WBBSE Solutions For Class 8 Maths Chapter 1 Algebraic Expression

Algebraic Expression

In our village Asadpur, the school building will be repaired. This year the ex-students have decided to take over this duty. There are two rooms of different shapes in this school building. Tathagatha and I calculated the expenditure of repairing the room of class-l.

Question 1. After taking measurements of the room of class-l, it has been found that its length, breadth and height are 5 m, 4 m and 3 m respectively. Let us calculate how much money will be required to cement the floor at the rate of Rs. 55 per square metre.

Solution:

Given

After taking measurements of the room of class-l, it has been found that its length, breadth and height are 5 m, 4 m and 3 m respectively.

The area of the floor of that room (5 x 4) square metres = 20 square metres.

∴ The total cost for cementing the floor of that room will be

Rs. (20 x 55)

= Rs 1100

Rs 1100 will be required to cement the floor at the rate of Rs. 55 per square metre

Read and Learn More WBBSE Solutions For Class 8 Maths

Question 2. There is a door of dimension of 2 m x 1.4 m and there are 2 windows of the dimension of 1.3 m x 1.2 m in that room. Let’s find out how much it will cost to colour the 2 windows and 1 door at Rs. 42/ sq.m.

Solution:

Given

There is a door of dimension of 2 m x 1.4 m and there are 2 windows of the dimension of 1.3 m x 1.2 m in that room.

The area of the door = 2 m. x 1.4 m = 2.8 sq.m.

The area of a window the room = 1.3 m x 1.2m = 1.56

∴ The area of two windows = 2  1.56 sq.m = 3.12 sq.m.

∴ The total area of 1 door and 2 windows is = (2.8 + 3.12) sq.m.. = 5.92 sq.m.

“WBBSE Class 8 Maths Chapter 1 solutions, Algebraic Expression”

The total expenditure to colour the doors and windows at Rs. 42 sq.m, will be

= Rs. 5.92 × 42

= Rs.248.64

Rs.248.64 it will cost to colour the 2 windows and 1 door at Rs. 42/ sq.m

Question 3. Now let’s calculate the total expenditure if I whitewash four walls and the roof of this room at Rs. 6 sq.m.

Solution:

Including the door and windows, the area of 4 walls and roof is

= 2 x length x height + 2 x breadth x height + length x breadth

= (2x5x3 + 2x4x3 + 5 x 4) sq.m.

= (30 + 24 + 20) sq.m..

= 74 sq.m.

∴ The total area of 4 walls and roof without door and windows is

=  74  sq.m. 5.92 sq.m.

= 68.08 sq.m.

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The total expenditure to whitewash the four walls and the roof is

= Rs.  68.06 × 6 =Rs.  404.4

=Total expenditure = Rs. 404.48 +Rs. 248.64 + Rs.1000

= Rs. 1653.12

Total expenditure = Rs. 1653.12

Question 4. There is a rectangular shaped playground at the back of our school. The length of this playground is 40 meter and the breadth is 20 meter. A road having 3 meter width is there all around inside the playground. Let’s work out how much money will be required to construct the road at Rs. 95/sq.m.

Solution:

Given

There is a rectangular shaped playground at the back of our school. The length of this playground is 40 meter and the breadth is 20 meter. A road having 3 meter width is there all around inside the playground.

The area of the rectangular playground including the road is (40 x 20) sq.m. =  800 sq.m.

The length of the rectangular playground excluding the road is

= 40 metre -2 x 3 metre = (40 – 6) metre

= 34 metre

The breadth of the rectangular playground excluding the road is

= 20 metre -2 × 3 metre = (20 – 6) metre

= 14 meter

∴ The area of the rectangular playground excluding the road is

= 34 sq.m, x 14 sq.m.

= 476 sq.m.

∴ The area of the road is = 800 sq.m.-476 sq.m. = 324 sq.m.

∴ Total amount required to construct this road at rate of Rs. 95 per sq.m, is

= Rs.  324 × 95 = Rs. 30780

Rs. 30780 will be required to construct the road at Rs. 95/sq.m.

“Class 8 WBBSE Maths Chapter 1 solutions, Algebraic Expression study material”

Question 5. If there is road all around outside the ground of width 3 meter, then let us work out the expenditure of constructing the road at the same rate above (let’s do).

Solution:

Given

If there is road all around outside the ground of width 3 meter,

The area of the playground = 40 x 20 sq.m. = 800 sq.m.

Length of the playground including the road = (40 + 2 x 3) m.

= (40 + 6) m. = 46 m.

Breadth of the playground including the road = (20 + 2 x 3 ) m.

= (20 + 6) m. = 26 m.

Area of the playground including the road = 46 x 26 sq.m. = 1196 sq.m.

Area of the road =(1196 – 800) sq.m. = 396 sq.m..

The cost of constructing this road at the rate of Rs. 95 per sq.m.

= Rs. 396 x 95 = Rs. 37,620

The cost of constructing this road at the rate of Rs. 95 per sq.m. = Rs. 37,620

Funny game with coloured paper

Today some of us decided that we will cut rectangular and square shaped coloured papers and paste some black paper on that coloured paper to from roads of same width.

Question 6. Let’s see how much region of the coloured paper is black.

Solution:

Tirthanker draws:

Area of the paper with black road = 12 x 8 sq. cm.

= 96 sq. cm.

Area of the paper without the black road = 32 sq. cm.

∴ Area of the black road = 64 sq. cm.

We stick the two black roads in the middle of the green paper and so we get four green rectangular regions with same area.

Titli draws:

∴ Length of a green rectangular region = \(\frac{28-4}{2}\) cm. = 12  cm.

Breadth of a green rectangular region = \(\frac{18-4}{2}\)cm. = [7] cm

∴ Area of a green rectangular region = 12 × 7 sq. cm

=84 sq. cm

Area of 4 green rectangular regions with black road = 4 × 84 sq. cm

= 336 sq. cm

“WBBSE Class 8 Maths Chapter 1, Algebraic Expression solved examples”

= Area of big rectangular region with black road 28  x 18 sq. cm

= 504 sq. cm.

∴ Area of black road drawn by Titli = 504 – 336= 168 sq. cm.

Question 7. Area of black road drawn by Sophia 42 sq. cm.

Solution:

It is seen that the black coloured road divides the square region into four equal squares.

Question 8. Let’s see the pictures drawn by David, Farukh and Mitali and then let’s find the area of the black coloured roads and note it down.

Solution:

Area of the black coloured road drawn by David :

Area of the rectangular area= (13 + 5) × (16 + 5) sq. cm.

= 18 x 21 sq. cm. = 378 sq. cm.

Area of the land = 2 (13 x 8). sq. cm.

= 2 x 104 sq. cm.

= 208 sq. cm

Area of the road = (378 – 208) sq. cm = 170

Area of the black coloured road draw by Farukh :

Area of the region including the road 20 × 20 sq. cm. = 400 sq. cm.

Area of the region excluding the road = (20-12) × (20 – 12) sq. cm.

= 8 x 8 = 64 sq. cm.

Area of the road = (400 – 64) sq. cm. = 336 sq. cm.

Area of the black coloured road draw by Mitali

Area of the region including the road = 24 x 24 sq. cm. = 576 sq. cm.

Area of a pink sub-division = \(\left(\frac{24-4}{2}\right) \times\left(\frac{24-4}{2}\right)\) sq. cm.

= 10 x 10 sq. cm.

= 100 sq.cm.

Area of 4 pink regions = 4 x 100 sq. cm.

= 400 sq. cm.

∴ Area of the road = (576 – 400) sq. cm.

= 176 sq. cm.

Algebraic Expression Exercise

Question 1. Let’s see the pictures in the graph paper below and let’s find the area of the figures and note it down.

Solution:

Question 2. The length and breadth of the rectangular courtyard of the house of Amina are 6 m. and 4.2 m. respectively. We put a mattress of measure 3.5 m x 2.5 m in the middle of the courtyard. Let’s find the area of the courtyard except the mattress and note it down.

Solution:

Given

The length and breadth of the rectangular courtyard of the house of Amina are 6 m. and 4.2 m. respectively. We put a mattress of measure 3.5 m x 2.5 m in the middle of the courtyard.

Area of the courtyard = 6 × 4.2 sq. m. = 25.2 sq.m.

Area of the matress = 3.5 m. × 2.5 m. = 8.75 sq.m.

Area of the courtyard without the mattress = (25.2 – 8.75) sq.m. = 16.45 sq. m.

The area of the courtyard except the mattress = 16.45 sq. m.

Question 3. There is a path of 3 m width all around outside the square-shaped park of Ajanta Housing Complex. The perimeter of the park including the path is 484 m. Let’s calculate the area of the parth.

Solution:

Given

There is a path of 3 m width all around outside the square-shaped park of Ajanta Housing Complex. The perimeter of the park including the path is 484 m.

Perimeter of the square shaped park including the road = 484 m.

Length of one side of the square shaped park including the road = 484 ÷ 4 m. = 121 m.

Area of the park including the road = (121 )² sq.m = 14641 sq.m.

= (121-6) m. = 115 m.

Area of the park including the road = (115)² sq.m = 13225 sq.m.

∴ Area of the road = (14641 – 13225) sq.m = 1416 sq.m.

The area of the parth= 1416 sq.m.

Question 4. The length and breadth of Mihir’s rectangular garden are 50 meters and 30 meters respectively. There is a road in the middle of the garden of width 4 meter parallel to the length of the garden. This road divides the garden into two rectangular regions of equal area. Let’s draw a picture and find the area of the road and note it down.

Solution:

Given

The length and breadth of Mihir’s rectangular garden are 50 meters and 30 meters respectively. There is a road in the middle of the garden of width 4 meter parallel to the length of the garden. This road divides the garden into two rectangular regions of equal area.

1. If the road of width 4 m passing through the middle of the garden is parallel to the breadth and this road divides the garden into two equal parts, then let’s draw a picture and find the area of the road
   and note it down.
2.  If there are 2 roads of width 4 m parallel to the length and breadth of Mihir’s garden passing through the middle and divide the garden into 4 equal parts, then let draw a picture to find the area of the road and note it down.

Solution: Length of the rectangular garden including road = 50 m.

Breadth of the rectangular garden with road = 30 m.

Area of the rectangular garden with road = 50 × 30 sq.m =1500 sq.m

Breadth of the garden without the road = (30 – 4) m. = 26 m.

Area of the garden without the road = 50 × 26 sq.m= 1300 sq.m

Area of the road = (1500 – 1300)sq.m = 200 sq.m

Breadth of the garden without the road = (50 – 4) m. = 46 m.

Area of the garden without the road = 46 x 30 sq.m = 1380 sq.m

Area of the road = (1500 – 1380) sq.m = 120 sq.m

Area of the garden including the road = 50 x 30 sq.m = 1500 sq.m

Length of each part of the garden =  \(\frac{50-4}{2}\)m. = 23 m.

Breadth of each part of the garden = \(\frac{30-4}{2}\)m. = 13 m.

Area of each part of the garden = 23 x 13 sq.m – 299 m.

Area of all 4 parts of the garden without the road = 4 x 299 sq.m = 1196 sq.m

Area of the road = (1500 – 1196) sq.m = 304 sq.m

“WBBSE Class 8 Algebraic Expression solutions, Maths Chapter 1”

Question 5. There is a rectangular field beside our house owned by Papia’s family. The length and breadth of this land are 48 meter and 26 meter respectively. Papia’s family has built their house on that land leaving a 4 meter width gap on all sides. Let’s find the area of the region in which they have built their house.

Solution : 

Given

There is a rectangular field beside our house owned by Papia’s family. The length and breadth of this land are 48 meter and 26 meter respectively. Papia’s family has built their house on that land leaving a 4 meter width gap on all sides.

Area of the rectangular field = 48 x 26 sq.m = 1248 sq. m.

Length of the house = (48 – 2 x 4) m. = 40 m.

Breadth of the house = (26 – 2 x 4) m. = 18 m.

Area of the house = 40 x 18 sq.m = 720 sq.m

Papia will make her home at a field of measure 720 sq. m.

Arrange the colour sticks in different forms:

I am sticking match sticks along with Amita, Satyaki and Ayan on a white paper in different order. We shall hand up this in our classroom. Satyaki has decided to write the number of sticks after counting from each arrangement.

I arrange

Sattaki write –

Sattaki write –

Amita arranged

Sattaki write –

Algebraic Expression Exercise

Question 1. Let’s find the total number of matchsticks required in the n-th position of the previous types of arrangements.

Solution:

Activity

Let’s make some square and rectangular cards where one side of them is blue and the other side is red.

Like the above picture,

blue (4 cm x 4 cm) square card → x²

blue (4 cm × 1 cm) rectangular card → x

blue (1 cm × 1 cm) square card →1

and red (4 cm x 4 cm) square card → x²

red (4 cm x 1 cm ) rectangular card → x

red (1 cm x 1 cm) square card → 1

Let us express the algebraic expression with blue and red cards :

2x²+ 4x – 3

– 2x² + 2x – 1

2x² – 3x + 5

-x² -8x + 6

4x² – 2x – 3

– 4x² + 7x – 4

(x² + 2x + 5) + (2x² + 2x + 1)

(3x² – 5x + 6) + (2x² + 8x – 4)

(8x² – 2x – 4) – (3x² + 4x + 2)

(- 2x² + 5x + 3) – (-4x² + 2x -2)

Hints: 2x² + 4x – 3

New Game With Fun Cards:

Uma Samir, Sudir and I have decided that we will make some square and rectangular cards and will write length and breadth. We will pick up cards one by one and will write area or length or breadth of the cards.

Let’s see the length and breadth of my card which I picked up.

Area of the card = (2x + 4)m x(x – 5)m

= (2x + 4)(x – 5) sq.m.

= (2x²+ 4x – 10x – 20)sq.m = (2x²– 6x – 20) sq.m

Let’s see the length and breadth of Lima’s card-

Area of Uma’s card =5x – 7m × \(\frac{x}{5}+2\) m

= \(x^2+10 x-\frac{7 x}{5} 14\) sq.m

However, area of Samir’s card is (7a2b – 35ab2 + 14abc) sq.m and breadth is 7ab m.

∴ Length of Samir’s card

= (7a²b – 35ab² + 14abc) sq. m. ÷ 7ab m.

= \(\left(\frac{{ }^a 7 a^2 b}{7 a b}-\frac{{ }^{5 b} 35 a b^2}{7 a b}+\frac{{ }^{2 c} 14 a b c}{7 a b}\right) m .\)

= (a – 5b + 2c) m.

Area of Subir’s card is (6x⁴y⁴ — 12x²y² + 30x²y⁴) sq. m. and length is 6x²y² metre.

Breadth of the card = 6x⁴y⁴ – 12x²y² + 30x²y⁴ sq. m. ÷ 6x²y²m.

= \(\frac{6 x^4 y^2-12 x^2 y^2+30 x^2 y^4}{6 x^2 y^2}\)

= \(\frac{6 x^4 y^2}{6 x^2 y^2}-\frac{12 x^2 y^2}{6 x^2 y^2}+\frac{30 x^2 y^4}{6 x^2 y^2}\)

= (x² – 2 + 5y²) m.

“Class 8 WBBSE Maths Chapter 1, Algebraic Expression easy explanation”

Let’s find the length of the card where the area of the card is (9p²-4q²) sq.m and breadth of the card is (3p-2q) m.

Solution : Length = Area ÷ Breadth = \(\frac{9 p^2-4 q^2}{3 p-2 q}\)

Let’s express 9p2 – 4q2 into product of two expressions, i.e., factorise it. ,

9p² – 4q² = (3p)² – (2q)²

= (3p + 2q)(3q – 2q)

∴ Length = \(\frac{9 p^2-4 q^2}{3 p-2 q}\) metre

= \(\frac{(3 p+2 q)(3 p-2 q)}{(3 p-2 q)}\)

= (3p + 2q) metre

Algebraic Expression Exercise

Length of the rectangular card And Breadth of the rectangular card:

Algebraic Expression Exercise

Question 1. Let’s write the number of matchsticks in the n-th position (n is a positive integer) in the arrangement :

Solution:

1. Figure …………5n + 1
2. Figure …………..5n + 2
3. Figure …………….4n + 1

Question 2. Let’s write the perimeter of an equilateral triangle if length of each side is (4y + 2) cm.

Solution:

Perimeter of an equilateral triangle = 3 x side cm.

= 3(4y + 2) cm.

= 12y + 6 cm.

Perimeter of an equilateral triangle = 12y + 6 cm.

Question 3. Let’s write the area of the rectangular region whose length is (8x + 3y) cm and breadth is (8x – 3y) cm.

Solution:

Area of the rectangular region = (8x + 3y)cm × (8x – 3y) cm.

= (8x + 3y) (8x – 3y) sq. cm.

= (8x)² – (3y)² sq. cm.

= (64x² – 9y²) sq.cm.

Area of the rectangular region = (64x² – 9y²) sq.cm.

Question 4. Let’s write the area of the square region in terms of m where length is (3m – 4) metre. Let’s calculate the value of m when the perimeter of the square is 8 metre.

Solution:

Area of the square = (3m – 4) m. × (3m – 4) m.

= (3m – 4)² sq. m.

= (3m)² – 2.3m.4 + (4)² sq m.

= (9m²– 12m + 16) sq. m.

The perimeter of the square =4 x (3m – 4) m.

= 12m – 16 metre.

According to the problem,

12m -16 = 8

or, 12m = 8+16

or, 12m = 24

or, m= \(\frac{24}{2}\) = 2

If m is 2, perimeter will be 8 m

The value of m is 2 when the perimeter of the square is 8 metre

“WBBSE Class 8 Maths Chapter 1 solutions, Algebraic Expression PDF”

Question 5. Let’s fill up the table below:

Solution :

1. Algebraic expression

x²+2y²……………… (1)

(-8y²+ 6x²+z²)……………… (2)

Let’s add:

(1 + 2) equations adding:

x²+2y² + (-8y²+6x²+z²)

= x²+2y²-8y²+6x²+z²

= 7x²-6y²+z²

Let’s Subtract: 

(1- 2) Equations substract:

x²+2y²-(-8y²+6x²+z²)

= x²+2y²+8y²-6x²-z²

= 10y²-5x²-z²

2. Algebraic expression

6a²+2 …………..(1)

-3a²+3a……….(2)

-2a+3…………….(3)

(1+2+3) equations adding:

= (6a²+2) + (-3a²+3a) + (-2a+3)

= 6a²+2 – 3a² + 3a -2a+3

= 3a² + a +5

(2-1) Equations subtract:

= (-3a²+3a) – (6a²+2)

= -3a²+3a – 6a² – 2

= – 9a² + 3a – 2

(3-1) Equations subtract:

= (-2a+3) – (6a²+2)

= -2a + 3 – 6a² – 2

= -6a² – 2a + 1

3. Algebraic expression.

9m²-2mn+n²………..(1)

m²+n²…………..(2)

m²-3mn-2n²…………..(3)

(1+2+3) equations adding:

= (9m²-2mn+n²) + (m²+n²) + (m²-3mn-2n²)

= 9m²-2mn+n² + m²+n² + m² – 3mn – 2n²

= 11m² – 5mn

(1-2) equations subtract:

= (9m²-2mn+n²) – (m²+n²)

= 9m²-2mn + n² – m² -n²

= 8m² – 2mn

“WBBSE Class 8 Maths Chapter 1, Algebraic Expression important questions”

(2-3) equations subtract:

= (m²+n²) – (m²-3mn-2n²)

= m²+n² – m² + 3mn + 2n²

= 3n² + 3mn

Question 6. Let’s fill up the table below

Solution :

1. Algebraic expression

9a³b²-15a²b³………….(1)

3ab………………….(2)

(1 × 2) equations multiply:

(9a³b²-15a²b³)x3ab

= 27a³⁺¹b2+1-45a²⁺¹b³⁺¹

= 27a⁴b³-45a³b⁴

(1÷2) equation divided:

= \(\frac{9 a^3 b^2-15 a^2 b^3}{3 a b}\)

= \(\frac{9 a^3 b^2}{3 a b}-\frac{15 a^2 b^3}{3 a b}\)

\(=3 a^{3-1} b^{2-1}-5 a^{2-1} b^{3-1}\)

= \(3 a^2 b-5 a b^2\)

2. Algebraic expression

x⁴-4x³+6x²………….(1)

x²………………………(2)

(1 × 2) equations multiply:

(x⁴-4x³+6x²) x x²

= x⁴⁺²– 4x³⁺²+ 6x²⁺¹

= x⁶ – 4x⁵+ 6x⁴

(1÷2) equation divided:

= \(\frac{x^4-4 x^3+6 x^2}{x^2}\)

= \(\frac{x^4}{x^2}-\frac{4 x^3}{x^2}-\frac{6 x^2}{x^2}\)

= \(x^{4-2}-4 x^{3-2}+6 x^{2-2}\)

= x² – 4x + 6

3. Algebraic expression

3m²n³+40m³n⁴-5m⁴n5……………(1)

10m²n²……………..(2)

(1 × 2) equations multiply:

(3m²n³+40m³n⁴-5m⁴n⁵) x 10m²n²

= 30m⁴n⁵+40m⁵n⁶-50m⁶n7

(1÷2) equation divided:

(3m²n³+40m³n⁴-5m⁴n⁵) ÷ 10m²n²

= \(\frac{3 m^2 n^3+40 m^3 n^4-5 m^4 n^5}{10 m^2 n^2}\)

= \(\frac{3 m^2 n^3}{10 m^2 n^2}+\frac{40 m^3 n^4}{10 m^2 n^2}-\frac{5 m^4 n^5}{10 m^2 n^2}\)

= \(\frac{3}{10} n+4 m n^2-\frac{1}{2} m^2 n^3\)

4. Algebraic expression

(49l²-100m²) + 10m)……………….(1)

(7I + 10m)………………. (2)

(1 × 2) equations multiply:

(49l²-100m²) x (7l + 10m)

343l³-700m²l+490ml²-1000m³

(1÷2) equation divided:

(49l²-100m²) ÷ (7l+10)

= \(\frac{\left.49\right|^2-100 m^2}{7 I-10 m}\)

= \(\frac{(7 \mathrm{l})^2-(10 m)^2}{7 \mathrm{I}-10 \mathrm{~m}}\)

= \(\frac{(7 \mathrm{I}+10 \mathrm{~m})(7 \mathrm{I}-10 \mathrm{~m})}{(7 \mathrm{I}-10 \mathrm{~m})}\)

= (71 + 10m)

4. Algebraic expression

625a⁴-81b⁴………………..(1)

5a+3b……………….(2)

(1 × 2) equations multiply:

= (625a⁴-81 b⁴) x (5a+3b)

= 3125a⁵-40⁵ab⁴+1875a⁴b-243b⁵

“Class 8 Maths Algebraic Expression solutions, WBBSE syllabus”

(1÷2) equation divided:

(625a⁴-81b⁴) ÷ (5a+3b)

= \(\frac{625 a^4-81 b^4}{5 a+3 b}\)

= \(\frac{\left(25 a^2\right)^4-\left(9 b^2\right)^4}{(5 a+3 b)}\)

= \(\frac{\left(25 a^2+9 b^2\right)\left(25 a^2-9 b^2\right)}{(5 a+3 b)}\)

= (25a2 + 9b2)(5a – 3b)

= 125a3 – 75a2b + 45ab2 – 27b3

Question 7. Let’s simplify :

1. (a-b) + (b-c) + (c-a)

Solution : (a-b) + (b-c) + (c-a)

=a – b + b – c + c – a

= 0

(a-b) + (b-c) + (c-a) = 0

2. (a+b)(a-b) + (b+c)(b-c) + (c+a)(c-a)

Solution:

(a+b)(a-b) + (b+c)(b-c) + (c+a)(c-a)

= a² – b² + b² – c²+ c² – a²

= 0

(a+b)(a-b) + (b+c)(b-c) + (c+a)(c-a)= 0

3.  \(x^2 \times\left(\frac{x}{y}-\frac{y}{x}\right) \times\left(\frac{y}{x}+\frac{x}{y}\right) \times y^2\)

Solution:

= \(x^2 \times\left(\frac{x}{y}-\frac{y}{x}\right) \times\left(\frac{y}{x}+\frac{x}{y}\right) \times y^2\)

= \(x^2 \times\left(\frac{x^2-y^2}{x y}\right) \times\left(\frac{y^2+x^2}{x y}\right) \times y^2\)

= \(\frac{x^2 \times\left(x^2-y^2\right) \times\left(x^2+y^2\right) \times y^2}{x^2 y^2}\)

= \(\left(x^2\right)^2-\left(y^2\right)^2\)

X⁴– Y⁴

4. a(b-c) + b(c-a) + c(a-b)

Solution :

a(b-c) + b(c-a) + c(a-b)

= ab – ac + be – ab + ac – be

= 0

a(b-c) + b(c-a) + c(a-b) = 0

5. x²(y²-z²) + y²(z²-x²) + z²(x²-y²)

Solution :

= x²(y²-z²) + y²(z²-x²) + z²(x²-y²)

= x²y² – x²z² + y²z² – x²y² + x²z² – y²z²

= 0

x²(y²-z²) + y²(z²-x²) + z²(x²-y²)= 0

“WBBSE Class 8 Chapter 1 Maths, Algebraic Expression step-by-step solutions”

6. (x³+y³)(x³-y³) + (y³+z³)(y³-z³) + (z³+x³)(z³-x³)

Solution :

= (x³+y³)(x³-y³) + (y³+z³)(y³-z³) + (z³+x³)(z³-x³)

= (x³)² – (y³)² + (y³)² – (z³)² + (z³)²– (x³)²

= x⁶-y⁶ +y⁶-z⁶+z⁶-x⁶

= 0

(x³+y³)(x³-y³) + (y³+z³)(y³-z³) + (z³+x³)(z³-x³)

(a+b) ²= a² + 2ab + b² and (a-b)² = a² – 2ab + b²

Question 8. Let’s form a whole square of the expressions given below usingthe above given identities –

1. 5x-2y
2. 7+2m
3. x+y+z
4. a+b-c-d

1. 5x-2y

Solution:

= (5x-2y)²

= (5x)²– 2.5x.2y + (2y)²

= 25x² – 20xy + 4y²

(5x-2y)²= 25x² – 20xy + 4y²

2. 7+2m

Solution:

= (7+2m)²

= (7)² + 2.7.2m + (2m)²

= 49 + 28m + 4m2

(7+2m)²  = 49 + 28m + 4m2

3. x+y+z

Solution:

= (x+y+z)²

= {(x+y)+z}²

= (x+y)²+2(x+y).z+(z)²

= x²+2xy+y²+2xz+2yz+z² .

= x²+y²+z²+2xy+2xz+2yz

x+y+z)² = x²+y²+z²+2xy+2xz+2yz

4. a+b-c-d

Solution:

= (a+b-c-d)²

= {(a+b) – (c+d)}²

= (a+b)²-2(a+b)(c+d)+(c+d)²

= a²+2ab+b²-2a(c+d)-2b(c+d)+c²+2cd+d²

= a²+2ab+b²-2ac-2ad-2bc-2bd+c²+2cd+d²

= a²+b²+c²+d²+2ab-2ac-2ad-2bc-2bd

(a+b)² = a² + 2ab + b² and (a-b)² = a²– 2ab + b²

(a+b-c-d)² = a²+b²+c²+d²+2ab-2ac-2ad-2bc-2bd

Question 9. Let’s form whole squares of the expressions given below using the above identities.

1.  \(9 x^2+\frac{9}{25 y^2}-\frac{18 x}{5 y}\)
2. 25m²-70mn+49n²
3. (2a-b)2+(4a-2b)(a+b)+(a+b)2
4. \(\frac{p^2}{q^2}+\frac{q^2}{p^2}-2\)

1. \(9 x^2+\frac{9}{25 y^2}-\frac{18 x}{5 y}\)

Solution:

= \((3 x)^2+\left(\frac{3}{5 y}\right)^2-2.3 x \cdot \frac{3}{5 y}\)

= \(\left(3 x-\frac{3}{5 y}\right)^2\)

2. 25m²-70mn+49n²

Solution:

Given 25m²-70mn+49n²

= (5m)²-2.5m.7n+(7n)²

= (5m-7n)²

25m²-70mn+49n² = (5m-7n)²

3. \(\frac{p^2}{q^2}+\frac{q^2}{p^2}-2\)

Solution:

Given \(\frac{p^2}{q^2}+\frac{q^2}{p^2}-2\)

= \(\left(\frac{p}{q}\right)^2+\left(\frac{q}{p}\right)^2-2 \cdot \frac{p}{q} \cdot \frac{q}{p}\)

= \(\left(\frac{p}{q}-\frac{q}{p}\right)^2\)

Question 10. Let’s express the expression as a difference of two squares :

1. 391× 409

Solution:

Given 391 x 409

= (400-9)(400+9)

= (400)²-(9)²

2. (4x+3y)(2x-3y)

Solution:

Given (4x+3y)(2x-3y)

= 8x²-12xy+6xy-9y²

= 8x²-6xy-9y²

=9x²-x²-6xy-9y²

=9×2-(x²+6xy+9y²)

=(3x)²-(x+3y)²

3. X

Solution:

Given X

= X x 1

= \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)

“WBBSE Maths Class 8 Algebraic Expression, Chapter 1 key concepts”

Question 11. Let’s factorize :

1. 225m²-100n²

Solution:

Given 225m²-100n²

= 25 (9m² – 4n²)

= 25 {(3m)² – (2n)²}

= 25 (3m + 2n) (3m – 2n)

225m²-100n² = 25 (3m + 2n) (3m – 2n)

2. 25x² – \(\frac{1}{9}\) y²z²

Solution:

Given 25x² – \(\frac{1}{9}\) y²z²

= (5x)2 – \(\left(\frac{1}{3} y z\right)^2\)

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

3. 7ax² + 14ax + 7a

Solution:

Given 7ax² + 14ax + 7a

= 7a (x² + 2x + 1).

= 7a {(x)² + 2.x.1 + (1)²}

= 7a (x + 1 )²

7ax² + 14ax + 7a = 7a (x + 1 )²

4. 3x⁴ – 6x²a² + 3a⁴

Solution:

Given 3x⁴ – 6x²a² + 3a⁴

= 3 (x⁴ – 2x²a² + a⁴) .

= 3 {(x²)² – 2.x².a² + (a²)²}

= 3 (x² – a²)²

= 3 {(x + a) (x – a)}²

= 3 (x + a)² (x – a)²

3x⁴ – 6x²a² + 3a⁴ = 3 (x + a)² (x – a)²

“WBBSE Class 8 Maths Chapter 1, Algebraic Expression summary”

5. 4b²c² – (b² + c² – a²)²

Solution:

Given 4b²c² – (b² + c² – a²)²

= (2bc)² – (b² + c² – a²)² = (2bc + b² + c² – a²) (2bc – b² – c² + a²)

= (b² + 2bc + c² – a²) (a² – b² + 2bc – c²)

= {(b + c)² – (a)²} {a²– (b² – 2bc + c²)}

= {(b + c)² – (a)²} {(a)² – (b-c)²}

= (b + c + a) (b + c – a) (a + b – c) (a – b + c)

4b²c² – (b² + c² – a²)² = (b + c + a) (b + c – a) (a + b – c) (a – b + c)

“WBBSE Class 8 Maths Chapter 1 Algebraic Expression, definitions and formulas”

6. 64ax² – 49a (x – 2y)²

Solution:

Given 64ax² – 49a (x – 2y)²

= a {64x² – 49 (x – 2y)²}

= a [64x² – {7(x – 2y)}²]

= a {(8x)² – (7x – 14y)²}

= a (8x + 7x – 14y) (8x – 7x + 14y)

= a (15x – 14y) (x + 14y)

64ax² – 49a (x – 2y)² = a (15x – 14y) (x + 14y)

7. x² – 9 – 4xy + 4y²

Solution:

Given x² – 9 – 4xy + 4y²

= x² – 4xy + 4y² – 9

= (x)² – 2.x.2y + (2y)² – 9

= (x – 2y)² – (3)²

= (x – 2y + 3) (x – 2y – 3)

x² – 9 – 4xy + 4y² = (x – 2y + 3) (x – 2y – 3)

8. x² – 2x – y² + 2y

Solution:

Given x² – 2x – y² + 2y

= x² – y² – 2x + 2y

= (x + y) (x – y) – 2(x – y)

= (x – y) (x + y – 2)

x² – 2x – y² + 2y = (x – y) (x + y – 2)

9. 3 + 2a – a²

Solution:

Given 3 + 2a – a²

= 4-1 + 2a – a²

= 4 – (1 – 2a + a2)

= (2)² – {(1 )2 – 2.1 .a + (a)²}

= (2)2 – (1 – a)2

= (2 + 1 – a) (2 – 1 + a)

= (3 – a) (1 + a)

3 + 2a – a² = (3 – a) (1 + a)

10. X⁴– 1

Solution:

Given X⁴– 1

= (x2)²-(1)²

= (x²+ 1) (X²– 1)

= (X²+ 1) {(X)² – (1)²}

= (x²+1) (x+1) (x-1)

X⁴– 1 = (x²+1) (x+1) (x-1)

11. a² – b² – c² + 2bc

Solution:

Given a² – b² – c² + 2bc

= a² – (b² – 2bc + c²)

= (a)² – (b – c)²

= (a + b – c) (a – b + c)

a² – b² – c² + 2bc = (a + b – c) (a – b + c)

12. ac + be + a + b

Solution:

Given ac + be + a + b

= c (a + b) + 1 (a + b)

= (a + b) (c + 1)

ac + bc + a + b = (a + b) (c + 1)

“WBBSE Class 8 Maths Algebraic Expression, revision notes”

13. x⁴+ x²y² + y⁴

Solution:

Given x⁴+ x²y² + y⁴

= (x²)² + 2.x²y² + (y²)² – x²y²

= (x² + y²) – (xy)²

= (x² + y² + xy) (x² + y² – xy)

= (x² + xy + y²) (x² – xy + y²)

x⁴+ x²y² + y⁴ = (x² + xy + y²) (x² – xy + y²)

Question 12. Let’s find the product by formulae:

1. (xy + pq) (xy – pq)

Solution:

Given (xy + pq)(xy – pq)

= (xy)² – (pq)²

= x²y² – p²q²

(xy + pq) (xy – pq) = x²y² – p²q²

2. 49 x 51

Solution:

Given 49 x 51

= (50- 1) (50 + 1)

= (50)2 – (1)2

= 2500 – 1

= 2499

49 x 51 = 2499

3. (2x – y + 3z) (2x + y + 3z)

Solution:

Given

= (2x + 3z – y) (2x + 3z + y)

= (2x + 3z)² – (y)²

= (2x)² + 2.2x.3z + (3z)² – y²

= 4x² + 12xz + 9z² – y²

(2x – y + 3z) (2x + y + 3z) = 4x² + 12xz + 9z² – y²

4. 1511 x 1489

Solution:

Given 1511 x 1489

= (1500 + 11) (1500 – 11)

= (1500)² – (11)²

= 2250000 – 121

= 2249879

1511 x 1489 = 2249879

5. (a – 2) (a + 2) (a² + 4)

Solution:

Given (a – 2) (a + 2) (a² + 4)

= {(a)² – (2)²} (a² + 4)

= (a² – 4) (a² +4)

= (a²)² – (4)²

= a⁴– 16

(a – 2) (a + 2) (a² + 4) = a⁴– 16

Question 13. (1). If x+\(\frac{1}{x}\)=4 then let’s show that x²+\(\frac{1}{x^2}\) and X⁴+\(\frac{1}{x^4}\)=194.

Solution:

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

=\(x^2+\frac{1}{x^2}\)

=\((x)^2+\left(\frac{1}{x}\right)^2\)

=\(\left(x+\frac{1}{x}\right)^2-2 \cdot x \cdot \frac{1}{x}\)

=(4)²-2

=16-2

=14

Again, \(x^4+\frac{1}{x^4}\)

=\(\left(x^2\right)^2+\left(\frac{1}{x^2}\right)^2\)

=\(\left(x^2+\frac{1}{x^2}\right)^2-2 \cdot x^2 \cdot \frac{1}{x^2}\)

=(14)²-2

=196-2

=194

2. If m+\(\frac{1}{m}\)=-5 then lets show that m²+\(\frac{1}{\mathrm{~m}^2}\)=23.

Solution: m+\(\frac{1}{m}\)=-5

= \(m^2+\frac{1}{m^2}\)

=\((m)^2+\left(\frac{1}{m}\right)^2\)

=\(\left(m+\frac{1}{m}\right)^2-2 \cdot m \cdot \frac{1}{m}\)

=(-5)²-2

=25-2

= 23

3. If p-\(\frac{1}{p}\)=m then let’s show that:

Solution:

1. p² +\(\frac{1}{p^2}\)= m² + 2; and,
2. \(\left(p+\frac{1}{p}\right)^2=m^2+4\)

p-\(\frac{1}{p}\) = m

1. \(p^2+\frac{1}{p^2}\)

Solution:

Given \(p^2+\frac{1}{p^2}\)

= \((p)^2+\left(\frac{1}{p}\right)^2\)

= \(\left(p+\frac{1}{p}\right)^2+2 \cdot p \cdot \frac{1}{p}\)

= (m)²+2

= m²+2

2. \(\left(p+\frac{1}{p}\right)^2\)

Solution:

Given \(\left(p+\frac{1}{p}\right)^2\)

= \(\left(p-\frac{1}{p}\right)^2+4 \cdot p \cdot \frac{1}{p}\)

= (m)2+4

= m2+4

3. If a + b = 5, a – b = 1 then let’s show that 8ab(a² + b²) = 624.

Solution :

= 8ab(a² + b²)

= 4ab. 2(a² + b²)

= {(a+b)²-(a-b)²}{(a+b)²+(a-b)²}

= {(5)²-(1)²}{(5)² + (1)²}

= (25-1) (25 + 1)

= (25)²-(1)²

= 625-1

= 624

8ab(a² + b²) = 624

4. If x – y = 3, xy = 28 then let’s find the value of (x² + y²).

Solution :

= x² + y²

= (x – y)² + 2xy

= (3)² + 2.28

= 9 + 56

= 65

(x² + y²) = 65

Question 14. Let’s express the expressions as the sum of two squares:

1. 2(a² + b²)
2. 50x²+18y²
3. a² + b² + c² + d²+ 2(ac -bd)

1. 2 (a² + b²)

Solution:

Given 2 (a² + b²)

= (a + b)² + (a – b)²

2. 50 x² + 18 y²

Solution:

Given 50 x² + 18 y²

= 2 (25x² + 9y²)

= 2 {(5x)² + (3y)²}

= (5x + 3y)² + (5x – 3y)²

3. a² + b² + c² + d² + (ac -bd)

Solution:

Given a² + b² + c² + d² + (ac -bd)

= a² + b² + c² + d² + 2ac – 2bd

= a² + 2ac + c² + b² – 2bd + d²

= (a + c)² + (b – d)²

Question 15.

Solution :

1. Let’s write for which value of t, x² – tx + \(\frac{1}{4}\) will be a whole square form.
2. Let’s write the expression which when added to a² + 4, gives a whole square.
3. If a and b are positive integers and a² – b² = 9 x 11, then let’s write the value of a & b.
4. Let’s write logically whether (x + y)² – (x – y)² = 4xy is an identity or an equation.
5. For each positive or negative value of x and y except zero, the value of (x² + y²) is always ( positive or negative ).

1.  x² – tx + \(\frac{1}{4}\)

= \((x)^2-t x+\left(\frac{1}{2}\right)^2\)

( ∴ 2ab = 2.x. \(\frac{1}{2}\) = x)

or, If =1 then x² – x + \(\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

or, If = -1 then x² – (-1)x +\(\frac{1}{4}\)

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

= \(\left(x+\frac{1}{2}\right)^2\)

The value of t being 1 or -1, x² – tx + \(\frac{1}{4}\) will be a whole square.

2. a² + 4

Solution :

Given a² + 4

= (a)² + (2)²

= (a + 2)² – 2.a.2.

= (a + 2)² – 4a

a² + 4 3 4a

3. a²– b2 = 9 x 11

Solution :

Given a²– b2 = 9 x 11

= (a + b) (a – b)

= (10 + 1) (10 -1)

or a = 10 and b = 1 then this equation will be established.

∴ a = 10, b= 1

4. (x + y)² – (x – y)² = 4xy

Solution :

Given (x + y)² – (x – y)² = 4xy

L.H.S = (x + y)²– (x -y)²

= x2 + 2xy + y² – (x² – 2xy + y²)

= x2 + 2xy + y² – (x² + 2xy + y²)

= 4xy

= R.H.S

∴ L.H.S = R.H.S.

∴ This is an identity.

5. Find for each positive or negative value x and y, except zero, the value of (x² + y²).

Question 16. Let’s solve :

1. 6x = 72

2. 9x + 2 = 20

3. 4x – 2x + 3 = 9 – 4x

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

5. 2x – 5 { 7 – (x – 6) + 3x } – 28 = 39

6. \(\frac{1}{3}\) (x-2) + \(\frac{1}{4}\) (x+3)= \(\frac{1}{5}\) (x+4) + 15

1. 6x = 72

Solution :

Given 6x = 72

or, x = \(\frac{72}{6}\)

or, x = 12

2. 9x + 2 = 20

Solution:

Given 9x + 2 = 20

or, 9x = 20 – 2

or, 9x = 18

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

or, x = 2

“Class 8 WBBSE Maths Chapter 1 Algebraic Expression, multiple-choice questions”

3. 4x – 2x + 3 = 9 – 4x

Solution:

Given 4x – 2x + 3 = 9 – 4x

or, 4x – 2x + 4x = 9 – 3

or, 8x – 2x = 6

or, 6x = 6

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

or, x = 1 Ans.

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

Solution :

Given \(\frac{x}{4}-\frac{x}{2}=3 \frac{1}{2}-\frac{x}{3}\)

or, \(\frac{x}{4}-\frac{x}{2}+\frac{x}{3}=\frac{7}{2}\)

or, \(\frac{3 x-6 x+4 x}{12}=\frac{7}{2}\)

or, \(\frac{7 x-6 x}{12}=\frac{7}{2}\)

or, \(\frac{x}{12}=\frac{7}{2}\)

or, \(x=\frac{7}{2} \times{ }^6 \not 2\)

or, x = 42 Ans.

5. 2x – 5 { 7 – (x – 6) + 3x} – 28 = 39

Solution:

Given 2x – 5 { 7 – (x – 6) + 3x} – 28 = 39

or, 2x – 5 (7 – x + 6 + 3x) – 28 = 39

or, 2x – 5 (2x + 13) -28 = 39

or, 2x – 10x – 65 – 28 = 39

or, – 8x = 39 + 65 + 28

or, -8x = 132

or, x = \(\frac{132}{-8}\)

or, x = \(-\frac{33}{2}=-16 \frac{1}{2}\) Ans.

6. \(\frac{1}{3}(x-2)+\frac{1}{4}(x+3)=\frac{1}{5}(x+4)+15\)

Solution:

Given \(\frac{1}{3}(x-2)+\frac{1}{4}(x+3)=\frac{1}{5}(x+4)+15\)

or, \(\frac{(x-2)}{3}+\frac{(x+3)}{4}-\frac{(x+4)}{5}=15\)

or, \(\frac{20(x-2)+15(x+3)-12(x+4)}{60}=15\)

or, 20x – 40 + 15x + 45 – 12x – 48 = 900

or, 35x – 12x = 900 + 40-45 + 48

= 23x = 988 – 45

or, 23x = 943

or, x = \(\frac{943}{23}\)

or, x = 41 Ans.

Algebraic Expression Geometric figures

Today we will play a game. We friends collect two blackboards. One of our friends will write several conditions and others will try to draw different geometrical figures according to the given conditions.

Seouli wrote, let’s draw a quadrilateral where length of four sides are 5 cm, 8 cm, 5 cm, and 8 cm respectively, i.e., opposite sides are equal in length.

Let’s see what kind of quadrilaterals are possible using two sticks of length 5 cm and two sticks of length 8 cm.’

Using sticks we see on fixed quadrilateral is possible. So to draw fixed quadrilateral another one condition is required. Now Seouli wrote, let’s draw a fixed quadrilateral whose opposite sides are equal in length and measurement of one angle is 90°. Hence we draw a Rectangle (rectangle square) figure where length is 8 cm and breadth is 5 cm.

Question 11. On another blackboard, Anita drew a rectangle ABCD of which the length is 8 cm and the breadth is 5 cm.
Solution :

each angle of the rectangle ABCD 90° and sum of 4 angles is  360° Measuring by scale, we see

AC = \(\sqrt{89}\)cm and BD = \(\sqrt{89}\)cm, so AC = BD [put=/≠] and AO = OC [put=/≠] BO = QD [put=/≠]. Let’s measure with a protractor, ∠AOD = 90° [put=/≠]

∴ Both the diagonals of rectangular figure ABCD bisect each other but not at right angle.

Question 12. Seoli wrote, I draw a quadrilateral PQRS, of which PQ=5cm QR=8 cm RS=5 cm, PS = 8 cm and ∠PQR = 45°; So PQRS is a parallelogram (parallelogram / rectangular region).

Solution :

Tuhin drew a parallelogram PQRS on another blackboard where PQ = 5 cm, QR = 8 cm, RS = 5 cm, PS = 8 cm, and ZPQR = 45° ∠PQR = 45°, ∠QRS =135°, ∠RSP = 45°, ∠SPQ = 135°

∴ The sum of measurement of 4 angles = 360°. Now measuring by scale we

PR =3.5 cm QS =5.5cm PO = OR [put=/≠] QO = OS [put=/≠]

∴ So both the diagonals bisect each other [ I also draw the diagram in exercise book and verify it measuring by myself].

Question 13. Now Tithi drew a square region of which each side is 4 cm.

Solution :

Each side of square LAND [4] cm measure of each angle 90 and LN and AD diagonals are equal [equal/unequal] LO = I ON [put =/≠ Put], AO = OD [ put=/≠] Measuring with the help of protractor we see ∠LOD = 90°

Both the diagonals of a square region bisect each other.

Question 14. Asif Iqbar drew a Rhombus LION of length of each side 5 cm and ∠LIO = 60°. Let’s measure with protractor.

Solution :

∠LIO = 60°, ∠ON = 120°, ∠ONL = 60°, ∠NLI = 120°, and ∠LXN = 90° The sum of measurement of four angles of the rhombus = 360°

Let’s see by measuring with scale, LX = XO [=/≠] and IX = XN [=/≠].

∴ Both the diagonals of a rhombus bisect each other perpendicularly. Let’s see what I found –

(Parallelogram):

1. Length of opposite sides are equal
2. Measurements of opposite angles are equal
3. Generally length of diagonals are: unequal
4. Diagonals bisect each other

Quadrilateral whose both pairs of opposite sides are parallel with each other.

(Rhombus):

1. Length of all sides are equal
2. Measurements of opposite angles are equal
3. Generally length of diagonals are unequal
4. Diagonals bisect each other at right angles

(Rectangle):

1. Generally length of all sides are equal
2. Generally, measurement of all angles are l 90° I
3. Length of diagonals are equal
4. Diagonals bisect each other

Parallelogram whose one angle is right angle

(Square):

1. Length of all sides are equal
2. Measurement of each angle is 90°
3. Length of diagonals are equal
4. Diagonals bisect each other

Parallelogram whose one angle is a right angle and one pair of adjacent sides are equal in length.

1. With two 45° – 90° sets square, I shall find square [rectangle/ square]
2. With two 30° – 60° – 90° sets square, I shall find [ [rectangle/ square]
3. With two measurement set square, I shall find parallelogram.

Activity

Many of us drew squares of various sizes in our exercise books. After that, we coloured the square shaped paper and cut them.

Question 1. Folding my blue square shaped paper, I made 90°, 45° and 22 \(\frac{1}{2}\)° angles.

Solution :

At first took a square shaped paper

Finding along with diagonal BD and then unfolding it, I got →

Mingling the sides BD and BC with each other then folding and after that unfolding it, I got

Measuring with the protractor I find ∠ABC = 90

Folding my green square shaped paper I try to make 15°, 30° and 60° angles.

At first, I take my green square shaped paper.

“WBBSE Class 8 Algebraic Expression solutions, Chapter 1 worksheet”

Question 2. With the help of scale, pencil and compass, draw 90°, 45°, 22 \(\frac{1}{2}^{\circ}\), 60°, 30°, 120°, 75°, 105°, 135°, 150° angles.

Solution :

Question 3. In quadrilateral PLAN opposite sides are equal in length, i.e., PL = AN = 6 cm and PN = LA = 5 cm. Let’s draw three types of quadrilateral PLAN and let’s see when it will be a rectangle.

Solution:  6cm 4 cm.

∠PLA = ∠LAN = ∠ANP = ∠NPL = 90°

When all angles are 90° then PLAN becomes a rectangle.

Question 4. Let’s write what conditions are required to draw a fixed parallelogram.

Solution:

To draw a parallelogram, at least the length of one side is required.

Question 5. Let’s see what conditions are required to draw a fixed square.

Solution:

To draw a fixed square, at least length of two sides and measure of one angle are required.

Question 6. Let’s draw a square DEAR where DE = 5.6 cm.

Solution:

DEAR is a square where DE = EA = AR = RD = 5.6 cm and each angle is 90°.

Question 7. Let’s draw a rectangle BEST where BE = 6 cm and ES = 4.8 cm.

Solution :

BEST is a rectangle where BE = 6 cm. and ES = 4.8 cm.

Question 8. Let’s draw a rhombus HOME where Z HOM = 60° and HO = 6 cm.

Solution :

HOME is a rhombus whose HO = 6 cm, ∠HOM = 60°,

Question 9. Let’s draw a parallelogram ROAD where RA = 8 cm and OD = 6 cm, 8cm 6 cm

Solution :

ROAD is a rhombus where RA = 8 cm and OD = 6 cm.

Question 10. Let’s draw a parallelogram GOLD where GO = 7 cm, OL = 5.8 cm and GL = 5.8 cm.

Solution:

ROAD is a rhombus where RA = 8 cm and OD = 6 cm.

Solution: 7cm 5.8cm

GOLD is a rhombus where GO = 7 cm, OL = 5.8 cm and GL = 5.8 cm.

Question 11.

1. ABCD is a rectangle. If AC = 5 cm, let’s write the length of BD.

Solution:

The diagonals of a rectangle are equal in length.

∴ ABCD is a rectangle whose length of one diagonal is AC = 5 cm.

∴ BD = 5 cm.

2. PQRS is a square two diagonals PR and QS intersect at O. PR = 5 cm Let’s write the length of QO.

Solution:

The diagonals of a square bisect each other at right angles. Length of one diagonal PR is 5 cm.

∴ Half of one diagonal OQ = \(\frac{5}{2}\) cm. = 2.5 cm.

3. Let’s write the measurement of ∠ABC in parallelogram ABCD where ∠ADC = 60°.

Solution:

Opposite angles of a parallelogram are equal.

∠ABC = ∠ADC = 60°

4. The diagonals AC and BD of rhombus ABCD intersect at O. Let’s write the measurement of ∠AOB.

Solution:

The diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = 90°

5. A square is always a rhombus but a rhombus is not always a square

6. A square is always a rectangle but a rectangle is not always a square