}\) part of the total work in* 1 day.

∴ Three men together plaster part of a room in 1 day

= \(\frac{1}{10}\) part + \(\frac{1}{10}\) part + \(\frac{1}{10}\) part

= \(\frac{6+5+4}{60}\) part

= \(\frac{15}{60}\) part

= \(\frac{1}{4}\) part

∴ Three of them together do \(\frac{1}{4}\) part of the total work in a day.

∴ The total work, i.e., 1 part is done in 1 ÷ \(\frac{1}{4}\) = 4 days.

∴ If three of them work, together, then they can complete the work in 4 days.

Question 12. Let’s calculate the number of days Mihlr uncle takes to complete the remaining \(\frac{1}{5}\) part of the work.

Solution:

Given

\(\frac{1}{5}\)

Mihir uncle does 1 part in 15 days.

Mihir uncle does \(\frac{1}{5}\) part in = \(15 \times \frac{1}{5}\) =3 days.

∴ Mihir uncle .alone completes the rest of the work in 3 days.

∴ If the work goes on such way, the required time to complete the work is = 2+2+4=7 days.

Question 13. Let’s calculate the part of the total work done by each of the if the work would go on such way.

Solution:

Harunchacha did the \(\left(\frac{1}{10} \times 2\right)\) part = \(\frac{1}{5}\) part of the total work.

Anawarabibi did the \(\left(\frac{1}{12} \times 4\right)\) part = \(\frac{1}{3}\) part of the total work.

Mihir uncle did the \(\left(\frac{1}{15} \times 7\right)\) part = \(\frac{7}{15}\) part of the total work. 15

Question 14. Bulu and Tathagata can do a work separately in 20 days and 30 days respectively. After working 7 days together both of them left away. Then Tatai came and completed the rest of the work alone in 10 days. Let’s make proportion and calculate how many days Tatai will take to complete the work alone.

Solution:

Given

Bulu and Tathagata can do a work separately in 20 days and 30 days respectively. After working 7 days together both of them left away. Then Tatai came and completed the rest of the work alone in 10 days.

Let’s calculate the remaining part of the total work after 7 days when Bulu and Tathagata left.

Bulu alone does \(\frac{1}{20}\) part in 1 day.

Tathagata alone does \(\frac{1}{30}\) part in 1 day.

∴ Together they do \(\left(\frac{1}{20}+\frac{1}{30}\right)\) part in 1 day.

= \(\frac{3+2}{60} \text { part }=\frac{5}{60} \text { part }=\frac{1}{12} \text { part }\)

∴ Bulu and Tathagata do =\(\frac{1}{12}\) part × 7 = \(\frac{7}{12}\) part in 7 days.

∴ After 7 days when Bulu and Tathagata left the remaining work is

= \(\left(1-\frac{7}{12}\right) \text { part }=\frac{5}{12} \text { part }\)

Tatai does the remaining \(\frac{5}{12}\) part in 10 days.

∴ The problem in mathematical language is

If the amount of work increases or decreases, the required time will (increase/decrease) Increase or Decrease.

∴ The amount of work and time are in Direct (direct/inverse)proportion.

The direct proportion is

= \(\frac{5}{12}\) : 1 : 10 : ?

∴ Required time days = \(\frac{10 \times 1}{\frac{5}{12}}\) days=  \(10 \times \frac{12}{5}\) = 24 days

∴ Tatai alone will do the work in 24 days.

“WBBSE Class 8 Time and Work solutions, Maths Chapter 17”

Question 15. An empty tank is filled up with two pipes separately in 12 minutes and 15 minutes respctively. Let’s calculate the time required to fill up the half full tank while two pipes are opened together.

Solution:

Given

An empty tank is filled up with two pipes separately in 12 minutes and 15 minutes respctively.

It takes 1 minute to fill up \(\frac{1}{12}\) part of the empty tank by the first pipe.

By the second pipe it takes 1 minute to fill up \(\frac{5}{12}\) part of the empty tank.

∴ If the two pipes are opened together, it takes 1 minute to fill up the \(\left(\frac{1}{12}+\frac{1}{15}\right)\) part

= \(\frac{5+4}{60} \text { part }\)

= \(\frac{9}{60} \text { part }=\frac{3}{20} \text { part }\)

Half of the tank is full.

∴ Two pipes together will fill  \(\left(1-\frac{1}{2}\right)\) part = \(\frac{1}{2}\) part of the tank.

Two pipes together fill  \(\frac{3}{20}\) part of the tank in 1 minute.

Two pipes together fill 1 part in 1× \(\frac{20}{3}\) minute.

Two pipes together fill \(\frac{1}{2}\) part in \(1 \times \frac{20}{3} \times \frac{1}{2}\) minute

= \(\frac{10}{3}\) minute = \(3 \frac{1}{3}\) minute

= 3 minute 20 seconds.

Let’s calculate using proportion. ,

∴ The problem in mathematical language is –

The amount of water and time are in Direct (direct/inverse) proportion.

∴ Using direct proportion, we get –

= \(\frac{3}{20}: \frac{1}{2}:: 1: ? \)(Required time) –

∴ Required time = \(1 \times \frac{1}{2} \times \frac{20}{3}\)

= \(\frac{10}{3}\) minutes = 3 minutes 20 seconds.

The time required to fill up the half full tank while two pipes are opened together = 3 minutes 20 seconds.

Time And Work Exercise 17.2

Question 1. Priya and Debu individually complete a work in 10 hours and 12 hours respectively. If they do the work together, let’s calculate in how many hours they will complete the work.

Solution:

Given

Priya and Debu individually complete a work in 10 hours and 12 hours respectively.

Priya can do \(\frac{1}{10}\) part of the in 1 hour. Debu cah do \(\frac{1}{12}\) part in 1 hour.

∴ Both can do \(\frac{11}{60}\) part of the work in 1 hour.

Both can do 1 part work in \(1 \div \frac{11}{60}\) hours.

= \(\frac{60}{11}\)

= \(5 \frac{5}{11}\)

If they work for \(5 \frac{5}{11}\) hours, the work will be completed.

Question 2. My elder brother, my elder sister and me together will paint our windows. My elder brother, my elder sister and me can complete the work separately in 12, 4 & 6 days respectively. If we do the work together then let’s calculate’ and write how many days we will take to complete the work.

Solution:

Given

My elder brother, my elder sister and me together will paint our windows. My elder brother, my elder sister and me can complete the work separately in 12, 4 & 6 days respectively.

My sister can do \(\frac{1}{12}\) part in 1 day.

My brother can do \(\frac{1}{4}\) part in 1 day. I can do \(\frac{1}{6}\) part in 1 day.

Three of us can do \(\left(\frac{1}{12}+\frac{1}{4}+\frac{1}{6}\right)\) part = \(\frac{1+3+2}{12}\) part in 1 day.

= \(\frac{6}{12} \text { part }\)

= \(\frac{1}{2} \text { part }\)

∵ Three of us can do \(\frac{1}{2}\) part in 1 day.

∴ Three of us can do 1 part in \(1 \div \frac{1}{2}\) days = 2 days.

If the three of us work together, then in 2 days we will complete the work.

Question 3. Abani and Anawar complete a work separately in 20 and 25 days respectively. After 10 days of their working together, they both left; then Sukhen came and completed the remaining work in 3 days. If Sukhen alone would do the work, let’s calculate how many days he would take to complete the work.

Solution:

Given

Abani and Anawar complete a work separately in 20 and 25 days respectively. After 10 days of their working together, they both left; then Sukhen came and completed the remaining work in 3 days.

Abani works \(\frac{1}{20}\) part in 1 day. Anawar works \(\frac{1}{25}\) part in 1 day.

Both of them together can do \(\left(\frac{1}{20}+\frac{1}{25}\right)\) part in 1 day

= \(\frac{5+4}{100} \text { part }\)

= \(\frac{9}{100} \text { part }\)

∴ Both of them together can do in 10 days = \(10 \times \frac{9}{100} \text { part }\)

= \(\frac{9}{10} \text { part }\)

Remaining work = \(\left(1-\frac{9}{10}\right) \text { part }\)

Remaining work = \(\frac{10-9}{10} \text { part }=\frac{1}{10} \text { part }\)

Remaining work =  Sukhen does \(\frac{1}{10}\)part in 3 days.

∴  Sukhen does 1 part in 3 x 10 days.

∴ Sukhen does 1 part in 30 days. If Sukhen works alone, then he will complete the work in 30 days.

Question 4. There are two pipes for taking water from the municipality water tank. The tank can be made empty in 4 hours by the two pipes separately. If both the pipes remain opened, let’s calculate when the full tank will be empty.

Solution:

Given

There are two pipes for taking water from the municipality water tank. The tank can be made empty in 4 hours by the two pipes separately. If both the pipes remain opened

Both pipes will together empty

Both pipes can empty part in 1 hour.

= \(\left(\frac{1}{4}+\frac{1}{4}\right) \text { part }\)

= \(\frac{1+1}{4} \text { part }\)

= \(\frac{2}{4} \text { part }\)

= \(\frac{1}{2} \text { part }\)

∵ Both pipes can empty \(\frac{1}{2}\)part in 1 hour.

∴ Both pipes can empty 1 part in 2 hours. If both pipes are opened then in 2 hours the full tank will be empty.

“Class 8 WBBSE Maths Chapter 17, Time and Work easy explanation”

Question 5. In our tank there are 3 pipes. With these three pipes separately the tank can be filled up in 18, 21 and 24 hours respectively

1. If the 3 pipes remain open together, let’s make proportion and calculate when the tank will be filled up with water.

Solution:

Given

In our tank there are 3 pipes. With these three pipes separately the tank can be filled up in 18, 21 and 24 hours respectively

If the three pipes remain open then in 1 hour the tank will fill up

= \(\left(\frac{1}{18}+\frac{1}{21}+\frac{1}{24}\right)\) part

= \(\frac{28+24+21}{504}\) part

= \(\frac{73}{504}\) part

In mathematical language the problem is

If part of the tank increases, time will increase.

∴ Here is direct proportion.

∴ \(\frac{\frac{73}{504}}{1}=\frac{1}{?}\)

or, \(\frac{73}{504}=\frac{1}{?}\) hrs.

∴ Required time = \(\frac{504}{73}\) hrs.

= \(6 \frac{66}{73}\) hrs.

If the three pipes remain open then in \(6 \frac{66}{73}\) hours the tank will be filled up.

2. If the first two pipes would remain open, let’s calculate the time required to fill up the tank with water.

Solution:

First two pipes will fill in 1 hour \(\left(\frac{1}{18}+\frac{1}{21}\right)\)

= \(\frac{7+6}{126}\)

= \(\frac{13}{126}\)

First two pipes fill up \(\frac{13}{126}\) part in 1 hour.

∴ First two pipes will fill 1 part in \(1 \div \frac{13}{126}\) hours

= \(\frac{126}{13}\) hours

= \(9 \frac{9}{13}\) hours

If first two pipes remain open, tank will require \(9 \frac{9}{13}\) hours to fill up.

Question 6. Rehana’s tank can be filled up in 30 minutes for by the municipality water supply pipe, opening all the taps, three hours. The can work for 4 hours with the water of the full tank. In one day the water supply pipe remains open for 25 minute. Let’s calculate how long they will work with that water.

Solution:

Given

Rehana’s tank can be filled up in 30 minutes for by the municipality water supply pipe, opening all the taps, three hours. The can work for 4 hours with the water of the full tank. In one day the water supply pipe remains open for 25 minute.

∵ Municipality pipe fills up 1 part in 30 minutes.

∴ Municipality pipe fills up \(\frac{1}{30}\) part in 1 minutes.

∴ Municipality pipe fills up \(\frac{25}{30}\) part in 25 minutes.

∴ Municipality pipe fills up \(\frac{5}{6}\) part in 25 minutes.

In mathematical language the problem is –

If part of tank decreases, time will decrease.

∴ Here is direct proportion.

∴ \(\frac{1}{\frac{5}{6}}=\frac{4}{?}\)

or, \(\frac{6}{5}=\frac{4}{?}\)

∴ Required time = \(\frac{4 \times 5}{6}\) hours

= \(\frac{10}{3}\) hours

= \(3 \frac{1}{3}\) hours

With that water they can work for \(3 \frac{1}{3}\) hours.

Question 7. Ruma and Rohit can complete a work in 20 days. Rohit and Sobha can complete that work in 15 days. Let’s calculate in how many days they will complete the work together.

Solution:

Given

Ruma and Rohit can do \(\frac{1}{20}\) part in 1 day. Rohit and Sobha can do in 1 day. Ruma and Sobha can do \(\frac{1}{15}\) part in 1 day. Ruma and Sobha can do \(\frac{1}{20}\) part in 1 day.

∴ 2 × (Ruma + Rohit + Sobha) can do \(\left(\frac{1}{20}+\frac{1}{15}+\frac{1}{20}\right)\)  part in 1 day part

= \(\frac{3+4+3}{60}\) part

= \(\frac{10}{60}\) part

= \(\frac{1}{6}\) part

∴ Ruma + Rohit + Sobha can do \(\frac{1}{2 \times 6}\) part in 1 day

= \(\frac{1}{12}\)

∵ Three of then can do \(\frac{1}{12}\) part in 1 day.

∴ All three together can complete the work in 12 day.

∴ Ruma can do \(\left(\frac{1}{12}-\frac{1}{15}\right)\) part in 1 day

= \(\frac{5-4}{60} \text { part }\)

= \(\frac{1}{60} \text { part }\)

Ruma can do \(\frac{1}{60} \text { part }\) part in 1 day.

∴ Ruma can complete the work alone in 60 days.

Ruma can do \(\left(\frac{1}{12}-\frac{1}{20}\right)\) part in 1 day

= \(\frac{5-3}{60}\) part

= \(\frac{2}{6}\) part

= \(\frac{1}{30}\) part

Rohit can do \(\frac{1}{30}\) part in 1 day.

∴ Rohit can complete the work alone in 30 days.

Sobha can do in 1 day \(\left(\frac{1}{12}-\frac{1}{20}\right)\) part

= \(\frac{5-3}{60}\)

= \(\frac{2}{6} \)

=\(\frac{1}{30} \)

Sobha can complete \(\frac{1}{30}\) part in 1 day.

∴ Sobha can do the work alone in 30 days.

Question 8. Alok, Kalam and Joseph individually can complete a work in 10, 12 and 15 days respectively. They started doing the work together. After 3 days Kalam had to go. Let’s make proportion and calculate in how many days Alok and Joseph will complete the remaining work.

Solution:

Given

Alok, Kalam and Joseph individually can complete a work in 10, 12 and 15 days respectively. They started doing the work together. After 3 days Kalam had to go.

Alok can do \(\frac{1}{10}\) part in 1 day.

Kalam can do \(\frac{1}{12}\) part in 1 day.

Joseph can do \(\frac{1}{15}\) part in 1 day.

All three together can do \(\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right)\)

= \(\frac{6+5+4}{60}\)

= \(\frac{15}{60}\) part

= \(\frac{1}{4}\) part

All three together can do 3 × \(\frac{1}{4}\) part =\(\frac{3}{4}\) part in 3 days.

Remaining work \(\left(1-\frac{3}{4}\right)\) part

= \(\frac{4-3}{4}\) part = \(\frac{1}{4}\)

Alok and Joseph can do together \(\left(\frac{1}{10}+\frac{1}{15}\right)\) part in 1 day

= \(\frac{3+2}{30}\) part

= \(\frac{5}{30}\) part

= \(\frac{1}{6}\) part

Alok and Joseph can together work \(\frac{1}{6}\) part in 1 day.

∴ Alok and Joseph can together work 1 part in 6 days.

∴ Alok and Joseph can together work \(\frac{1}{4}\)  part in \(\frac{1}{6}\)×6 days

= \(\frac{3}{2}\) days

= \(1 \frac{1}{2}\) days

Alok and Joseph will complete the remaining work in \(1 \frac{1}{2}\)days.

Question 9. Mary and David can do a work individually in 10 days and 15 days respectively. At first Mary alone worked for 4 days, then David alone worked for 5 days and left. Maria came and completed the remaining work in 4 days. If Mary, David and Maria would work together, let’s calculate in how many days they woild complete the work.

Solution:

Given

Mary and David can do a work individually in 10 days and 15 days respectively. At first Mary alone worked for 4 days, then David alone worked for 5 days and left. Maria came and completed the remaining work in 4 days.

Mary can do \(\frac{1}{10}\) part in 1 day.

∴ Mary can do \(\frac{4}{10}\) part = \(\frac{2}{5}\) part in 4 days.

David alone can do \(\frac{1}{15}\) part in 1 day.

∴ David alone can do \(\frac{5}{15}\) part = \(\frac{1}{3}\) part in 5 days.

Both can do \(\left(\frac{2}{5}+\frac{1}{3}\right)\) part of the total work

= \(\frac{6+5}{15}\) part

= \(\frac{11}{15}\) part

Remaining work = \(\left(1-\frac{11}{15}\right)\)

= \(\frac{15-11}{15}\) part

= \(\frac{4}{15}\) part

Maria can do \(\frac{4}{15}\) part in 4 days.

∴ Maria can do  \(\frac{4}{4 \times 15}\) part = \(\frac{1}{15}\) part in 1 day.

Mary, David and Maria together can do \(\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{15}\right)\) part in 1 day.

= \(\frac{3+2+2}{30}\) part

= \(\frac{7}{30}\) part

All three can do \(\frac{7}{30}\) part of the work in 1 day.

∴ All three can do 1part of  the work in \(\frac{30}{7}\) days.

∴ All three can complete the work in \(4 \frac{2}{7}\) days.

If Mary, David and Maria do the work together, then in \(4 \frac{2}{7}\) days it will be completed.

Question 10. A municipality made a tank for water preservation and added pumps with it. The pumps separately can fill the empty tank in 16, 20 & 30 hours respectively. Today when the 3 pumps started together at 7 am, \(\frac{1}{3}\) part of the tank was filled with water. After 1 hour 36 minutes the first pump, and after 2 more hours, the third pump were stopped.

Solution:

Given

A municipality made a tank for water preservation and added pumps with it. The pumps separately can fill the empty tank in 16, 20 & 30 hours respectively. Today when the 3 pumps started together at 7 am, \(\frac{1}{3}\) part of the tank was filled with water. After 1 hour 36 minutes the first pump, and after 2 more hours, the third pump were stopped.

First pump can fill \(\frac{1}{16}\) part in 1 hour.

Second pump can fill \(\frac{17}{20}\) part in 1 hour.

Third pump can fill \(\frac{1}{30}\) part in 1 hour.

All three pumps can fill

= \(\left(\frac{1}{16}+\frac{1}{20}+\frac{1}{30}\right)\) part in 1 hour

= \(\frac{15+12+8}{240}\) part

= \(\frac{35}{240}\) part

= \(\frac{7}{48}\) part

1 hour 36 minutes = 96 minutes

∵ All three pumps fill = \(\frac{7}{48}\) part in 60 minutes.

∴ All three pumps fill \(\frac{7 \times 96}{48 \times 60}\) part in 96 minutes.

= \(\frac{7}{30}\)

Second and third pump fill = \(\left(\frac{1}{20}+\frac{1}{30}\right)\) part in 1 hour.

= \(\frac{3+2}{60}\) part

= \(\frac{5}{60}\) part

= \(\frac{1}{12}\) part

∴ Second and third pump fill = \(2 \times \frac{1}{12}\) part part in 2 hours.

= \(\frac{1}{6}\) part

Now, the filled part of the tank = \(\left(\frac{1}{3}+\frac{1}{30}+\frac{1}{6}\right)\) part

= \(\frac{10+7+5}{30}\) part

= \(\frac{22}{30}\) part

= \(\frac{11}{15}\) part

Remaining to be filled = \(\left(1-\frac{11}{15}\right)\) part

= \(\frac{15-11}{15}\) part

= \(\frac{4}{15}\) part

First pump fills 1 part in 20 hours.

∴ Second pump will fill

= \(\frac{4}{15}\) part in

= \(\frac{4}{15} \times 20\) hours

= \(\frac{16}{3}\) hours

= 5 hours 20 minutes

1. Let’s calculate when the tank was totally filled up.

Solution:

Time required to fill the tank fully

= 1 hour 36 minutes + 2 hours + 5 hours 20 minutes

= 8 hours 56 minutes

Morning 7 am + 8 hours 56 minutes = Evening 3 hours 56 minutes

The tank will be filled completely at 3 o’clock 56 minutes.

2. Let’s calculate how much of the tank the second pump filled up.

Solution:

Time of second pump

= 1 hours 36 minutes + 2 hours + 5 hours 20 minutes

= 8 hours 56 minutes

= (8 x 60 + 56) minutes (480 + 56) minutes

= 536 minutes

Second pump can fill up = \(\frac{1}{20}\) part in 60 minutes.

∴ Second pump can fill up = \(\frac{1}{20 \times 60}\) part in 1 minutes.

∴  Second pump can fill up = \(\frac{1 \times 536}{20 \times 60}\) part in 536 minutes.

∴ Second pump can fill up = \(\frac{67}{150}\) part in 536 minutes.

Second pump filled up = \(\frac{67}{150}\) part of the tank.

“WBBSE Class 8 Maths Chapter 17 solutions, Time and Work PDF”

3. When the third pump is stopped, let’s calculate how much of the tank was filled with water.

Solution:

When third pump was was stopped, \(\frac{4}{15}\) part was empty.

∴ Filled up part = \(\left(1-\frac{4}{15}\right)\)

= \(\frac{15-4}{15}\)

= \(\frac{11}{15}\)

When the third pump was stopped, \(\frac{11}{15}\) part of the tank was filled with water.

Question 11. My friend Rina can do garden work alone in 4 hours. I can do that work alone in 9 hours. But if we together do that work, let’s calculate how much time will be needed with number.

Solution: 

Given

My friend Rina can do garden work alone in 4 hours. I can do that work alone in 9 hours.

Rina can do \(\frac{1}{4}\) part of the work in 1 hour.

I can do\(\frac{1}{3}\) part of the work in 1 hour.

Both can do together

= \(\left(\frac{1}{4}+\frac{4}{15}\right)\) part of the work in 1 hour

= \(\frac{3+4}{12}\) part

= \(\frac{7}{12}\) part

Both can do \(\frac{7}{12}\) part of the work in 1 hour.

∴ Both can do the work in \(\frac{12}{7}\) hours

= \(1 \frac{5}{7}\) hours.

Both can do the work in = \(1 \frac{5}{7}\) hours.

Construction Of Triangles

If I draw a triangle, lengths of whose two sides are 5 cm and 4 cm and the measurement of the angle opposite to the side of length 4 cm is 45°, then let’s see what type of triangle I shall get.

It is seen that it is not possible to draw any triangle with these conditions.

1. Firstly, 30° arid 45° are drawn and a straight line 6 cm long is drawn.
2. Now straight line AX is drawn and from it 6 cm long is cut off.
3. On points, A and B of straight line AB, two angles equal to 45° ∠YAB and ∠ZBX are drawn respectively.
4. Now on point B equal to 30° is drawn AY straight line on the same side of BZ ZPBZ. PB and AY cut each other at C point

In ΔABC = 6 cm

Read and Learn More WBBSE Solutions For Class 8 Maths

∠CAB = 45 and ∠ACB = 30°

Proof: ∠XBZ = ∠XAY (By construction)

∴ ∠BZ//AY (∵ Corresponding angles)

∴ ∠ZBP = Alternate angle ∠BCA

∵ ∠ZBP = 30°

∴∠BCA = 30°

∴ In ΔABC ∠CAB = 45°, ∠ACB = 30°

and the opposite side of 30° is AB = 6 cm

“Class 8 WBBSE Maths Chapter 21 solutions, Construction of Triangles study material”

Construction Of Triangles Exercise 21.1

Question 1. If I draw a triangle whose 2 sides are 5 cm and 4 cm and the measurement of the angle opposite to the side of length 4 cm is 45°, then let’s see what type of triangle I shall get. It is seen that it isn’t possible to draw any triangle with these conditions.

Solution:

Given

If I draw a triangle whose 2 sides are 5 cm and 4 cm and the measurement of the angle opposite to the side of length 4 cm is 45°

1. With the help of scale two 5 cm long straight lines are drawn.
2. With the help of a scale, a ray AX is drawn. On the ray off. Taking AX at point A an angle of 45° ∠XAY is drawn.
3. From the ray A, X a straight line AB equal to 5 cm is cut off. Taking B as the centre, taking a radius equal to the straight line 4 cm, an arc is drawn which cuts ray AX at point C. B, and C are joined. An ABC is drawn whose AB = 5 cm, ∠BAC = 45° and BC = 4 cm and the angle opposite to the side BC ∠BAC = 45°.

Question 2. But why is it so? Sometimes we are getting one triangle, sometimes two triangles and sometimes no triangle. It is seen that the perpendicular distance from B on the AX line segment is BM = cm = I say.

Solution:

We see that perpendicular distance AX from point B to the straight line

BM = 3.6 cm l = let

a = 5cm

b = 4cm

We see that if b > a then |a  1 | triangle can be drawn.

If l < b < a then 2 triangles can be drawn.

If b = a then 1 triangle can be drawn.

If b < l then no triangle can be drawn.

If b, = l then 1  triangle can be drawn.

“WBBSE Class 8 Maths Chapter 21, Construction of Triangles solved examples”

If a = b, i.e., try to form a triangle whose two sides a = 5 cm, b = 5 cm and the opposite angle to the 5 cm long side is ∠x= 100°

Question 3. Let’s see whether such a triangle can be drawn.

Solution:

straight line AC is drawn and on point A equal to angle 100°, ∠BAC is drawn.

From the AD side, a 5 cm long AB is cut off. Now taking B as the centre 5 cm straight line equal to taking as radius, on a point on AC an arc is drawn then it is seen that cuts at only on one point of AC, A. So it is not possible to draw a triangle.

Question 4. If a < b, i.e., try to form a triangle whose two sides a = 5 cm, b = 4 cm and the opposite angle of 4the  cm long side is x = 100°. Let’s see if such a triangle can be drawn.

Solution:

Given

if a < b, i.e., try to form a triangle whose two sides a = 5 cm, b = 4 cm and the opposite angle of 4 cm long side is x = 100°.

A straight line AC is drawn and on point A equal to angle 100°, ∠BAC is drawn. From the AD side, a 5 cm long AB is cut off.

Now taking point B as a centre and taking a radius length equal to a 4 cm straight line AC cut an arc on any point then it is seen that the AC arc doesn’t cut at any point.

So it is not possible to construct a triangle.

Construction Of Triangles Exercise 21.2

Question 1. Let’s draw a triangle whose two sides are 6 cm and 7 cm and the measurement of the angle opposite to the side of length 7 cm is 60°. Let’s write what will be the measurement of the sides to form a triangle.

Solution:

Measurement of the sides to form the triangle

A straight line AX is drawn and on point A equal to angle 60°, ∠DAX is drawn. From side AD, AB is cut equal to 6 cm long. Taking B as the centre and equal to 7 cm long taking as radius point AX is cut at C. B, and C are joined. ABC is the required triangle.

From point B to AC, if the length of the side is more than the length of the perpendicular drawn then two triangles will be formed.

“WBBSE Class 8 Construction of Triangles solutions, Maths Chapter 21”

Question 2. Let’s construct a triangle whose lengths of two sides are 6 cm and 9 cm and the measurement of the angle opposite to the side of length 9 cm is 105°. Let’s write for what length of sides we will not be able to construct two triangles.

Solution:

“Class 8 WBBSE Maths Chapter 21, Construction of Triangles easy explanation”

A straight line AX is drawn and at AX on point A equal to angle 105°, ∠DAX is drawn. From side AD AB is cut equal to 6 cm long. Taking B as centre equal to 9 cm length taking as radius AX is cut at point C. B, and C are joined. ABC is the required triangle.

Geometrical Proofs

Mitali gave me two sticks of length 3 cm and 5 cm respectively.

I took a stick of length 2 cm and attached it with sticks given by Mitali and we got a figure like the given figure beside.

But it is seen that we are not able to form a triangle. Here it is seen – that 3 cm + 2 cm [I] 5 cm (put =/</> ). Jasmin replaced the stick of length 2 cm by a stick of length 3 cm and she tried to form a figure like a triangle and she is able to make the figure.

Here it is seen that 3 cm + 3 cm > 5 cm ( put =/</>)

Sneha made a figure Triangle with three sticks 3 cm, 5 cm and 4 cm respectively.

Here it is seen that 4 cm + 3 cm g] 5 cm ( put =/</>).

“WBBSE Class 8 Maths Chapter 22 solutions, Construction of Parallel Lines”

I am able to form a shape like a triangle with two sides of lengths 3 cm, 5 cm and a third stick of length 6 cm or 7 cm, i.e., it is seen that the figures like triangles are formed with the sticks of length (3 cm,6 cm & 5 cm) and (3 cm, 7 cm & 5 cm).

Read and Learn More WBBSE Solutions For Class 8 Maths

Sneha, by placing the shapes like this triangle on the exercise book, she got some triangles, of which the sum of length of third sides Big (big / small).

Now Sucheta tried to make a shape like a triangle with the sticks of lengths 9 cm, 3 cm, and 5 cm and she got –

Here the sum of lengths of any two sides is not greater than the length of the third side.

We got,

1. The length of the sticks which makes a figure like a triangle with the sticks of length 5 cm and 3 cm is Big (less / more) than 5 cm – 3 cm = 2 cm and then small (less/more) (5 + 3) cm = 8 cm

2. The sum of the length of any two sides of a triangle is then the length of the third side Big (less/more)

Now Anik gave two sticks of length 2 cm and 4 cm.

“Class 8 WBBSE Maths Chapter 22 solutions, Construction of Parallel Lines study material”

Now let’s try to make a figure like a triangle with sticks of different lengths and let’s see what I get.

It is seen that

1. It is possible to construct a figure like a triangle with the (4-2) cm = 2 cm Big (less/more) long stick and (4 + 2) cm = 6 cm small (less/more) long stick 2 cm and 4 cm long sticks.

2. Again it is also seen that the sum of the lengths of any two sides is Big (less/more) than the length of the third side.

Geometrical Proofs Exercise

Question 1. Let’s see what the length of the sticks can make a figure like a triangle with the sticks of length 4 cm and 6 cm.

Solution:

It is possible to make a triangle with sticks of length greater than (6-4) cm = 2 cm and less than less than (6+4) cm = 10, i.e., from 3 to 9 cm triangle can be made with sticks of any length.

Question 2. If the length of two sides of triangle are 3 cm and 6 cm then let’s find the length of the third side that will lie between which two numbers. Let’s write the relation between the sum of the lengths of any two sticks with the length of the third stick.

Solution:

Given

If the length of two sides of triangle are 3 cm and 6 cm then let’s find the length of the third side that will lie between which two numbers.

(6-3) cm = 3 cm greater than

∴ The length of the third side can be 4 cm or 5 cm.

Aminur has drawn some triangles on the exercise books. He said let’s try to find the relation between the length of each side of these triangles.

AB = 2.6 cm, BC = 4.6 cm, CA = 3.4 cm

AB+BC > CA [Put >/<]

BC+CA > BC [Put >/<]

CA+AB > [put >/>]

Similarly measuring the length of the sides of  ΔPQR & ΔXYZ, it is seen that the sum of the lengths of two sides is Big than the length of third side.

Question 3. Megha has drawn PQR triangle. I will prove with the help of facts that PQ + QR > PR ; OR + RP > PQ and RP + PQ > QR.

Solution:

Given : Let the greatest side of PQR triangle is QR.

We have to prove that RP + PQ > QR.

Construction : From P, the highest point of ΔPQR, PS a perpendicular is drawn on OR which cuts, QR at S i.e., PS ⊥ QR.

Proof : In ΔPSR ∠PSR=1 right angle (Constructed)

∴ ∠PSR is a right angle and ∠SPR is an acute angle.

∴ ∠PSR> ∠SPR

So, PR > SR …………. (1)

In ΔPSQ < PSQ = 1 right angle (∵ PS⊥QR)

∴ ∠PSQ right angle and ∠QPS acute angle ∠

∴ PSQ>∠QPS

So : PQ > QS …………… (2)

by adding equations (1) and (2)

PR + PQ > SR + QS

or, PR + PQ > QR Proved

“WBBSE Class 8 Maths Chapter 22, Construction of Parallel Lines solved examples”

Geometrical Proofs Exercise

Question 1. Let’s prove logically step by step that the difference of the lengths of any two sides of triangle is less than the length of the third side.

Solution:

Let the greatest side of ΔABC is BC.

We have to prove that AB-AC<BC.

Construction : From A, the highest point of ΔABC, AD a perpendicular is drawn on BC which cuts BC at D, i.e., AD ⊥ BC

Proof : In ΔABD ∠ADB=1 right angle (∵ AD ⊥ BC)

∴ ∠ADB > ∠BAD

∴ AB > BD ………………..(1)

In ΔADC ∠ADC=1 right angle

∴ ∠ADC is a right angle and ∠DAC is an acute angle.

∴ ∠ADC > ∠DAC

∴ AC > DC ………………(2)

(1) — (2) AB – AC < BD + DC

or, AB – AC < BC Proved

Question 2. Let’s observe the lengths of the following sides and let’s write in which cases it is possible to draw a triangle –

1. (3 cm, 6 cm and 8 cm)

Solution:

Given

3 cm + 6 cm > 8 cm

∴ It is possible to construct a triangle.

2. (8 cm, 6 cm and 15 cm)

Solution:

Given

8 cm + 6 cm < 15 cm

∴ It is not possible to construct a triangle.

3. (2.7 cm, 6.1 cm and 8.8 cm) .

Solution:

Given

2.7 cm + 6.1 cm – 8.8 cm

∴ It is possible to construct a triangle.

4. (2.5 cm, 8 cm and 6 cm) .

Solution:

Given

2.5 cm + 6 cm > 8 cm

∴ It is possible to construct a triangle.

Geometrical Proofs Exercise

Question 1. D is any, point on the side BC of A ABC. Let’s prove that AB + BC + CA > 2AD.

Solution:

Given

In A ABC, D is a point on BC.

Prove that AB + BC + CA > 2AD

Construction : A straight line AD is drawn.

Proof : In ΔABD,

AB + BD > AD …………(1)

(∴ The sum of two sides of a triangle is greater than the third side.)

Again, In ΔADC,

DC + AC > AD …………….(2)

by . adding equations (1) and (2) .

AB + BD + DC + AC > AD + AD

or, AB + BC + AC > 2AD Proved

Question 2. O is any point inside ΔABC. Let’s prove that:

1. AB + AC > OB + OC
2. AB + BC + AC > OA + OB + OC

Construction : AO is extended which cuts BC at Q. Similarly BO and CO are extended which cut AC and AB at points R and P respectively.

Proof : In ΔABQ, AB + AQ > BQ >

or, AB + AQ > OB + OQ ………….(1)

In A COQ, OQ + CQ > OC ………..(2)

By adding equations (1) and (2)

AB + AQ + OQ + CQ > OB > OB + OQ + OC

or, AB + (AQ + CQ) > OB + OC

or, AB + AC > OB + OC ……………(3)

∵ BC > OA ………………………(4)

By adding equations (3) and (4)

AB + AC + BC > OB + OC + OA Proved

Question 3. Let’s prove that the perimeter of a triangle is more than the sum of lengths of the medians.

Solution:

Let the three medians of A ABC are AD, BE and CF.

We have to prove that AB + BC + CA > AD + BE + CF.

Proof :  ∵ BA + AC > 2AD …………….(1)

BC + AC > 2CF …………………..(2)

and AB + BC > 2BE ………………….(3)

By adding (1), (2) and (3)

AB + AC + BC + AC + AB +BC > 2AD + 2CF + 2BE

or, 2(AB + BC + AC) > 2 (AD + CF + BE)

or, AB + BC + AC > AD + BE + CF Proved

Question 4. P is any point inside A ABC. Let’s prove that,

1. AB + BP > AB
2. AB + BC + AC < 2 (AP + BP + CP)

Solution:

Given

In ΔABC, P is any point. Prove that (1) AP + BP > AB (2) AB + BC + AC < 2 (AP + BC + CP)

Construction: A, P ; B, P and C, P are joined.

Proof: In A ABP

(1) AP + BP > AB (∵ The sum of any two sides of a triangle is greater than the third side.)

In A ABP, AB<AP + BP……………………….. (1)

In A BPC, BC < BP + PC………………. (2)

and in A APC, AC < AP + PC …………….. (3)

by adding (1), (2) and (3)

AB + BC + AC < 2AP + 2BP + 2PC

or, AB + BC + AC < 2 (AP + BP + PC) (2) Proved

“WBBSE Class 8 Construction of Parallel Lines solutions, Maths Chapter 22”

Question 6. Let’s prove that the sum of lengths of two diagonals is greater than the sum of the lengths of any two opposite sides of a quadrilateral.

Solution:

Let ABCD is a quadrilateral whose two diagonals are AC and BD. Let’s prove that AC + BD > AD + BC.

Proof : In AOD, OA + OD > AD …………..(1)

In BOC, OB + OC > BC ………….(2)

by adding (1) and (2)

OA + OD + OB + OC > AD + BC or, (OA + OC) + (OB + OD) > AD + BC

or, AC + BD > AD + BC Proved

Question 7. Let’s prove that the sum of lengths of two diagonals of a quadrilateral is more tham the semi-perimeter of the quadrilateral.

Solution:

Let ABCD is a quadrilateral whose two diagonals AC and BD cut each other at 0.

Proof: In AOD, AO + OD > AD ………………… (1)

In AOB, AO + OB > AB ………………… (2)

In BOC, OB + OC > BC ………………… (3)

In COD, OC + OD > CD ………………… (4)

by adding (1), (2), (3) and (4)

AO + OD + AO + OB + OB + OC + OC + OD > AD + AB + BC + CD

or, 2A0 + 20C + 20D + 20B > AB + BC + CD + AD

or, 2 (A0 + OC) + 2(0D + OB) > AB + BC + CD + AD

or, 2AC + 2BD > AB + BC + CD + AD

or, 2 (AC + BD) > AB + BC + CD + AD

or, AC + BD > (AB + BC + CD + AD) Proved

Question 8. From a point inside a quadrilateral (not on any diagonal) we join the vertices of the quadrilateral. Let’s prove that the sum of lengths of these line segments is greater than the sum of lengths of the diagonals. Now let’s see for which position of the point inside the quadrilateral the sum of lengths of the line segment obtained by the vertices of the quadrilateral with that point will be the smallest.

Solution:

Given

From a point inside a quadrilateral (not on any diagonal) we join the vertices of the quadrilateral.

Let inside ABCD quadrilateral 0 is any point and AC and BD are the two diagonals.

Let’s prove that: OA + OB + OC + OB > AC + BD

Construction : O, A; O, B; C; and O, D

Proof: From figure (1) In ΔAOC, OA + OC > AC ……………….(1)

and in ΔBOD, OB + OD > BD ……………….(2),

From (1) + (2), OA + OC + OB + OD > AC + BD (Proved)

From figure (2) In ΔAOB, AO + OB > AB ………………….(1)

(The sum of lengths of any two sides of a triangle is greater than length of the third side.)

In A AOC, BO + OC > BC ……………….. (2) (For the above reason)

In A GOD, OC + OD > CD ………………. (3) (For the above reason)

In A ODA, OD + OA > AD ……………………. (4) (For the above reason)

Now, from A + B + C + D

AO+OB+BO+OC+OC+OD+OD+OA>AB+BC+AD

or, 2A0+20C+20B+20D>AB+BC+CD+CA

or, AO+OG+OB+OD> \(\frac{1}{2}\) (AB+BC+GD+GA)………….(A)

From figure (3) In ΔAOD, AO + OD > AD ……………(1)

In ΔAOB, AO + OB > AB ………..(2)

In ΔBOC, OB + OC > BC ………………(3)

In ΔCOD, OC + OD > CD ……………. (4)

by adding (1+2+3+4)

AO+OD+AO+OB+OB+OC+OC+OD>AD+AB+BC+CD

or, 2AC + 20C + 20D+20B>AD+AB+BC+CD

or, 2(AO+OC)+2(OD+OB)>AD+AB+BC+CD

or, 2AC =2BD>AD+AB+BC+CD

or, AC+BD> (AB+BC+CD+AD)

AC+BD> (AB+BC+CD+AD) ………………….(B)

From (A) and (B)

AC + BD < OA + OB + OC + OD

So we see that if the point inside the quadrilateral is the point of intersection of two diagonals then the sum of lengths of line segments obtained by vertices of quadrilateral with that point will be the smallest.

Salema and Bibhas have made-

I drew these figures on a thick art paper and named them. I measured the interior angles b>Ca protractor and got :

Measuring by protractor, we see, In ΔABC ∠ABC = 60 degree ∠ABC = 60 degree and ∠ACB= 60 degree

By measuring angles of quadilateral HAND we see ∠AHD = 90 degree ∠HAN = 90 degree ∠AND = 90 degree ∠HDN = 90 degree

By measuring angles of pentagon CAMEL we see ∠ACL = 90 degree ∠CLE = 110 degree ∠LEM =100 degree ∠EMA = 90 degree and ∠MAC = 140 degree

By measuring angles of quadrilateral REST we see

∠ERT = 100 degree ∠RES = 80 degree ∠EST = 110 degree and ∠RTS = 70 degree

Now I add the measurement of the interior angles of each polygon and then let’s try to find a general formula.

Sum of the measurement of the interior angles of ΔABC = ∠BAC + ∠ABC +∠ACB = 180 degree = 2 right angles

“Class 8 WBBSE Maths Chapter 22, Construction of Parallel Lines easy explanation”

Sum of the measurement of the interior angles of HAND = ∠AHD + ∠HAN + ∠AND + ∠BAC = 360 degree = 4 right angles

Sum of the interior angles of CAMEL = ∠ACL + ∠CAM + ∠AME + ∠MEL + ∠ELC = 540 degree = 6 right angles

Sum of the interior angles of REST = ∠ERS + ∠RES + ∠EST + ∠RTS = 360° degree= 4 right angles

Let’s try to find out the relation between the interior angles and the number of sides of a polygon.

Question 9. The number of sides of a regular polygon is 18. Let’s find the measurement of each interior angle and each exterior angle.

Solution:

Given

The number of sides of a regular polygon is 18.

Sum of measurements of 18 interior angles of the polygon is = 2 (18-2) × 90°

= 2 × 16 × 90°= 2880°

∴ Measure of each interior angle is = 2880° -18 = 160°

Sum of measurements of all exterior angles is = 360°

∴ Measurement of each exterior angle is = 360, 18=20°

Alternatively, measurement of 1 interior angle + 1 measurement of exterior angle = 180°

∴ Measurement of 1 exterior angle = 360° 18 = 20°

Measurement of 1 interior angle = 180° – 20° = 160°

Question 10. Measurement of each exterior angle of a regular polygon is 144°. Let’s find the number of sides of the polygon.

Solution:

Given

Measurement of each exterior angle of a regular polygon is 144°.

Measurement of 1 interior angle + measurement of 1 exterior angle = 180

∴ Measurement of 1 exterior angle = 180°

-144° = 36°

∴ Number of sides of the regular polygon = 360° ÷ 360° = 10

Geometrical Proofs Exercise

Question 1. Let’s prove logically the sum of four interior angles of a quadrilateral is 360°.

Solution:

Let ABCD is a quadrilateral.

Let’s prove that ∠A + ∠B + ∠C + ∠D = 360°

Construction : Diagonal BD is drawn.

Proof : In ΔABD, ∠A + ∠ABD + ∠ADB = 180°

(∵ Sum of all angles of a triangle is 180°)

Again, In ΔBCD ZBDC + ZCBD + ZC = 180°

ΔABD+ ABCD= ∠A+ ∠ABD+ ∠ADB+ ∠BDC+ ∠CBD+ ∠C= ∠A+ ∠B+ ∠C+ ∠D

∴ ΔABD + ABCD = ∠A+∠B+∠C+∠D = 180°+180°

∴ ABCD =∠A+ ∠B+ ∠C+ ∠D= 360° Proved

Question 2. Let’s write by calculation the sum of the measurements of eight interior angles of an octagon.

Solution:

Sum of measurements of the interior angles of an octagon = 2 (8-2) × 90°

Sum = 2 (8-2) × 90°

= 2 × 6 × 90°

= 10802°

The sum of the measurements of eight interior angles of an octagon = 10802°

Question 3. Let’s write the measurement of each interior and enterior angle of a regular polygon with 10 sides.

Solution:

Sum of measurements of 10 interior angles of a decagon = 2 (10-2) × 90° .

= 2 × 8 × 90° = 1440°

∴ Measurement of each interior angle = 1440° ÷ 10 = 144°

Measurement of each exterior angle = \(\frac{360^{\circ}}{10}\)= 36°

“WBBSE Class 8 Maths Chapter 20, Geometrical Proofs solved examples”

Question 4. Let’s write by calculation the number of sides of a regular polygon where measurement of each angle is 120°.

Solution:

Each interior angle of the polygon = 120°

∴ Measurement of each exterior angle of the polygon = 180° – 120°

Number of sides of the regular polygon = \(\frac{360^{\circ}}{10}\) = 6

Geometrical Proofs Exercise

Question 1. Let’s write the sum of the interior angles of the following polygons:

1. Pentagon
2. Hexagon
3. Heptagon
4. Octagon
5. Decagon
6. A polygon with 12 sides.

1. Pentagon

Solution:

Sum of measurements of the interior angles of a pentagon

= 2 (5-2) × 90°

= 2 × 3 × 90° = 540°

pentagon = 540°

2. Hexagon

Solution:

Sum of measurements of the interior angles of a hexagon

= 2(6-2) × 90°

= 2 × 4 × 90°

= 720°

Hexagon = 720°

3. Heptagon

Solution:

Sum of measurements of the interior angles of a heptagon = 2

(7-2) × 90°

= 900°

Heptagon = 900°

4. Octagon

Solution:

Sum of measurements of the interior angles of an octagon

= 2 (8-2) × 90°

= 2 × 6 × 90° = 1080°

Octagon = 1080°

5. Decagon

Solution:

Sum of measurements of the interior angles of a decagon = 2

(10-2) × 90°

= 2×8×90°

= 1440°

Decagon = 1440°

6. Polygon with 12 sides

Solution:

Sum of measurements of the interior angles of a polygon with 12 sides

= 2 (12-2) × 90°

= 1800°

Polygon with 12 sides = 1800°

Question 2. Three angles of a quadrilateral are 104.5°, 65° and 72.5°; find the measurement of the fourth angle.

Solution:

Given

Three angles of a quadrilateral are 104.5°, 65° and 72.5°

Sum of measurements of three angles of a quadrilateral.

= 104.5° + 65° + 72.5°

= 242°

Sum of measurements of all four angles of a quadrilateral = 360°

∴ Measurements of the 4th angle of the quadrilateral = 360° – 242°

= 118°

Question 3. Four angles of a pentagon are respectively 65°, 89°, 132° and 116°. Let’s write the measurement of the fifth angle.

Solution:

Given

Four angles of a pentagon are respectively 65°, 89°, 132° and 116°.

Sum of 4 angles of a pentagon = 65° + 89° + 132° + 116° = 402°

Sum of all the interior angles of a pentagon = 2 (5- 2) × 90° .

= 2×3×90° =540°

Measurement of the 5th angle of the pentagon = 540°-402° = 138°

Question 4. Let’s write whether the measurement of 3 angles of a convex v quadrilateral being 68°, 70° and 75° is possible or not.

Solution:

Sum of 3 angles of a quadrilateral = 68° + 70° + 75° = 213°

∵ Sum of 4 angles of a quadrilateral = 360°

∵ 213° < 360°

∴ The three angles of a convex quadrilateral can be respectively 68°, 70° and 75°

Question 5. Let’s write whether the five angles of a convex hexagon 120°, 70°, 95°, 78° and 160° is possible or not.

Solution:

Sum of 5 angles of a hexagon = 120°+70°+95°+78°+160° = 523°

Sum of six angles of a hexagon = 2 (6-2) × 90°

= 2 × 4 × 90°

= 720°

∵ Measure of the 6th angle of a hexagon = 720° – 523° = 197°

∵ 197° > 180°, which is impossible.

∴ The five angles of a convex hexagon can’t be respectively 120°,

70°, 95°, 78° and 160°.

“WBBSE Class 8 Maths Chapter 22 solutions, Construction of Parallel Lines PDF”

Question 6. Let’s write the measurement of each interior and exterior angle of the following polygons :

1. Pentagon

Solution:

Sum of measurements of the interior angles of a pentagon

= 2 (5-2) × 90°

= 2 × 3 × 90°

= 540°

∴ Measurement of each interior angle of a regular pentagon

= 54° ÷ 5

= 108°

Interior angle + Exterior angle = 180°

Measurement of each interior angle of a reguiar pentagon= 180°

108° =72°

2. Hexagon

Solution:

Sum of measurements of the interior angles of a hexagon = 2 (6-2) × 90°

= 2 × 4 × 90°

= 720°

∴ Measurement of each interior angle of an regular hexagon = 720°÷6 = 120°

∴ Measurement of each exterior angle of a regular hexagon = 180° -120° = 60°

3. Octagon

Solution:

Sum of measurements of the interior angles of an Octagon

= 2 (8-2) x 90° = 2 × 6 × 90° = 1080°

∴ Measurement of each interior angle = 1080° + 8 = 135°

∴ Measurement of each exterior angle of the octagon = 180° – 135° = 45°

4. Polygon with 9 sides

Solution:

Sum of the interior angles of a polygon with 9 sides = 2 (9-2) × 90° = 1260° .

∴ Measurement of each interior angle of a polygon with 9 sides = 1260°-9= 140° –

∴ Measurement of each exterior angle of a polygon with 9 sides = 180°-120° = 40°

5. Decagon

Solution:

Sum of measurements of the interior angles of a Decagon

= 2 (10-2) x 90° = 2 × 8 × 90° = 1440°

∴ Measurement of each interior angle of a polygon with 10 sides = 1440° – 10 = 144°

∴ Measurement of each exterior angle of a regular polygon with 10 sides

= 180°-144° = 36°

6. Polygon with 18 sides

Solution:

Sum of measurements of the interior angles of a regular polygon with 18 sides.

= 2 (18-2) × 90°

= 2 × 16 × 90°

= 2880°

∴ Measurement of each interior angle of a polygon with 18 sides = 2880° ÷ 18 = 160°

∴ Measurement of each exterior angle of a polygon with 18 sides = 180°-160°=20°

Question 7. Let’s find whether the following measurements are possible or not for an exterior angle of a regular polygon – (write yes /no):

1. 6°
2. 10°
3. 13°
4. 18°
5. 35°

1. 6°

Solution:

The measurement of each exterior angle of a regular polygon can be 6°

2. 10°

Solution:

The measurement of each exterior angle of a regular polygon can be 10°.

3. 13°

Solution:

The measurement of each exterior angle of a regular polygon can be .13°.

4. 18°

Solution:

The measurement of each exterior angle of a regular polygon can be 18°.

5. 35°

Solution:

The measurement of each exterior angle of a regular polygon can be 35°.

“WBBSE Class 8 Maths Chapter 22, Construction of Parallel Lines important questions”

Question 8. Let’s find whether the following measurement of each interior angle is possible or not for a regular polygon

Solution:

1. 80°
2. 100°
3. 120°
4. 144°
5. 155°
6. 160°

1. 80°

Solution:

The measurement of each interior angle of a regular polygon can be 80° .

2. 100°

Solution:

The measurement of each interior angle of a regular polygon can be 100°.

3. 120°

Solution:

The measurement of each interior angle of a regular polygon can be 120°.

4. 144°

Solution:

The measurement of each interior angle of a regular polygon can be 144°.

5. 155°

Solution:

The measurement of each interior angle of a regular polygon can be 155°.

6. 160°

Solution:

The measurement of each interior angle of a regular polygon can be 160°.

Question 9. Each exterior angle of a regular polygon is 60°. What is the number of sides of it?

Solution:

Given

Each exterior angle of a regular polygon is 60°.

The measurement of each exterior angle of a regular polygon can be = 60°

∴ Number of sides of the regular polygon = \(\frac{360^{\circ}}{60^{\circ}}\)= 6

Question 10. Each interior angle of a regular polygon is 135°. What is the number of sides of that polygon?

Solution:

Given

Each interior angle of a regular polygon is 135°.

The measurement of each interior angle of a regular polygon can be = 135°

∴ The measurement of each exterior angle of a regular polygon can be = 180° – 135° = 45°

∴ Number of sides of the regular polygon = \(\frac{360^{\circ}}{45^{\circ}}\)= 8

Question 11. The ratio of measurements of an interior angle and an exterior angle of a polygon is 3:2. Let’s write the number of sides of the polygon.

Solution:

Given

Ratio of measurements of each interior and exterior angles of a regular polygon = 3:2

Sum of ratio = 3 + 2 = 5

∴ Measurement of exterior angle = \(\frac{2}{5}\) × 180° = 72°

∴ Number of sides of the regular polygon = \(\frac{360^{\circ}}{72^{\circ}}\) = 5

Question 12. The sum of measurements of all exterior angles of a polygon is 18000°. Let’s write the number of sides of the polygon.

Solution:

Given

The sum of measurements of all exterior angles of a polygon is 18000°.

Let the number of sides’ of the polygon = n

Sum of measurements of all the interior angles of the polygon = 2 (n-2) × 9

B.T.P.,

2 (n-2) × 90° = 1800°

or’ \(n-2=\frac{1800^{\circ}}{2 \times 90^{\circ}}\)

or, n – 2 = 10

or, n -2 = 10 + 2 = 12

∴ Number of sides of the polygon = 12

Question 13. Measurement of each of 5 interior angles is 172° and the measurement of each of the other interior angles of a polygon is 160°. Let’s write number of sides of the polygon.

Solution:

Given

Measurement of each of 5 interior angles is 172° and the measurement of each of the other interior angles of a polygon is 160°.

Let the number of sides of the polygon be n.

∴ Sum of measurements of the interior anqles of the polyqon = 2 (n-2) × 90°

Total of five interior angles ofthe polygon = 5 × 172° = 860°

∴ Measurements of the remaining interior angles = (n- 5) × 160°

2(n-2)×90° = 860°-(n-5)×160°

or, 2(n-2)×90°-(n-5)×160°= 860°

or, 20°{9(n-2)-8(n-5)} = 860°

or, 9x-18-8n+40 = 43

or, n = 43+18-40

or, n = 21

∴ n = 21

∴ Number of sides of polygon = 21

“Class 8 Maths Construction of Parallel Lines solutions, WBBSE syllabus”

Question 14. ABODE is a regular pentagon. Let’s prove the AABC is an isosceles triangle, BE and CD are two parallel line segments.

Solution:

Given : ABODE is a regular pentagon.

We have to prove that :

1. ABC is an isoceles triangle and

2. BE//CD.

Proof : In ΔABC

AB = BC (∵ABCDE is a regular pentagon)

∴ ΔABC is an isoceles triangle.

Question 15. ABCDEF is a regular hexagon. The bisector of ∠BAF interects DE at X. Let’s write the measurement of ∠AXD.

Solution:

Given

∴ ABCDEF is a regular hexagon.

∴ Measurement of each of its interior angles

= \(\frac{2(6-2) \times 90^{\circ}}{6}\) = \(\frac{2 \times 4 \times 90^{\circ}}{6}=120^{\circ}\)

∴ ∠B= ∠C= ∠D = 120°

∴ ∠B + ∠C + ∠D = 3 x 120° = 360°

∠BAX = \(\frac{120^{\circ}}{2^{\circ}}\) = 60° (∵ ∠A is the bisector of AX.)

∵ ABCDX is a pentagon

∴ Sum of measurements of the interior angles of ABCDX = 2(5-2) × 906 = 2 × 3 × 90° = 540°

Sum of measurements of the 4 angles of ABCDX = 360° + 60° = 420°

∴ Pentagon AXD = 540° – 420° = 120°

Manasi constructs some triangles with some sticks like the figure beside. The bases of these triangles are in a same straight line and have same vertex.

Question 17. Let’s try to measure the lengths of the sides of the triangle, i.e., the lengths of the sides only from the measurement of the angles of the triangle.

Solution:

It is seen that in ΔABC, ∠ACB is an obtuse angle.

∴ ABC is an acute angle (obtuse angle/acute angle)

We know that the opposite side of greater angle is than the opposite side of smaller angle Big (less/more), i.e, AB > AC [ Put >/<]

Again, measuring by protractor we see, side AD is perpendicular to FC, i.e, AD⊥FC

∴ ∠ADB = 90°

∴ In ΔADC,∠ACD acute angle (is an obtuse/acute angle), i.e, AC>AD [Put >/< ]

∴ It is seen the relation between sides, AC, AD and AB is AD < AC < AB.

Geometrical Proofs Exercise

Question 1. Two persons, one of them is coming along south direction to reach to east-west road and the other person is coming along the south-east direction, starting from the same place. Let’s calculate which person reaches first to the road.
Solution:

Given

Two persons, one of them is coming along south direction to reach to east-west road and the other person is coming along the south-east direction, starting from the same place.

Let BC is an east-west straight line.

Both persons started journey together from point A to C. First person traversed in the southern direction distance AB, while the second traversed BC distance in the eastern direction.

Second person traversed AC distance in south-east.

Distance covered by the first person = AB + BC

Distance covered by the second person = AC

∵ AB + BC > AC (∵ The sum of measurements of two sides of a triangle is greater than that of its third side.)

∴ The second person reached first.

Question 2. In a quadrilateral ABCD, AB=AD and BC = DC; and DP is the smallest distance drawn from D on AC. Let’s prove that B5 P and D are collinear.

Solution:

Given : In quadrilateral ABCD, AB = AD and BC = DC, the shortest distance from point D to side AC is DP.

Let’s prove that three points B, P, D are collinear.

Construction : DB is extended B to point forwards.

Proof: ∵ DP is at the shortest distance from AC.

∴ DP ⊥ AC

∴ ∠APD = 90° and ∠DPC = 90°

∴ APB = 90° (∵ the Vertically Opposite Angle of ∠DPC is ∠APB)

∴ APD + APB = 90° = 180°

∴ BPD = 180°

“WBBSE Class 8 Chapter 22 Maths, Construction of Parallel Lines step-by-step solutions”

Question 3. AD is a median of the triangle ABC. The smallest distance of AD from B and C are BP and CQ respectively. Let’s prove that BP = CQ.

Solution:

Given: ABC is the median of AD. The shortest distance of AD From points B and C is BP and CQ.

∴ BP ⊥ AD and CQ ⊥ AD

∴ ∠BPD = ∠CQD = 90°

In equiangular ΔBPD and equiangular ΔCQD, angle BD = angle DC (∵ AD is the median ∴  BD = DC)

And PD is a common side.

∴  ΔBPD = ΔCQD (R-H-S congruency)

∴  BP = CQ Proved

Graphs

Let’s find out the points in the graph paper.

We can see many points in the graph paper. Let’s try to write the co-ordinates of these points.

We see, A point is 4 units away from the origin on the x-axis.

Question 1. What are the coordinates of point A ?
Solution:

The coordinates of point A

The distance of point A from the y-axis is 4 units and from the x-axis is 0 unit.

∴ The coordinates of point A are (4, 0).

Understand, the coordinates of the point B are (9,0); the coordinates of point C are (14,0)

The coordinates of the point (D) are (18, 0).

“WBBSE Class 8 Maths Chapter 18 solutions, Graphs”

So we got, the y- coordinate of any point on x-axis is 0 (zero).

We see that the point E is situated at 3 units away from the origin and it is on the y-axis.

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. What are the coordinates of point E ?
Solution:

The coordinates of point E

Point E lies on y-axis and 3 units away from x-axis but it is at 0 unit away from y-axis.

∴ The coordinates of point E are (0, 3).

Now we understand, the coordinates, of point F are (0.7).

The coordinates of point G are (0,10).

The coordinates of point H are (0, 14).

So we got, the x-coordinate of any point on the y-axis is 0

Let’s write the coordinates of the points I and J from the graph paper.

We see, the point I is 5 units away from the y-axis and 3 units away from the x-axis.

∴ The coordinates of point I are (5, 3).

The point J is 10 units away from the y-axis and 3 units away from the x-axis.

∴ The coordinates of J are (10,3).

The coordinates of K are (13,3)

The coordinates of L are (17,3)

Similarly let’s write the coordinates of the points M, N, P, Q, R, S and T.

“Class 8 WBBSE Maths Chapter 18 solutions, Graphs study material”

The coordinates of M are (3,4)

The coordinates of N are (3,6)

The coordinates of P are (3,9)

The coordinates of Q are (3,13)

The coordinates of R are (8,9)

The coordinates of S are (10,11)

The coordinates of T are (14,12)

Now let’s join the points of the graph paper with a pencil and see which three or more than three points are collinear.

We see that the points A, B, C, and D lie on the X axis and they are collinear (collinear/noncollinear).

The Points E,F,G and H are on the y-axis and they are collinear.

The points I, J, K, and L are collinear and M, N,P and Q are collinear. However, the points R, S and T are noncollinear.

Graphs Exercise 18.1

Question 1. Let’s plot the points A (4, 0), B (0, 6), C (2, 5), D (7, 1), E(6, 5) and F (10, 5) on a graph sheet.

Given

A (4, 0), B (0, 6), C (2, 5), D (7, 1), E(6, 5) and F (10, 5)

Coordinates of point A are (4, 0)

Coordinates of point B are (0, 6)

Coordinates of point C are (2, 5)

Coordinates of point D are (7, 1)

Coordinates of point E are (6, 5)

Coordinates of point F are (10, 5)

Question 2. Let’s plot the points (1, 1), (3, 7), (9, 1) and (12, 1) on a graph sheet and verify whether they lie on a line.
Solution:

Given

(1, 1), (3, 7), (9, 1) and (12, 1)

Question 3. On a graph sheet let’s write 4 collinear points and write the coordinates of the points. Similarly, plotting the points on the graph sheet and joining them we get some pictures. What will we call that?
Solution:

I took on the graph paper 4 collinear points A, B, C and D.

Plotting the points on the graph sheet from the coordinates of the points and joining the points we get a picture which is called a (Graph).

If joining the points we get a straight line segment, then this graph is called a linear graph.

“WBBSE Class 8 Maths Chapter 18, Graphs solved examples”

1. Let’s plot the points A(2,0), B(6, 0) and C(4, .3) on the graph paper and join the points A, B ; B, C and C, A. Let’s find out what we have got.

Joining the points we have got a [triangle (triangle/quadrilateral).

2. Joining the points P(2,2) Q(4,4), R(6,6) we got a (collinear/non collinear) graph.

3. Let’s join the points P(1,2), A (2,3), T(3,4), and H(4,5) and see whether we have got a linear graph.
Solution:

Given

P(1,2), A (2,3), T(3,4), and H(4,5)

Path is a linear graph

Coordinates of P are (1, 2)

Coordinates of A are (2, 3)

Coordinates of T are (3, 4)

Coordinates of H are (4, 5)

Graphs Exercise 18.2

Question 1. I took 4 copies for Rs. 20. Let’s write the number and price of exercise books on the following table and draw a graph from the data.

Let’s find out from the graph the price of 6 exercise books and the number of exercise books for Rs. 45.
Solution:

Given

I took 4 copies for Rs. 20

1. First draw x- and y-axes.

2. Let us consider the length of one smallest square along x-axis = copy and the length of one smallest square along the y-axis = 1 copy and the length of one smallest square along the y-axis = Re. 1

3. From the data of graph plot points. (4, 20), (8, 40), (10, 50) and (12, 60)

4. Joining the points got AB straight line. So it is a linear graph.

5. On x-axis took the 6 exercise books and drew a ray parallel to the y-axis which cuts straight line at P point. Coordinates of P are (6, 30).

∴ Coordinates of point P are (6,30).

“WBBSE Class 8 Graphs solutions, Maths Chapter 18”

6. On the y-axis from the point of Rs. 45 drew a ray parallel to x- axis which cuts AB at Q. Coordinates of Q are (9, 45)

∴ Got from the graph that 9 exercise books are got in 45 rupees.

Graphs Exercise 18.3

Question 1. Let’s draw a graph of time and distance on the graph paper based on the information given below and find the distance covered in 4 hours and find how long it will take to cover 150 km.

Solution:

1. First of all, draw on the graph paper x- and y-axes.

2. Let distance is measured on x-axis and time along y-axis. Let the length of one smallest square along x-axis = 5 km and the length of one smallest square along y-axis = 1 hour

3. Plot the points on the graph paper (50, 2), (75, 3) and (125, 5) from the data given above.

4. Joining the above points got AB straight line. So it is a linear graph.

5. On the y-axis from the point of 4 hours drew a ray parallel to the x-axis which cuts AB at P.

Coordinates of point P are (100, 4)

∴ Got from the graph that in 4 hours, 100 km distance is covered.

6. On x-axis from the point of 150 km drew a ray parallel to y-axis which cuts AB at point Q. Coordinates of point Q are (150, 6).

∴ To cover 150 distance 6 hours are required.

Graphs Exercise 18.3

Question 1. For the graph paper beside we consider: 1 unit = length of 2 sides of the smallest squares along both the axes. Let’s write the coordinates of the points.
Solution:

Given

For the graph paper beside we consider: 1 unit = length of 2 sides of the smallest squares along both the axes.

Coordinates of A are (2, 0)

Coordinates of B are (5, 0)

Coordinates of C are (0, 4)

Coordinates of D are (0, 6)

Coordinates of E are (0, 8)

Coordinates of F are (4, 2)

Coordinates of G are (7, 2)

Coordinates of-H are (10, 2)

Coordinates of I are (2, 3)

Coordinates of J are (3, 4)

Coordinates of K are (4, 5)

2. Let’s find which three points are collinear.
Solution:

CDE, UK, FGH, CIF, and JFB are collinear.

3. Let’s find which three points are non-collinear.
Solution:

A, B and C are non-collinear.

Question 2. Let’s plot the points (1,0), (0, 5), (2, 1), (3,3), (1,3), (2,5) and (0,0) on the graph paper.
Solution:

Given

(1,0), (0, 5), (2, 1), (3,3), (1,3), (2,5) and (0,0)

O = (0, 0)

A = (1,0)

B = (0; 5)

C = (2, 1)

D = (3, 3)

E = (1, 3)

F = (2, 5)

Question 3.1. Let’s put the points (1,1), (2,2) and (6,6) on the graph paper and see whether they are collinear.

2. Let’s put three non-collinear points on the graph paper.Let’s put three more

3. collinear points other than the above three collinear points on the graph paper and write their coordinates.

A=(1,1)

B=(2,2)

C=(6,6)

A,B and C are collinear.

2. Lets plot any three non-collinear are points on the graph paper.
Solution:

D = (13, 5)

E = (15, 10)

F = (20, 10)

D, E and F are three non-collinear points.

3. After the three points above, we plot three collinear points and write their coordinates.
Solution:

P = (5, 10)

Q = (10, 15)

R = (15, 20)

P,Q and R are three collinear points.

Let 1 guava = length of 2 sides of the smallest square along x- axis and 1 rupee = length of a side of one smallest square along y-axis.

1. Let’s write the relation between the number of guava and the price of guava from graph.
Solution: There is direct relation between the number of guavas and price.

2. Let’s write the price of 4 guavas.
Solution:

4 guavas cost Rs. 12.

3. Let’s write the number of guavas in Rs. 30 from the graph.
Solution:

10 guavas are bought in Rs. 30.

4. Let’s write the number of guavas in Rs. 9.
Solution:

3 guavas are bought in Rs.9.

Question 5. Let’s see the graph beside and find the answers to the questions given below

Let 1 hdur = length of the 2 sides of the smallest squares along x- axis and 5 km = length of a side of the smallest square along y-axis.

1. Let’s write the relation between time and distance.
Solution:

Distance is in direct relation with time.

2. Let’s write the distance covered in 3. hours.
Solution:

90 km distance is covered in 3 hours.

3. Let’s write the time required to cover 120 km.
Solution:

To cover 120 km distance 4 hours are required.

4. Let’s find the velocity in km per hour.
Solution:

Velocity’is 30 kmph.

“Class 8 WBBSE Maths Chapter 18, Graphs easy explanation”

5. Let’s find and write the distance covered in 2 hours 30 minutes from the graph.
Solution:

In 2 hours 30 minutes, 75 km distance is covered.

6. Let’s find and write the time required to cover 45 km.
Solution:

To cover 45 km distance 1 hour 30 minutes are-required.

Question 6. Let’s draw the graph on the graph paper of the following data and see whether it is a linear graph.

Solution:

1. Let on the x-axis the length of one smallest square = 1. Pencil. Let along the y-axis the length of one smallest square = Rs. 1
2. Plot the points (3, 6), (5, 10), (7, 14) and (8, 16) on graph from the above data.
3. Joining the points got AB straight line. So it is a linear graph.

“WBBSE Class 8 Maths Chapter 18 solutions, Graphs PDF”

Question 7. Let’s draw the graph on the graph paper of the foilwing data and see whether it is a linear graph.

Solution:

1. Let the length of the smallest square on the axis = 1 hour; on the y-axis the length of one smallest square = 5 km.
2. Let on the plot the points (2, 40), (4, 80), (6, 120) and (8, 160)on the graph from the above data.
3. Joining the points got AB straight line. So it is a linear graph.

Question 8. Let’s draw the graph of the following data on the graph paper and see whether it is a linear graph.

Solution:

1. Let on the the x-axis the length of 5 sides of the smallest square =1 bag. Let on the y-axis the length of one smallest square = Rs. 5
2. Plot the points (1, 50), (2,100), (3, 150) and (4, 200) on the graph paper from the above data.
3. Joining the points got AB straight line. So the graph is linear.

“WBBSE Class 8 Maths Chapter 18, Graphs important questions”

Question 9. Let’s, draw the graph of the following data on the graph paper and see whether it is a linear graph.

Solution:

1. Let on the x-axis the length of 5 sides of the smallest square = 1 book and on the y-axis the length of one side of the smallest square = Rs. 5.
2. Plot the points (2, 50), (3, 75), (5, 125) and (8, 200) on the
   graph paper from the above data.
3. Joining the points got AB straight line. So the graph is linear.

Question 10. Let’s draw the graph on the graph paper of the following data and see whether it is a linear graph.

Solution:

1. Let on the x-axis the length of 5 sides of the smallest square = 1 over. On the y-axis the length of one side of the smallest square = 1 run.
2. Plot the points (1, 4), (3, 12), (5, 20) and (7, 24) on the graph paper from the above data.
3. Joining the points got AB graph. So the graph is not linear.

Verification Of The Relation Between The Angles And The Sides Of A Triangle

Today Tania, Kuntal, Tulika and I have decided that we will try to make different types of shapes like triangles with different colored straws and pins.

I formed by three straws-

The shape ABC is like a triangle. It has three sides AB,BC and three angles  and CA and three angles ∠BAC, ∠ABC and ∠ACB

l also formed a shape of a triangle like Tutika –

The exterior angle of ΔUVW is ∠UWX (∠UWX/∠UWV). In ΔUVW the interior opposite angles of the exterior angle ∠UWX are ∠UVW and ∠VUW

In figure1 the exterior angle is ∠ACD = 140 degree. So interior opposite angle ∠ABC = 50 degree and ∠BAC = 90 degree.

Now I will find the relation between the exterior angle and the interior opposite angle.

Read and Learn More WBBSE Solutions For Class 8 Maths

It is seen by measuring by a protractor, ∠ABC + ∠BAC= 140 degree = ∠ACD (approx).

My brother wrote by measuring three angles of each triangle by a protractor. In ΔABC the exterior angles are 90°, 50° and 40°; ∠ABC=50 degree, ∠BCA=40 degree, and ∠CAB=90 degree.

It is seen that ∠ABC + ∠BCA + ∠CAB =180 degree.

“WBBSE Class 8 Maths Chapter 16 solutions, Verification of the Relation Between the Angles and the Sides of a Triangle”

It is seen by measuring with a protractor, the sum of measurement of three angles of each triangle in figure 2 and 3 is 180 degree

4 Let’s cut off two angles ∠A and ∠B from another triangle and paste those two angles on the exterior angles of the first triangle as shown in the figure below :

It is seen that the exterior angle ∠ACD = ∠A + ∠B = ∠BAC + ∠ABC

Theorem : We try to identify which postulates are required in the construction of the theorem. We also write which postulates are required for this theorem.

Tania drew triangle PUT and produced the side UT to R for this one exterior angle (∠PTR; and two interior opposite angles ∠PUT and  produced. Let’s prove logically step-by-step that ∠PTR= ∠PUT + ∠UPT.

Given : PUT is a triangle whose side UT is extended to R. As a result an exterior angle ∠PTR and two opposite interior angles ∠PUT and ∠UPT are formed.

We have to prove that ∠PTR = ∠PUT + ∠UPT.

Construction : In ΔPUT from T parallel to UP a straight line TM is drawn.

Proof: UP//TM and UR is a transversal.

∴ ∠MTR = Corresponding ∠PUT

Again : UP//TM and PT is a transversal.

∴ ∠PTM = Alternate ∠UPT

∴ ∠MTR + ∠PTM = ∠PUT + ∠UPT

∴ ∠PTR = ∠PUT + ∠UPT Proved

The Sides Of A Triangle Exercise

Question 1. Let’s calculate and write the measurement of each of the exterior angles x° in each of the following triangles –

Solution:

The measurement of each of the exterior angles x° in each of the following triangles are

∠X = 35°+45°= 80°

∠X = 50°+35°= 85°

∠X = 50°+40°= 90°

“Class 8 WBBSE Maths Chapter 16 solutions, Relation Between Angles and Sides of a Triangle study material”

Question 2. Let’s write the relation between the exterior angle ∠PRS and the interior opposite angles –

Solution:

∠PRS = ∠PQR + ∠QPR

Pallabi and Kuntal drew some triangles of different shapes.

In is seen by measuring by a protractor that in ΔABC ∠BAC = 85 degree ∠ABC =60 degree and ∠ACB =35 degree. Again ∠BAC + ∠ABC + ∠ACB =180 degrees.

Kuntal measured with a protractor that in ΔACT ∠CAT = 145 degree ∠ACT = 20 degree and ∠CTA =15 degree. Again ∠ACT + ∠QAT + ∠CTA = 180 degree.

Pallabi measured and saw in ΔAMN that the sum of measurement of the three angles is =180 degrees.

The Sides Of A Triangle Exercise

I will find the value of missing angles from the triangles below.

x = 180° – (40° + 55°)

= 180° – 95° = 85°

x = 85°

x = 180° – (30°+30°)

= 180° – 60° = 120°

x = 120°

x = 180° – (30°+90°)

= 180° – 120° = 60°

x = 60°

The Sides Of A Triangle Exercise 16.3

Question 1. Find the value of ∠x in each of the following regions :

Solution:

x = 60° + 40° + 20° =120

x =120

∠PQR = 180° – (50° + 60°) = 180° – 110° = 70

∴ ∠SQT = 180° -70° = 110°

∴ x = ∠SQT+ ∠QST =110° + 30°= 140°

x = 180°-( ∠S+ ∠T) = 180° – (60°+55°) = 180°-115°= 65°

The value of ∠x = 65°

Question 2. From the figure beside find the value of the angles of ΔEHG.

Solution: 

∠AFH = Corresponding ∠CHE = 110° (∵ AB//CD and EF is a transveral).

ΔEHG∠EHG = 180°- 110° = 70°

HGE = 60° (Given) and HEG = 180° – (70°+60°)

= 180° – 130° = 50°

The value of the angles of ΔEHG = 50°

Question 3. From the figure beside find the value of ∠A+ ∠B+ ∠C+ ∠D+ ∠E+ ∠F.

Solution:

In ΔAOB ∠A+∠B+∠AOB=180° (1)

In ΔCOD ∠C+ ∠D+ ∠COD = 180° (2)

In ΔEOF ∠E + ∠F + ∠EOF-180° (3)

Adding (1), (2) and (3)

∠A + ∠B + ∠C + ∠D +∠E + ∠F + ∠AOB + ∠COD + ∠EOF = 540°

∵ ∠AOB = ∠DOE (∵ Vertically opposite angles)

∠COD = ∠AOF (∵ Vertically opposite angles)

and ∠EOF = ∠BOC (∵Vertically opposite angles)

∵ ∠AOB + ∠DOE + ∠COD + ∠AOF + ∠EOF + ∠BOC = 360°

∵ ∠AOB + ∠COD + ∠EOF = \(\frac{1}{2}\) × 360°= 180°

∵ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 540° – 180° = 360°

The value of ∠A+ ∠B+ ∠C+ ∠D+ ∠E+ ∠F = 360°

Question 4. If AB = AC then find the value of ∠ABC,∠ACB +∠ACD = 180°

Solution:

∠ACB +∠ACD = 180°

∴ ∠ACB + 112° = 180°

∴ ∠ACB =180° – 112° = 68°

∵ AB = AC

∴ ∠ACB = ∠ABC = 68°

∠BAC = 180° – (68° + 68°) = 180° – 136° = 44°

∴ ∠ABC = 68°,∠ACB = 68°, ∠BAC = 44°

The value of ∠ABC,∠ACB +∠ACD = 180°

Question 5. If AB = AC then find the value of ∠ABC and ∠ACB.

Solution:

In ΔABC ∠BAC = 80°

∠ABC + ∠ACB = 180° – 80° = 100°

∵ AB = AC

∴ ∠ABC = ∠ACB =\(\frac{100^{\circ}}{2}\) 50°

∴ ∠ABC = 50°, ∠ACB = 50°

Question 6. If AB = AC then find the value of ∠ACB and ∠BAC.

Solution:

In ΔABC ∠ABC = 70c

∵ AB = AC

∠ABC = ∠ACB = 70°

∠BAC = 180° – (70° + 70°) = 180° – 140° = 40°

∴ ∠ACB = 70° and ∠BAC = 40°

“WBBSE Class 8 Maths Chapter 16, Verification of Angles and Sides of a Triangle solved examples”

Question 7. If AB = BC and ∠BAC + ∠ACB =50°; then find the value of the angles of ΔABC.

Solution:

∵ ∠BAC + ∠ACB = 50°

∵ AB = BC

∠BAC = ∠ACB = \(\frac{50^{\circ}}{2}\) = 25°

∴ ∠ABC = 180°-50°= 130°

If ΔABC ∠ABC = 130°, ∠BAC = 25°, ∠ACB = 25°

Question 9. If we produce the side BC on both sides, the two exterior angles are fomed. Let’s prove that the sum of the measurement of these two exterior angles is more than 2 right angles.

Solution:

Given: In ΔABC side BC is extended on both sides.

We have to prove that the sum of the exterior angles is more than 2 right angles, i.e., ∠ABD + ∠ACE > two right angles.

Proof : According to the relation between the exterior angle and the interior opposite angles, ∠ABD = ∠BAC + ∠ACB and ∠ACE = ∠BAC + ∠ABC

∴ ∠ABD + ∠ACE = ∠BAC + ∠ACB + ∠BAC + ∠ABC

∠ABD + ∠ACE = ∠BAC + ∠ACB + ∠ABC + ∠BAC

∠ABD + ∠ACE = 2 right angles + ∠BAC

(∵ The sum of all three angles of a triangle is equal to two right angles.)

∴ ∠ABD + ∠ACE > two right angles.

Question 9. Two straight lines parallel to sides BC and BA respectively through two vertices A and C of A ABC meet at D. Let’s prove that ∠ABC = ∠ADC.

Solution:

Given: Two straight lines parallel to sides BC and BA respectively through two vertices A and C of A ABC are drawn, which meet at D.

We have to prove that ∠ABC = ∠ADC.

Proof: In ΔABC and ΔADC

∠BAC = Alternate ∠ACD (∴ BA//CD and AC is a transversal)

∠ACB = Alternate. ZCAD (∴ BC//AD and AC is a transversal)

and AC is a common side.

∴ ΔABC = ΔADC(A-A-S)

∴ ∠ABC = ∠ADC Proved.

Question 10 .In ΔABC the internal bisectors of ∠ACB and ∠ACB angles meet at O. Let’s prove that ∠BOC = 90°-\(\frac{1}{2}\) ∠BAC.

Solution:

Given: In A ABC the internal bisectors of angles ∠ABC and ∠ACB meet at O.

Let’s prove that ∠BOC = ∠90° +\(-\frac{1}{2}\) ∠BAC.

Proof: In A ABC

∠A + ∠B + ∠C = 180°

∴ \(\frac{1}{2}\) ∠A + \(\frac{1}{2}\) ∠B + \(\frac{1}{2}\) ∠C × 180° = 90°

∴ \(\frac{1}{2}\) ∠B + \(\frac{1}{2}\) ∠C = 90° – \(\frac{1}{2}\) ∠A

Now, in ΔBOC

= \(\frac{1}{2}\) ∠B+ \(\frac{1}{2}\) ∠C + ∠BOC = 180°

∴  ∠BOC = 180° – (\(\frac{1}{2}\) ∠B+ \(\frac{1}{2}\) ∠C)

∴ ∠BOC=180°-(90°-\(\frac{1}{2}\) ∠A)

∴ ∠BOC = 180° – 90° + \(\frac{1}{2}\) ∠A = 90° + \(\frac{1}{2}\) ∠A Proved

Question 11. In A ABC the external bisectors of ∠ABC and ∠ABC meet at O. Let’s prove that ∠BOC = 90° \(-\frac{1}{2}\) ∠BAC.

Solution:

Given

In ΔABC the external bisectors of ∠ABC and ∠ACB meet at O.

We have to prove that ∠BOC = 90° – \(\frac{1}{2}\)∠BAC.

Proof: ∵ ∠ABC + ∠OBC + ∠OBP = 180°

or, ∠ABC + ∠OBC + ∠OBC = 180° (∵ ∠OBC = ∠OBP)

or, ∠ABC + 2 ∠OBC = 180°

or, 2 ∠OBC = 180°- ∠ABC

or, ∠OBC = 90° –\(\frac{1}{2}\) ∠ABC

Similarly, ∠OCB = 90°-\(\frac{1}{2}\) ∠ACB

Now, in ΔBOC,

∠OBC + ∠OCB + ∠BOC = 180°

or, 90° – \(\frac{1}{2}\) ∠ABC + 90° – \(\frac{1}{2}\) ∠ACB + ∠BOC = 180°

or, 180° – \(\frac{1}{2}\) (∠ABC + ∠ACB) + ∠BOC = 180°

or, ∠BOC = \(\frac{1}{2}\) (∠ABC + ∠ACB)

or, ∠BOC = \(\frac{1}{2}\)(180°- ∠BAC) (∵ ∠A+ ∠B + ∠C = 180°)

∴ ∠B + ∠C = 180° -∠A

or, ∠BOC = 90°\(\frac{1}{2}\) ∠BAC Proved.

Question 12. One base angle of the isosceles triangle ABC is twice of the vertical angle. Let’s write the measurement of the three angles of the triangle.

Solution:

Given

One base angle of the isosceles triangle ABC is twice of the vertical angle.

Let ABC is a triangle where AB = AC and ∠B = 2 ∠A

∵ In ΔABC, AB = AC

∴ ∠B= ∠C

Let ∠A = X

∴ ∠B= ∠C = 2X

Proof: In ΔABC,

∠A + ∠B + ∠C = 180°

or, X + 2X + 2X = 180°

or, 5X = 180°

or X = \(\frac{180^{\circ}}{5}\)

or, X = 36°

∴ ∠A = X = 36°

∴ ∠B = ∠C = 2 x 36° = 72°

∴ The angles of the triangle are 36°, 72°, 72°

Question 13. In ΔABC, ∠BAC = 90° and ∠BCA = 30°. Let’s prove that AB: \(\frac{1}{2}\) BC.

Solution:

Given

In ΔABC, ∠BAC = 90° and ∠BCA = 30

Let’s prove that AB = \(\frac{1}{2}\) BC.

Proof: In A ABC

∠ABC = 180° – (90° + 30°) = 180° – 120° = 60°

∵ ∠C= 30° and ∠B= 60° ,

∴ ∠B = 2∠C

∵ The opposite side is proportional to the angles of a triangle. The opposite side of ZC is AB and the opposite side of ∠B is AC.

∵ The opposite side of the largest angle is the largest and that of the smallest angle is the smallest. Here the value of ∠B is double than that of ∠C.

∴ AB = \(\frac{1}{2}\) AC Proved.

Question 14. In ΔXYZ, ∠XYZ = 90° and XY=XZ. Let’s prove that ∠YXZ = 60°

Solution:

Given : ∠XYZ = 90° and XY = XZ

Here ∠Y is the largest.

∴ XZ is the largest side

∴ XY≠XZ The question is fallacious.

“WBBSE Class 8 Verification of Relation Between Angles and Sides of a Triangle solutions, Maths Chapter 16”

Question 15. Let’s prove that the measurement of-each angle of an equilateral triangle is 60°.

Solution:

Let ABC is a triangle where AB = BC = AC.

Let’s prove that ∠A = ∠B = ∠C = 60°.

Proof : The sides of a triangle are proportional to the angles.

∵ AB = BC = AC

∴ ∠A = ∠B = ∠C

∵ The sum of the three angles of a triangle is 180°.

∴ ∠A + ∠B + ∠C = 180°

or, ∠A + ∠A + ∠A = 180°

or, 3∠A = 180°

or, ∠A =\(\frac{180^{\circ}}{3}\) = 60°

∴ ∠A = ∠B = ∠C = 60° Proved.

Question 16. The bisector of ∠BAC of ΔABC and the straight line through the mid point D of the side AC and parallel to the side AB meet at a point E, outside BC. Let’s prove that ∠AEC = 1 right angle.

Solution:

Given

The bisector of ∠BAC of ΔABC and the straight line through the mid point D of the side AC and parallel to the side AB meet at a point E, outside BC.

Diptarka and Puja have made different triangles of different shapes and colours. Any two sides of these triangles are unequal. They formed-

I measure the length of each side of each triangle and let’s compare which side is smaller and which one is greater.

Measuring we see, in ΔABC, the length of AC > the length of AB.

In ΔPQR, the length of PR > the length of QR.

In ΔXYZ, the length of XZ > the length of XY

Now, let’s measure each angle triangle with a protractor and compare.

Measuring we see, in ΔABC ∠ACB [>] ∠ACB [ Put</> ]

In ΔPQR, ∠PQR [>] ∠QPR [ Put</> ]

In ΔXYZ, ∠XYZ > ∠YZX ∠YXZ /∠YZX Put ]

But it is seen that in ΔABC, AC is opposite to the side ∠ABC, and AB is opposite to the side ∠ACB.

Again, in ΔPQR, the opposite angle to the side PR is ∠PQR and the opposite angle to the side QR is ∠QPR

And, in ΔXYZ, the opposite angle to the side XZ is ∠XYZ and the opposite angle to the side XY is ∠XYZ

“Class 8 WBBSE Maths Chapter 16, Verification of Angles and Sides easy explanation”

Measuring with a protractor I got the measurement of the angle opposite to the greater side of each triangle than the angle opposite to the smaller side is greater [greater/smaller]. Pallabi and Sirap drew some triangles whose two sides are unequal.

Measuring the sides of the triangle with a scale and the angles of the triangle with a protractor we see that the measurement of the angle opposite to the greater side is greater (smaller/greater) than the measurement of the angle opposite to the smaller side .

The Sides Of A Triangle Exercise

Question 1. Let’s measure the length of the sides of the following triangles and let’s compare the angles :

Solution:

1. ∠P>∠R(∠R/∠Q)
2. ∠X>∠Z(∠Y/∠Z)
3. ∠C>∠B(∠A/∠B)

Pallabi and Siraj drew some triangles, each of whose two angles are unequal.

Now I measure the angles with a protractor and let’s compare the angles of each triangles.

Measuring I see, in ΔMAN, ∠AMN > ∠MAN [</>Put]

In ΔPAN, ∠PAN > ∠PNA [∠PNA / ∠APN Put ]

In ΔFAN, ZFNA < FAN [ ∠FAN / ZAFN Put ]

Let’s measure each side of each triangle with a scale and let’s compare the length of the sides.

Measuring I see, in ΔMAN, MN < AN [ >/< Put ]

In ΔPAN, PN > PA [ PA / AN Put ]

In ΔFAN, AN < FN [ FA / FN Put ]

But it is seen that in ΔMAN the opposite side of ∠AMN is AN

and the opposite side of ∠MAN is MN

In ΔPAN the opposite side of ∠PAN is PN

and the opposite side of ∠PNA is AP

In ΔFAN the opposite side of ∠FNA is FA

and the opposite side of ∠FAN is FN

I got by measuring with a scale and a protractor that in each triangle the length of the side opposite is to greater side than the length of the side opposite |greater| (greater/smaller).

The Sides Of A Triangle Exercise

Question 1. Let’s observe the angles of the following triangles and let’s compare the length of the sides which one is smaller and which one is greater:

Solution:

Side BC > side AB [Put >/<]

Side YZ > side XY [Put XZ/XY]

Side PR > side PQ [Put up side]

The Sides Of A Triangle Exercise 16.6

Question 1. In figure ∠QPR > ∠PQR. Let’s write the relation between PR and QR.

Solution:

Given ∠QPP > ∠PQR

The opposite side of ∠QPR is QR and the opposite side of ∠PQR is PR.

∵  The side of the largest angle of a triangle is greater than that of the smallest angle.

∴ QR>PR

Question 2. In ΔABC, AC>AB. D is any point on the side AC such that ∠AOB = ∠ABD. Prove that ∠ABC > ∠ACB.

Solution:

Given

In ΔABC, AC > AB; D is any point on the side AC such that ∠ADB = ∠ABD.

Prove that ∠ABC > ∠ACB.

Proof : ∠ADB = ∠DBC + ∠BCD (∵ The exterior angle of A BCD is ZADB)

or, ∠ABD = ∠DBC + ∠BCD (∵ ∠ADB = ∠ABD)

or, ∠ABD = ∠DBC + ∠ACB (∵ ∠BCD and ∠ACB are the same angle)

Again, ∠ABC = ∠ABD + ∠ACB

Question 3. In triangle ABC, AD is perpendicular to BC, and AD, BC, AC > AB. Let’s prove that :

Solution:

1. ZCAD > ZBAD
2. DC > BD.

Given

In ΔABC, AD is perpendicular to BC and AC > AB.

Prove that: 1. ∠CAD > ∠BAD; 2.  DC > BD.

Proof: In Δ ∠ABC ∠BAC + ∠ABC + ∠ACB = 180°

In ΔADC ∠DAC + ∠ADC +∠ACD = 180°

or, ∠DAC + 90° + ∠ACD = 180° [∵ AD⊥C]

or, ∠DAC + ∠ACD = 180° – 90° = 90°

In ΔABD ∠BAD + ∠ABD + ∠ADB = 180°

or, ∠BAD + ∠ABD + 90° = 180° [∵ AD⊥C]

or, ∠BAD + ∠ABD = 180° – 90° = 90°

∴  ∠BAD +∠ABD = ∠DAC + ∠ACD = 90°

∵ AC > AB

∴  ∠B > ∠C

∴ ∠CAD > ∠BAD

(1) (Proved)

∴ In ΔABD ∠D= 90° and ∠A + ∠B = 90°, ∠B> ∠A

∴ AD > BD

Similarly in ΔADC, DC > AD

∴ DC > AD > BD

∴ BD > BD (ii) (Proved)

“WBBSE Class 8 Maths Chapter 16 solutions, Verification of Relation Between Angles and Sides PDF”

Question 4. In figure AB < OB and CD > OD. Let’s prove that ∠BAO > ∠OCD.

Solution:

Given

AB < OB and CD > OD

Prove that ∠BAO > ∠OCD

Proof : ∵ CD > OD

∴  ∠COD >∠OCD

and ∠COD = ∠BOA (Vertically opposite angles)

Again, AB < ∠OB

∠BOA < ∠BAO

or, ∠COD < ∠BAO (∵ ∠BOA = ∠COD)

In triangle OCD, ∠COD > ∠OCD (∵ CD > OD)

and ∠BAO > ∠COD –

∴ ∠BAO > ∠OCD (Proved)

Simplification Of Algebraic Expressions

Sumita has a 20 meter long red ribbon. Santanu and I have made some cards. We have decided that we will cover the four sides of the cards by the red ribbon.

If Sumita has a 4x²meter long red ribbon and Santanu and I took ax meter and 2xb meter respectively from Sumita’s ribbon, then let’s calculate how much portion we took from Sumita’s ribbon.

1 took \(\frac{2 x b}{4 x^2}\) Part = \(\frac{b}{2 x}\)2 Par

I have seen that \(\frac{2 x b}{4 x^2}\) and \(\frac{b}{2 x}\) are same. What are these called?

Santanu took \(\frac{a x}{4 x^2}\) part = \(\frac{a}{4 x}\) part.

“WBBSE Class 8 Maths Chapter 15 solutions, Simplification of Algebraic Expressions”

We took, \(\frac{b}{2 x}\) part + \(\frac{a}{4 x}\) part.

= \(\left(\frac{b}{2 x}+\frac{a}{4 x}\right)\) part = \(\left(\frac{2 b+a}{4 x}\right)\) part of the total ribbon. Let’s simplify the algebraic expression.

Question 1. \(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\)

Solution:

Given

\(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\)

First we see the L.C.M. of (b-c) (c-a), (c-a) (a-b) and (a-b) (b-c) (b-c) (c-a), (c-a) (a-b) and (a-b) (b-c) = (a-b) (b-c) (c-a)

= \(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\)

= \(\frac{a-b}{(a-b)(b-c)(c-a)}+\frac{b-c}{(a-b)(b-c)(c-a)}+\frac{c-a}{(a-b)(b-c)(c-a)}\)

(a-b)(b-c)(c-a)÷(b-c)(c-a)=(a-b)

(a-b)(b-c)(c-a)÷(c-a)(a-b)=(b-c)

(a-b)(b-c)(c-a)÷(a-b)(b-c)=(c-a)

= \(\frac{a-b+b-c+b-a}{(a-b)(b-c)(c-a)}=\frac{0}{(a-b)(b-c)(c-a)}=0\)

\(\frac{1}{(b-c)(c-a)}+\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}\) = 0

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. Let the length of the ribbon is 8a²x²meter. I took 2ax meter ribbon.

Solution:

Given

The length of the ribbon is 8a²x²meter. I took 2ax meter ribbon

I took \(\frac{2 a x}{8 a^2 x^2}\) part = \(\frac{1}{4 a x}\) part

∴ The reduced from of \(\frac{2 a x}{8 a^2 x^2}\) is \(\frac{1}{4 a x}\)

Question 3. Let’s express \(\frac{(a+1)}{a+2} \times \frac{a^2-a-2}{a^2+a}\) in reduced form-

Solution:

Given

\(\frac{(a+1)}{a+2} \times \frac{a^2-a-2}{a^2+a}\)

= \(\frac{a+1}{a+2} \times \frac{a^2-a-2}{a^2+\bar{a}}=\frac{a+1}{a+2} x\)

= \(=\frac{(a-2) \times(a+1)}{a(a+1)}\)

= \(\frac{(a+1)(a-2)}{a(a+2)}\)

\(\frac{(a+1)}{a+2} \times \frac{a^2-a-2}{a^2+a}\) = \(\frac{(a+1)(a-2)}{a(a+2)}\)

Algebraic Expressions Exercise

Question 1. Let’s express \(\frac{a(a+b)}{(a-b)} \times \frac{(a-b)}{b(a+b)} \times \frac{a}{b}\) in reduced form.

Solution:

Given

\(\frac{a(a+b)}{(a-b)} \times \frac{(a-b)}{b(a+b)} \times \frac{a}{b}\)

= \(\frac{a(a+b)}{a-b} \times \frac{a-b}{b(a+b)} \times \frac{a}{b}\)

= \(\frac{a^2(a+b)(a-b)}{b^2(a-b)(a+b)}=\frac{a^2}{b^2}\)

\(\frac{a(a+b)}{(a-b)} \times \frac{(a-b)}{b(a+b)} \times \frac{a}{b}\) = \(\frac{a^2(a+b)(a-b)}{b^2(a-b)(a+b)}=\frac{a^2}{b^2}\)

“Class 8 WBBSE Maths Chapter 15 solutions, Simplification of Algebraic Expressions study material”

Question 2. Let’s express \(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}\) in reduced form.

Solution:

Given

\(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}\)

= \(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}=\frac{x^2+x-2}{x^2-2 x-8} \times \frac{x^2-3 x-4}{x^2-x-6}\)

= \(\frac{x^2+2 x-x-2}{x^2-4 x+2 x-8} \times \frac{x^2-4 x+x-4}{x^2-3 x+2 x-6}\)

= \( \frac{x(x+2)-1(x+2)}{x(x-4)+2(x-4)} \times \frac{x(x-4)+1(x-4)}{x(x-3)+2(x-3)}\)

= \(\frac{(x+2)(x-1)}{(x-4)(x+2)} \times \frac{(x-4)(x+1)}{(x-3)(x+2)}\)

= \(\frac{(x+2)(x-1)(x-4)(x+1)}{(x-4)(x+2)(x-3)(x+2)}\)

[In numerator and in denominator [x+2×x-4] is the common factor, so dividing numerator and denominator by (x+2) (x-4) we get the reduced form.]

= \(\frac{(x+2)(x-1)(x-4)(x+1)}{(x-4)(x+2)(x-3)(x+2)}\)

= \(\frac{(x-1)(x+1)}{(x-3)(x+2)}\)

\(\frac{x^2+x-2}{x^2-2 x-8} \div \frac{x^2-x-6}{x^2-3 x-4}\) = \(\frac{(x-1)(x+1)}{(x-3)(x+2)}\)

Question 3. Let’s express \(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\) in reduced form and let’s write the common factor of the numerator and the denominator.

Solution:

Given

\(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\)

= \(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\)

= \(\frac{p^2-q^2}{x-y} \div \frac{x^2-y^2}{p+q}\)

= \(\frac{(p+q)(p-q)}{(x-y)} \times \frac{(x-y)(x+y)}{p+q}\)

\(\frac{p^2-q^2}{x-y} \div \frac{p+q}{x^2-y^2}\) = (p-q)(x+y)

The common factor of the numerator and the denominator is (p-q)(x+y)

Algebraic Expressions Exercise

Let’s express the algebraic expressions given below in reduced form:

1. \(\frac{a^2 \times c^2}{c^2 \times d^2} \div \frac{b c}{a d}\)

Solution:

Given

= \(\frac{a^2 \times c^2}{c^2 \times d^2}\)

= \(\frac{a^3}{b c d}\)

\(\frac{a^2 \times c^2}{c^2 \times d^2} \div \frac{b c}{a d}\) = \(\frac{a^3}{b c d}\)

2. \(\frac{x^2 y-x y^2}{x^2-x y}\)

Solution:

Given

= \(\frac{x^2 y-x y^2}{x+y}\)

= \(\frac{x y(x-y)}{x(x-y)}\)

= y

\(\frac{x^2 y-x y^2}{x^2-x y}\) = y

“WBBSE Class 8 Maths Chapter 15, Simplification of Algebraic Expressions solved examples”

3. \(\frac{p^2-q^2}{x^2-x y} \div \frac{p-q}{x^2-y^2}\)

Solution:

Given

= \(\frac{p^2-q^2}{x^2-x y} \div \frac{p-q}{x^2-y^2}\)

= \(\frac{(p+q)(p-q)}{(x+y)} \times \frac{(x+y)(x-y)}{(p-q)}\)

= (p+q)(x-y)

\(\frac{p^2-q^2}{x^2-x y} \div \frac{p-q}{x^2-y^2}\) = (p+q)(x-y)

Solution:

Algebraic Expressions Exercise

Question 1. Let’s see the relations below and find which one is true and which one is false :

1. \(\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\)

Solution:

L.H.S = \(\)

= \(\frac{a+b}{c}\)

= R.H.S

∴ True

2. \(\frac{a}{x+y}=\frac{a}{x}+\frac{a}{y}\)

Solution:

L.H.S = \(\frac{a}{x+y}\)

R.H.S = \(\frac{a}{x}+\frac{a}{y}=\frac{a y+a x}{x y}\)

L.H.S≠R.H.S

∴ False

3. \(\frac{x-y}{a-b}=\frac{y-x}{b-a}\)

Solution:

= \(\frac{x-y}{a-b}\)

= \(\frac{-(x-y)}{-(a-b)}\)

= \(\frac{x-y}{a-b}\)

∵ L.H.S=R.H.S

∴ True

“WBBSE Class 8 Simplification of Algebraic Expressions solutions, Maths Chapter 15”

4.  \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y}\)

Solution:

L.H.S = \(\frac{1}{x}+\frac{1}{y}\)

= \(\frac{y+x}{x y}\)

R.H.S = \(\frac{1}{x+y}\)

∵ L.H.S≠R.H.S

∴ False

Question 2. Let’s express the following algebraic fractions in reduced form:

Solution:

1. \(\frac{63 a^3 b^4}{77 b^5}\)

Solution:

= \(\frac{63 a^3 b^4}{77 b^5}\)

= \(\frac{63 a^3 b^4}{77 b^4}{\prime\cdot b}\)

= \(\frac{9 a^3}{11 b}\)

\(\frac{63 a^3 b^4}{77 b^5}\) = \(\frac{9 a^3}{11 b}\)

2. \(\frac{18 a^4 b^5 c^2}{21 a^7 b^2}\)

Solution:

= \(\frac{18 a^4 b^5 c^2}{77 a^7 b^2}\)

= \(\frac{18 \times a^4 \times b^2 \times b^3 \times c^2}{21 \times a^4 \times a^3 \times b^2}\)

= \(\frac{6 b^3 c^2}{7 a^3}\)

\(\frac{18 a^4 b^5 c^2}{21 a^7 b^2}\) = \(\frac{6 b^3 c^2}{7 a^3}\)

3. \(\frac{x^2-3 x+2}{x^2-1}\)

Solution:

= \(\frac{x^2-3 x+2}{x^2-1}\)

= \(\frac{x^2-2 x-x+2}{(x)^2-(1)^2}\)

= \(\frac{x(x-2)-1(x-2)}{(x+1)(x-1)}\)

= \(\frac{(x-2)(x-1)}{(x+1)(x-1)}\)

= \( \frac{x-2}{x+1}\)

\(\frac{x^2-3 x+2}{x^2-1}\) = \( \frac{x-2}{x+1}\)

4.  \(\frac{a+1}{a-2} \times \frac{a^2-a-2}{a^2+a}\)

Solution:

= \(\frac{a+1}{a-2} \times \frac{a^2-a-2}{a^2+a}\)

= \(\frac{a+1}{a-2} \times \frac{a^2-2 a+a-2}{a(a+1)}\)

= \( \frac{(a+1)}{(a-2)} \times \frac{a(a-2)+1(a-2)}{a(a+1)}\)

= \(\frac{(a+1)}{(a-2)} \times \frac{(a-2)(a+1)}{a(a+1)}\)

= \(\frac{a+1}{a}\)

\(\frac{a+1}{a-2} \times \frac{a^2-a-2}{a^2+a}\) = \(\frac{a+1}{a}\)

5.  \(\frac{p^3+q^3}{p^2-q^2} \div \frac{p+q}{p-q}\)

Solution:

= \(\frac{p^3+q^3}{p^2-q^2} \div \frac{p+q}{p-q}\)

= \(\frac{(p+q)\left(p^2-p q+q^2\right)}{(p+q)(p-q)} \times \frac{(p-q)}{(p+q)}\)

= \(\frac{p^2-p q+q^2}{p+q}\)

\(\frac{p^3+q^3}{p^2-q^2} \div \frac{p+q}{p-q}\) = \(\frac{p^2-p q+q^2}{p+q}\)

6.  \(\frac{x^2-x-6}{x^2+4 x-5} \times \frac{x^2+6 x+5}{x^2-4 x+3}\)

Solution:

= \(\frac{x^2-x-6}{x^2+4 x-5} \times \frac{x^2+6 x+5}{x^2-4 x+3}\)

= \(\frac{x^2-3 x+2 x-6}{x^2+5 x-x-5} \times \frac{x^2+5 x+x+5}{x^2-3 x-x+3}\)

= \(\frac{x(x-3)+2(x-3)}{x(x+5)-1(x+5)} \times \frac{x(x+5)+1(x+5)}{x(x-3)-1(x-3)}\)

= \(\frac{(x-3)(x+2)}{(x+5)(x-1)} \times \frac{(x+5)(x+1)}{(x-3)(x-1)}\)

= \(\frac{(x+2)(x+1)}{(x-1)(x-1)}\)

\(\frac{x^2-x-6}{x^2+4 x-5} \times \frac{x^2+6 x+5}{x^2-4 x+3}\) = \(\frac{(x+2)(x+1)}{(x-1)(x-1)}\)

“Class 8 WBBSE Maths Chapter 15, Simplification of Algebraic Expressions easy explanation”

7. \(\frac{a^2-a b+b^2}{a^2+a b} \div \frac{a^2+b^3}{a^2-b^2}\)

Solution:

= \(\frac{a^2-a b+b^2}{a^2+a b} \div \frac{a^2+b^3}{a^2-b^2}\)

= \(\frac{a^2-a b+b^2}{a(a+b)} \div \frac{(a+b)\left(a-a b+b^2\right)}{(a+b)(a-b)}\)

= \(\frac{\left(a^2-a b+b^2\right)}{a(a+b)} \times \frac{(a+b)(a-b)}{(a+b)\left(a^2-a b+b^2\right)}\)

= \(\frac{a-b}{a}(a+b)

= [latex]\frac{a-b}{a^2+a b}\)

\(\frac{a^2-a b+b^2}{a^2+a b} \div \frac{a^2+b^3}{a^2-b^2}\) = \(\frac{a-b}{a^2+a b}\)

Question 3. Let’s simplify the following algebraic expressions:

Solution:

= \(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\)

= \(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\)

= \(\frac{c+a+b}{a b c}\)

= \(\frac{a+b+c}{a b c}\)

2. \(\frac{a-b-c}{a}+\frac{a+b+c}{a}\)|

Solution:

= \(\frac{a-b-c}{a}+\frac{a+b+c}{a}\)

= \(\frac{a-\not b-\not c+a+\not b+\not c}{a}\)

= \(\frac{2 a}{a}\)

= 2

\(\frac{a-b-c}{a}+\frac{a+b+c}{a}\)| = 2

3. \(\frac{x^2+a^2}{a b}+\frac{x-a}{a x}-\frac{x^3}{b}\)

Solution:

= \(\frac{x^2+a^2}{a b}+\frac{x-a}{a x}-\frac{x^3}{b}\)

= \(\frac{x\left(x^2+a^2\right)+b(x-a)-a x^4}{a b x}\)

= \(\frac{x^3+a^2 x+b x-a b-a x^4}{a b x}\)

\(\frac{x^2+a^2}{a b}+\frac{x-a}{a x}-\frac{x^3}{b}\) = \(\frac{x^3+a^2 x+b x-a b-a x^4}{a b x}\)

4. \(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \div \frac{4 b c^3}{9 a^2}\)

Solution:

= \(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \div \frac{4 b c^3}{9 a^2}\)

= \(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \times \frac{9 a^2}{4 b c^3}\)

= \(\frac{a}{2 b^2}\)

\(\frac{2 a^2 b}{3 b^2 c} \times \frac{c^4}{3 a^3} \div \frac{4 b c^3}{9 a^2}\) = \(\frac{a}{2 b^2}\)

“WBBSE Class 8 Maths Chapter 15 solutions, Simplification of Algebraic Expressions PDF”

5. \(\frac{1}{x^2-3 x+2}+\frac{1}{x^2-5 x+6}+\frac{1}{x^2-4 x+3}\)

Solution:

= \(\frac{1}{x^2-3 x+2}+\frac{1}{x^2-5 x+6}+\frac{1}{x^2-4 x+3}\)

= \(\frac{1}{x^2-2 x-x+2}+\frac{1}{x^2-3 x-2 x+6}+\frac{1}{x^2-3 x-x+3}\)

= \(\frac{1}{x(x-2)-1(x-2)}+\frac{1}{x(x-3)-2(x-3)}+\frac{1}{x(x-3)-1(x-3)}\)

= \(\frac{1}{(x-2)(x-1)}+\frac{1}{(x-3)(x-2)}+\frac{1}{(x-3)(x-1)}\)

= \(\frac{x-3+x-1+x-2}{(x-1)(x-2)(x-3)}\)

= \(\frac{3 x-6}{(x-1)(x-2)(x-3)}\)

= \(\frac{3(x-2)}{(x-1)(x-2)(x-3)}\)

= \(\frac{3}{(x-1)(x-3)}\)

=\(\frac{3}{x^2-4 x+3}\)

\(\frac{1}{x^2-3 x+2}+\frac{1}{x^2-5 x+6}+\frac{1}{x^2-4 x+3}\) =\(\frac{3}{x^2-4 x+3}\)

“WBBSE Maths Class 8 Simplification of Algebraic Expressions, Chapter 15 key concepts”
“WBBSE Class 8 Maths Chapter 15, Simplification of Algebraic Expressions summary”

6. \(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)

Solution:

= \(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)

= \(\frac{x+1+x-1}{(x-1)(x+1)}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)

= \(\frac{2 x}{x^2-1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\)

= \(\frac{2 x\left(x^2+1\right)+2 x\left(x^2-1\right)}{\left(x^2 1\right)\left(x^2+1\right)}+\frac{4 x^3}{x^4+1}\)

= \(\frac{2 x^3+2 x+2 x^3-2 x}{x^4-1}+\frac{4 x^3}{x^4+1}\)

= \(\frac{4 x^3}{x^4+1}+\frac{4 x^3}{x^4-1}\)

= \(\frac{4 x^3\left(x^4+1\right)+4 x^3\left(x^4-1\right)}{\left(x^4 1\right)\left(x^4+1\right)}\)

= \(\frac{4 x^7+4 x^2+4 x^7-4 x^8}{x^8-1}\)

= \(\frac{8 x^7}{x^8-1}\)

\(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2 x}{x^2+1}+\frac{4 x^3}{x^4+1}\) = \(\frac{8 x^7}{x^8-1}\)

7.  \(\frac{b^2-5 b}{3 b-4 a} \times \frac{9 b^2-16 a^2}{b^2-25} \div \frac{3 b^2+4 a b}{a b+5 a}\)

Solution:

= \(\frac{b^2-5 b}{3 b-4 a} \times \frac{9 b^2-16 a^2}{b^2-25} \div \frac{3 b^2+4 a b}{a b+5 a}\)

= \(\frac{b(b-5)}{3 b-4 a} \times \frac{(3 b)^2-(4 a)^2}{(b)^2-(5)^2} \div \frac{b(3 b+4 a)}{a(b+5)}\)

= \(\frac{b(b-5)}{(3 b-4 a)} \times \frac{(3 b+4 a)(3 b-4 a)}{(b+5)(b-5)} \times \frac{a(b+5)}{b(3 b+4 a)}\)

= a

\(\frac{b^2-5 b}{3 b-4 a} \times \frac{9 b^2-16 a^2}{b^2-25} \div \frac{3 b^2+4 a b}{a b+5 a}\) = a

“WBBSE Class 8 Chapter 15 Maths, Simplification of Algebraic Expressions step-by-step solutions”

8.  \(\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}\)

Solution:

= \(\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}\)

= \(\frac{b+c}{-(a-b)(c-a)}+\frac{c+a}{-(a-b)(b-c)}+\frac{a+b \cdot}{-(c-a)(b-c)}\)

= \( -\left\{\frac{b+c}{(a-b)(c-a)}+\frac{c+a}{(a-b)(b-c)}+\frac{a+b}{(c-a)(b-c)}\right\}\)

= \(-\left\{\frac{(b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b)}{(a-b)(b-c)(c-a)}\right\}\)

= \(-\left\{\frac{b^2-e^2+e^2-a^2+a^2-b^2}{(a-b)(b-c)(c-a)}\right\}\)

= \(\frac{0}{(a-b)(b-c)(c-a)}=0\)

\(\frac{b+c}{(a-b)(a-c)}+\frac{c+a}{(b-a)(b-c)}+\frac{a+b}{(c-a)(c-b)}\) = 0

9. \(\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)}\)

Solution:

= \(\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)}\)

= \(\frac{b+c-a}{-(a-b)(c-a)}+\frac{c+a-b}{-(b-c)(a-b)}+\frac{a+b-c}{-(c-a)(b-c)}\)

= \(-\left\{\frac{b+c-a}{(a-b)(c-a)}+\frac{c+a-b}{(b-c)(a-b)}+\frac{a+b-c}{(c-a)(b-c)}\right\} \)

= \(-\left\{\frac{(b-c)(b+c-a)+(c-a)(c+a-b)+(a-b)(a+b-c)}{(a-b)(b-c)(c-a)}\right\}\)

= \(-\left\{\frac{b^2+b c-a b-b c-c^2+a c+c^2+a c-b c-a c-a^2+a b+a^2+a b-a c-a b-b^2+b c}{(a-b)(b-c)(c-a)}\right\}\)

= \(-\frac{0}{(a-b)(b-c)(c-a)}=0\)

\(\frac{b+c-a}{(a-b)(a-c)}+\frac{c+a-b}{(b-c)(b-a)}+\frac{a+b-c}{(c-a)(c-b)}\)= 0

“Class 8 Maths Simplification of Algebraic Expressions solutions, WBBSE syllabus”

10.  \(\frac{\frac{a^2}{x-a}+\frac{b^2}{x-b}+\frac{c^2}{x-c}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)

Solution:

= \(\frac{\frac{a^2}{x-a}+\frac{b^2}{x-b}+\frac{c^2}{x-c}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)

= \(\frac{\frac{a^2+a x-a^2}{x-a}+\frac{b^2+b x-b^2}{x-b}+\frac{c^2+c x-c^2}{x-c}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)

= \(\frac{\frac{a x}{x-a}+\frac{b x}{x-b}+\frac{c x}{x-c}}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)

= \(=\frac{x\left(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}\right)}{\left(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}\right)}=x\)

\(\frac{\frac{a^2}{x-a}+\frac{b^2}{x-b}+\frac{c^2}{x-c}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}\)=x

11.  \(\left(\frac{a^2+b^2}{a^2-b^2}-\frac{a^2-b^2}{a^2+b^2}\right) \div\left(\frac{a+b}{a-b}-\frac{a-b}{a+b}\right) \times\left(\frac{a}{b}+\frac{b}{a}\right)\)

Solution:

= \(\left(\frac{a^2+b^2}{a^2-b^2}-\frac{a^2-b^2}{a^2+b^2}\right) \div\left(\frac{a+b}{a-b}-\frac{a-b}{a+b}\right) \times\left(\frac{a}{b}+\frac{b}{a}\right)\)

= \(\left\{\frac{\left(a^2+b^2\right)^2-\left(a^2-b^2\right)^2}{\left(a^2+b^2\right)\left(a^2-b^2\right)}\right\} \div\left\{\frac{(a+b)^2-(a-b)^2}{(a+b)(a-b)}\right\} \times\left(\frac{a^2+b^2}{a b}\right)\)

= \(\frac{a^4+2 a^2 b^2+b^4-\left(a^4-2 a^2 b^2+b^4\right)}{\left(a^2+b^2\right)(a+b)(a-b)} \div \frac{a^2+2 a b+b^2-\left(a^2-2 a b+b^2\right)}{(a+b)(a-b)} \times \frac{\left(a^2+b^2\right)}{a b}\)

= \(\frac{a^4+2 a^2 b^2+b^4-a^4+2 a^2 b^2+b^4}{\left(a^2+b^2\right)(a+b)(a-b)} \div \frac{a^2+2 a b+b^2-a^2+2 a b-b^2}{(a+b)(a-b)} \times \frac{\left(a^2+b^2\right)}{a b}\)

= \(=\frac{4 a^2 b^2}{\left(a^2+b^2\right)(a+b)(a-b)} \times \frac{(a+b)(a-b)}{4 a b} \times \frac{\left(a^2+b^2\right)}{a b}\)

= \(\frac{4 a^2 b^2}{4 a^2 b^2}\)

= 1

\(\left(\frac{a^2+b^2}{a^2-b^2}-\frac{a^2-b^2}{a^2+b^2}\right) \div\left(\frac{a+b}{a-b}-\frac{a-b}{a+b}\right) \times\left(\frac{a}{b}+\frac{b}{a}\right)\)=1

“WBBSE Class 8 Maths Chapter 15, Simplification of Algebraic Expressions important questions”

12.  \(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\)

Solution:

= \(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\)

= \(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\)

= \(-\left\{\frac{y^2+y z+z^2}{(x-y)(z-x)}+\frac{z^2+z x+x^2}{(y-z)(x-y)}+\frac{x^2+x y+y^2}{(z-x)(y-z)}\right\}\)

= \(-\left\{\frac{(y-z)\left(y^2+y z+z^2\right)+(z-x)\left(z^2+z x+x^2\right)+(x-y)\left(x^2+x y+y^2\right)}{(x-y)(y-z)(z-x)}\right\}\)

= \(\frac{y^3-z^3+z^3-x^3+x^3-y^3}{(x-)(y-z)(z-x)}\)

= \(-\frac{0}{(x-y)(y-z)(z-x)}\)

= 0

\(\frac{y^2+y z+z^2}{(x-y)(x-z)}+\frac{z^2+z x+x^2}{(y-z)(y-x)}+\frac{x^2+x y+y^2}{(z-x)(z-y)}\) = 0