Maths WBBSE Class 9 Solutions Chapter 5 Linear Simultaneous Equations Exercise 5.1

Let me frame the simultaneous equations in each of the following cases and see whether they can be solved or not.

Question 1. The sum of my elder sister’s present age and my father’s present age is 55 years. By calculating, I observed that after 16 years, my father’s age will be two times more than my sister’s.

1. Let me draw the graph after framing simultaneous equations.
2. With the help of a graph, let me find out whether the general solution of two equations can be determined.
3. Let me write the present age of my elder sister and father from the graph.

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Solutions: Let the elder sister’s present age x years and the father’s present age be y years.

According to the conditions,

x + y = 55……(1)
2(x+16) = y + 16 …(2)

From equation…(1)

→AB

x	25	35	40
y	30	20	15

 

From equation …(2)

2x-y = -16

→CD

x	-5	0	-8
y	6	16	0

 

General solution x = 13, 4 = 42
∴ The present age of sister (x) = 13 years and the present age of father (y) = 42 years.

 

 

Question 2. Mita bought 3 pens and 4 pencils at Rs. 42 from the shop of Jadabkaku. I bought 9 pens and 1 dozen pencils at the same rate to give gifts to my friends at the cost of Rs. 126.

1.  Let me draw the graph after forming the simultaneous equations.
2. With the help of a graph, let me find whether the general solution of two equations can be determined.
3. Let me write the price of 1 pen and 1 pencil separately from the graph. Ans. Let the cost price of one pen = Rs. x and the cost price of one pencil = Rs. y.

∴ Required simultaneous equations are
3x + 4y = 42…..(1)
9x+12y= 126…(2)

From equation ….(1)
3x + 4y = 42

or, \(x=\frac{42-4 y}{3}\)

→AB

x	14	10	6
y	0	3	6

 

From equation…………. (2)
9x=126-12y

or, \(x=\frac{126-12 y}{9}\)

→CD

x	10	14	6
y	3	0	6

 

Class IX Maths Solutions WBBSE

2. Infinite number of general solutions x = 10, y = 3 or, x = 6, y = 6

3. price of 1 pen = Rs. 10 & price of 1 pencil = Rs. 3
or, price of 1. pen = Rs. 6 & price of 1 pen = Rs. 6.

 

 

Question 3. Today we shall draw the pictures as we like in our school. For this, I bought 2 art paper and 5 sketch pens at Rs. 16. But Dola has bought 4 art paper and 10 sketch pens of the same rate and from the same shop.

1. Let me form simultaneous equations and draw the graph.
2. Let me see whether the general solution of the equations can be found out from the graph.
3. Let me write whether I can get the price of 1 art paper and 1 sketch pen.

Solution: Let the price of art paper = Rs. x & that of one sketch pen = Rs. y.

Class IX Maths Solutions WBBSE

∴ Required equations,
2x+5y= 16

\(x=\frac{16-5 y}{2}\)

From equation….(2)
4x + 10y = 28

or, x = \(x=\frac{28-10 y}{4}\)

2. will have no general solutions

3. No definite value of one art paper & one sketch pen.

 

 

Class IX Maths Solutions WBBSE  Chapter 5 Linear Simultaneous Equations Exercise 5.6

Let us solve the following linear equations in two variables by applying the cross-multiplication method :

Question 1. 8x+5y= 11, 3x – 4y = 10

Solution:
8x+5y = 11 …(1)
3x-4y = 10…..(2)

From equation (1) 8x+5y-11=0
From equation (2) 3x-4y-10=0

or,\(\frac{x}{5 \times(-10)-(-11) \times(-4)}=\frac{y}{(-11) \times 3-8(-10)}=\frac{1}{8(-4)-5 \times 3}\)

or,\(\frac{x}{-50-44}=\frac{y}{-33+80}=\frac{1}{-32-15}\)

or,\(\frac{x}{-94}=\frac{y}{47}=\frac{1}{-47}\)

∴ \(\frac{x}{-94}=\frac{1}{-47} \text { and } \frac{y}{47}=\frac{1}{-47}\)

or,\(x=\frac{-94}{-47} \text { and } y=\frac{47}{-47}\)

or, X = 2  and y = -1

∴ y=-1 , x = 2

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Question 2. 3x4y = 1, 4x = 3y+6

Solution: 3x-4y = 1 ….(1)

4x = 3y+ 6 …(2)

From equation (1) 3x-4y-1-0
From equation (2) 4x-3y-6=0

or, \(\frac{x}{(-4) \times(-6)-(-1)(-3)}=\frac{y}{(-1) \times 4-3(-6)}=\frac{1}{3(-3)-(-4) \times 4}\)

or, \( \quad \frac{x}{24-3}=\frac{y}{-4+18}=\frac{1}{-9+16}\)

or, \(\quad \frac{x}{21}=\frac{y}{14}=\frac{1}{7}\)

∴ \(\frac{x}{21}=\frac{1}{7} \text { and } \frac{y}{14}=\frac{1}{7}\)

or, \(x=\frac{21}{7} \text { and } y=\frac{14}{7}\)

or, x=3 and y = 2

∴ X = 3
y = 2

Class IX Maths Solutions WBBSE

Question 3. 5x + 3y = 11, 2x-7y = – 12

Solution: 5x + 3y = 11….(1)

2x-7y=-12….(2)

From equation (1) 5x + 3y-11=0
From equation (2) 2x-7y+12=0.

∴ \(\frac{x}{3 \times 12-(-11) \times(-7)}=\frac{y}{-11 \times 2-5 \times 12}=\frac{1}{5 \times(-7)-3 \times 2}\)

or, \(\frac{x}{36-77}=\frac{y}{-22-60}=\frac{1}{-35-6}\)

or, \(\frac{x}{-41}=\frac{y}{-82}=\frac{1}{-41}\)

or, \(\frac{x}{41}=\frac{y}{82}=\frac{1}{41}\)

∴ \(\frac{x}{41}=\frac{1}{41} \text { and } \frac{y}{82}=\frac{1}{41}\)

or, \(x=\frac{41}{41} \text { and } y=\frac{82}{41}\)

∴ x= 1 and y = 2

Question 4. 7x-3y-31=0, 9x-5y-41=0

Solution: 7x-3y31= 0 ….(1)

9x-5y-41=0 …..(2)

or, \(\frac{x}{(-3) \times(-41)-(-31)(-5)}=\frac{y}{(-31) \times 9-7 \times(-41)}=\frac{1}{7 \times(-5)-(-3) \times 9}\)

or, \(\frac{x}{123-155}=\frac{y}{-279+287}=\frac{1}{-35+27}\)

or, \(\frac{x}{-32}=\frac{y}{8}=\frac{1}{-8}\)

∴ \(\frac{x}{-32}=\frac{1}{-8} \text { and } \frac{y}{8}=\frac{1}{-8}\)

or,\(x=\frac{-32}{-8} \text { and } y=\frac{8}{-8}\)

∴ X = 4 and y = -1

Class IX Maths Solutions WBBSE

Question 5. \(\frac{x}{6}-\frac{y}{3}=\frac{x}{12}-\frac{2 y}{3}=4\)

Solution: \(\frac{x}{6}-\frac{y}{3}=4\) …(1)

\(\frac{x}{12}-\frac{2 y}{3}=4\) …(2)

 

From equation (1) \(\frac{x}{6}-\frac{y}{3}=4\)

or, \(\frac{x-2 y}{6}=4\)

or, x-2y = 24…(3)

From equation (2) \(\frac{x}{12}-\frac{2 y}{3}=4\)

or, \(\frac{x-8 y}{12}=4\)

or,X-8y = 48 ….(3)

From equation (1) x-2y-24=0
From equation (4) x-8y-48=0

∴ \(\frac{x}{(-2) \times(-48)-(-24)(-8)}=\frac{y}{(-24) \times 1-1 \times(-48)}=\frac{1}{1 \times(-8)-(-2) \times 1}\)

or,\(\frac{x}{96-192}=\frac{y}{-24+48}=\frac{1}{-8+2}\)

or, \(\frac{x}{-96}=\frac{y}{24}=\frac{1}{-6}\)

∴ \(\frac{x}{-96}=\frac{1}{-6} \text { and } \frac{y}{24}=\frac{1}{-6}\)

or, \(x=\frac{-96}{-6} \text { and } y=\frac{24}{-6}\)

or, x =16, y= -4

∴ x= 6
y = -4

Question 6. \(\frac{x}{5}+\frac{y}{3}=\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)

Solution: \(\frac{x}{5}+\frac{y}{3}=0\) …(1)

\(\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)…(2)

Class IX Maths Solutions WBBSE

From equation (1) \(\frac{x}{5}+\frac{y}{3}=0\)

or, \(\frac{3 x+5 y}{15}=0\)

or, 3x+5y =0 ….(3)

From equation (2) \(\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)

or, \(\frac{15 x-20 y-9}{60}=0\)

or, 15x-20y -9 =0 …(4)

From equation (3) 3x+5y+0=0
From equation (4) 15x-20y-9=0

∴ \(\frac{x}{5 \times(-9)-0 \times(-20)}=\frac{y}{0 \times 15-3 \times(-9)}=\frac{1}{3 \times(-20)-5 \times 15}\)

or, \(\frac{x}{-45-0}=\frac{y}{0+27}=\frac{1}{-60-75}\)

or,\(\frac{x}{-45}=\frac{y}{27}=\frac{1}{-13}\)

∴\( \frac{x}{-45}=\frac{1}{-135} \text { and } \frac{y}{27}=\frac{1}{-135}\)

or,\(x=\frac{-45}{-135} \text { and } y=\frac{27}{-135}\)

or,\(x=\frac{1}{3} \text { and } y=-\frac{1}{5}\)

∴ \(x=\frac{1}{3} \text { and } y=-\frac{1}{5}\)

Question 7. \(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8, \quad \frac{2 y-3 x}{3}+2 y=3 x+4\)

Solution: \(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8\) ….(1)

\(\frac{2 y-3 x}{3}+2 y=3 x+4\) ….(2)

Class IX Maths Solutions WBBSE

From equation(1) \(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8\)

or, \(\frac{4 x+8+7 y-7 x}{28}=2 x-8\)

or, \(\frac{-3 x+7 y+8}{28}=2 x-8\)

or,56-224= -3x + 7y+8
or,56x+3x-7y = 8 +224

or,59x-7y = 232…(3)

From equation (2) \( \frac{2 y-3 x}{3}+2 y=3 x+4 \)

Or, \( \frac{2 y-3 x+6 y}{3}=3 x+4 \)

or, 9x+12=8y-3x

or, 9x + 3x – 8y = – 12
or,12x-8y=-12
or,3x-2y=-3 ….(4)

From equation (3) 59x-7y-232 = 0
From equation (4) 3x-2y+3-0

\( \frac{x}{(-7) \times 3-(-232) \times(-2)}=\frac{y}{(-232) \times 3-59 \times 3}=\frac{1}{59 \times(-2)-(-7) \times 3} \)

Class 9 Mathematics West Bengal Board

or, \( \frac{x}{-21-464}=\frac{y}{-696-177}=\frac{1}{-118+21} \)

or, \( \frac{x}{-485}=\frac{y}{-873}=\frac{1}{-97} \)

\( \frac{x}{-485}=\frac{1}{-97} \text { and } \frac{y}{-873}=\frac{1}{-97} \)

or, \( x=\frac{-485}{-97} \text { and } y=\frac{-873}{-97} \)

or, x = 5 and y = 9
x = 5,y=9

Question 8. x+5y = 36, \( \frac{x+y}{x-y}=\frac{5}{3} \)

Solution: x+5y = 36 ….(1)

\( \frac{x+y}{x-y}=\frac{5}{3} \) …(2)

 

or, 5x-5y=3x+3y
or, 5x-3x-5y-3y=0
or, 2x – 8y = 0
or,X-4y=0 …(4)

From equation (1) x+5y-36=0
From equation (3) x-4y+0=0

\( \frac{x}{5 \times 0-(-36) \times(-4)}=\frac{y}{(-36) \times 1-1 \times 0}=\frac{1}{1 \times(-4)-5 \times 1} \)

 

or, \( \frac{x}{0-144}=\frac{y}{-36-0}=\frac{1}{-4-5} \)

or, \( \frac{x}{-144}=\frac{y}{-36}=\frac{1}{-9} \)

or, \( \frac{x}{144}=\frac{y}{36}=\frac{1}{9} \)

\( \frac{x}{144}=\frac{1}{9} \text { and } \frac{y}{36}=\frac{1}{9} \)

Class 9 Mathematics West Bengal Board

or, \( x=\frac{144}{9} \text { and } y=\frac{36}{9} \)

or,  x=16 and y = 4

Question 9. 13x12y+15=0, 8x-7y = 0

Solution: 13x-12y+15=0 …(1)
8x-7y+0=0…(2)

\( \frac{x}{-12 \times 0-15 \times(-7)}=\frac{y}{15 \times 8-13 \times 0}=\frac{1}{13 \times(-7)-(-12) \times 8} \)

 

Or, \( \frac{x}{0+105}=\frac{y}{120-0}=\frac{1}{-91+96} \)

Or, \( \frac{x}{105}=\frac{y}{120}=\frac{1}{5} \)

\( \frac{x}{105}=\frac{1}{5} \text { and } \frac{y}{120}=\frac{1}{5} \) \( x=\frac{105}{5} \text { and } y=\frac{120}{5} \)

Or, x=21 and y = 24
x = 21,
y = 24

Question 10. x + y = 2b, x – y = 2a

Solution: x + y = 2b …(1)
x – y = 2a …(2)

From equation (1) x+y-2b=0
From equation (2) xy- 2a =0

\( \frac{x}{1 \times(-2 a)-(2 b)(-1)}=\frac{y}{(-2 b) \times 1-1 \times(-2 a)}=\frac{1}{1 \times(-1)-1 \times 1} \)

Class 9 Mathematics West Bengal Board

Or, \( \frac{x}{-2 a-2 b}=\frac{y}{-2 b+2 a}=\frac{1}{-1-1} \)

Or, \( \frac{x}{-2(a+b)}=\frac{y}{-2(b-a)}=\frac{1}{-2} \)

Or, \( \frac{x}{2(a+b)}=\frac{y}{2(b-a)}=\frac{1}{2} \)

\( \frac{x}{2(a-b)}=\frac{1}{2} \text { and } \frac{y}{2(b-a)}=\frac{1}{2} \)

 

Or,\( x=\frac{2(a-b)}{2} \text { and } y=\frac{2(b-a}{2} \)

X= a-b, y= b-a

Question 11. x – y = 2a, ax + by = a² + b²

Solution: x – y = 2a …(1)
ax + by = a² + b² …(2)

From equation (1) x-y-2a=0
From equation (2) ax + by-(a²+b²)=0

\( \frac{x}{(-1)\left\{-\left(a^2+b^2\right)\right\}-(-2 a) \times b}=\frac{y}{-2 a \times a-1 \times\left\{-\left(a^2+b^2\right)\right\}}=\frac{1}{1 \times b-(-1) \times a} \)

Class 9 Maths WB Board

Or, \( \frac{x}{a^2+b^2+2 a b}=\frac{y}{-2 a^2+a^2+b^2}=\frac{1}{b+a} \)

Or, \( \frac{x}{(a+b)^2}=\frac{y}{b^2-a^2}=\frac{1}{b+a} \)

Or, \( \frac{x}{(a+b)(a+b)}=\frac{y}{(b+a)(b-a)}=\frac{1}{b+a} \)

\( \frac{x}{(a+b)(a+b)}=\frac{1}{b+a} \text { and } \frac{y}{(b+a)(b-a)}=\frac{1}{b+a} \)

Or, \( x=\frac{(a+b)(a+b)}{(a+b)} \text { and } y=\frac{(b+a)(b-a)}{(b+a)} \)

or, x = a + b and y = b-a
x = a + b
y = b-a

Question 12. \( \frac{x}{a}+\frac{y}{b}=2 \), ax-by=a²-b²

Solution: \( \frac{x}{a}+\frac{y}{b}=2 \) ….(1)
ax-by=a²-b² …(2)

From equation (1) \( \frac{x}{a}+\frac{y}{b}=2 \)

Or, \( \frac{b x+a y}{a b}=2 \)

Or, bx+ay = 2ab ….(3)

From equation (3) bx+ay-2ab = 0
From equation (4) ax + by – (a² + b²) = 0

\( \frac{x}{a\left(b^2-a^2\right)-(-2 a b) \times x(-b)}=\frac{y}{(-2 a b) \times a-b\left(b^2-a^2\right)}=\frac{1}{b \times(-b)-a \times a} \)

 

Or, \( \frac{x}{a b^2-a^3-2 a b^2}=\frac{y}{-2 a^2 b-b^3+a^2 b}=\frac{1}{-b^2-a^2} \)

Or, \( \frac{x}{-a^3-a b^2}=\frac{y}{-a^2 b-b^3}=\frac{1}{-a^2-b^2} \)

Or, \( \frac{x}{-a\left(a^2+b^2\right)}=\frac{y}{-b\left(a^2+b^2\right)}=\frac{1}{-\left(a^2+b^2\right)} \)

Or,\( \frac{x}{a\left(a^2+b^2\right)}=\frac{y}{b\left(a^2+b^2\right)}=\frac{1}{\left(a^2+b^2\right)} \)

\( \frac{x}{a\left(a^2+b^2\right)}=\frac{1}{a^2+b^2} \text { and } \frac{y}{b\left(a^2+b^2\right)}=\frac{1}{a^2+b^2} \)

Class 9 Maths WB Board

Or, \( x=\frac{a\left(a^2+b^2\right)}{\left(a^2+b^2\right)} \text { and } y=\frac{b\left(a^2+b^2\right)}{\left(a^2+b^2\right)} \)

X=4

Y= b

Question 13. ax + by = 1, \( b x+a y=\frac{2 a b}{a^2+b^2} \)

Solution: ax + by = 1 ….(1)

\( b x+a y=\frac{2 a b}{a^2+b^2} \)…..(2)

 

From equation (1) ax + by – 1 = 0
From equation (2) b(a²+ b²)x + a(a²+ b²)y – 2ab = 0

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

Class 9 Maths WB Board

Or,\( \frac{x}{-2 a b^2+a^3+a b^2}=\frac{y}{-a^2 b-b^3+2 a^2 b}=\frac{1}{a^4+a^2 b^2-a^2 b^2-b^4} \)

Or,\( \frac{x}{a\left(a^2-b^2\right)}=\frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

Or,\( \frac{x}{a\left(a^2-b^2\right)}=\frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

\( \frac{x}{a\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \text { and } \frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

 

Or, \( x=\frac{a\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \text { and } y=\frac{b\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

Or,\( x=\frac{a}{a^2+b^2} \text { and } y=\frac{b}{a^2+b^2} \)

\( \begin{aligned}
& therefore x=\frac{a}{a^2+b^2} \\
& y=\frac{b}{a^2+b^2}
\end{aligned} \)

 

 

Class IX Maths Solutions WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.5

Question 1. Let us express the value of the variable x from the equation \(\frac{2}{x}+\frac{3}{y}=1\) in terms of the variable y.

Solution: \(\frac{2}{x}+\frac{3}{y}=1\)

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

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or, (y-3)x = 2y

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

Question 2. Let us write the value of x by putting \(\frac{7-4 x}{-5}\) -4x -5 instead of y in the equation 2x + 3y = 9.

Solution: 2x+3y=9

\(or, \quad 2 x+\frac{3(7-4 x)}{-5}=9
or, \quad 2 x-\frac{3(7-4 x)}{5}=9
or, \quad \frac{10 x-21+12 x}{5}=9
or, \quad 22 x-21=45\)

Class IX Maths Solutions WBBSE

or, 22x = 45 +21
or,22x=66

or, \(x=\frac{66}{22}\)

∴ x = 3.

Question 3. Let us solve the following equations in two variables by substitution method and check them graphically.

1. 3x – y = 7, 2x + 4y = 0

Solutioin: 3x – y = 7 …(1)
2x + 4y = 0…(2)

From equation (1)-y=-3x+7
or, y = 3x-7 …..(3)

Putting the value of y in equation (2),
2x + 4(3x-7)= 0

or, 2x+12x-28=0
or, 14x= 28

or, \(x=\frac{28}{14}\)

x = 2

Putting the value of y in equation (4),
y=3×2-7

or, y = 6-7
or,y = -1

∴ x = 2
y=-1

From equation (3)
y = 3x-7

x	2	4	-1
y	-1	5	-10

 

From equation (2) 4y = -2x

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

x	2	6	-2
y	-1	-3	1

 

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

Solution:

\(\frac{x}{2}+\frac{y}{3}=2\) …(1)

\(\frac{x}{4}+\frac{y}{2}= \) …(2)

Class IX Maths Solutions WBBSE

From equation (1) \(\frac{x}{2}=2-\frac{y}{3}\)

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

or, \(x=\frac{12-2 y}{3}\) …(3)

Putting the value of y in equation (4),

\(\frac{\frac{12-2 y}{3}}{4}+\frac{y}{2}=2\)

 

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

or, \(\frac{6-y}{6}+\frac{y}{2}=2\)

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

or, 6+ 2y = 12
or, 2y=12-6

or, y = 6/2

∴ y = 3

Putting the value of y in equation (4),

\(x=\frac{12-2 \times 3}{3}\)

 

or, \(x=\frac{12-6}{3}\)

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

∴ x = 2, y=2

⇒ y = 3

From equation (3)

\(x=\frac{12-2 y}{3}\)

 

x	4	2	-6
y	0	3	-3

 

From equation (2)

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

 

or, x = 8- 2y

x	4	0	12	2
y	2	4	-2	3

 

Class IX Maths Solutions WBBSE

Question 4. Let us solve the following equations in two variables by substitution method and check whether the solutions satisfy the equations:

1. \(2 x+\frac{3}{y}=1,5 x-\frac{2}{y}=\frac{11}{12}\) 

Solution:

\(2 x+\frac{3}{y}=1\) …(1)

\(5 x-\frac{2}{y}=\frac{11}{12}\) …..(2)

From equation (1) \(\frac{3}{y}=1-2 x\)

or, (1-2x)y = 3

or, \(y=\frac{3}{1-2 x}\)

Putting the value of y in equation (2),

\(5 x-\frac{2(1-2 x)}{3}=\frac{11}{12}\)

 

or, \(\frac{15 x-2+4 x}{3}=\frac{11}{12}\)

or, \(\quad 19 x-2=\frac{11}{4}\)

or, \(19 x=\frac{11}{4}+2\)

or, \(19 x=\frac{11+8}{4}\)

or, \(x=\frac{19}{19 \times 4}\)

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

Putting the value of x in equation (3)

\(y=\frac{3}{1-2 \times \frac{1}{4}}\)

 

or, \(y=\frac{3}{1-\frac{1}{2}}\)

or, y = 6

⇒ \(x=\frac{1}{4}\)

y = 6

From equation (1) L. H.S. = \(2 x+\frac{3}{y}\)

\(\begin{aligned}
& =2 \times \frac{1}{4}+\frac{3}{6} \\
& =\frac{1}{2}+\frac{1}{2} \\
& =\frac{1+1}{2} \\
& =\frac{2}{2}
\end{aligned}\)

Class IX Maths Solutions WBBSE

= 1
= R. H. S.

Again from equation (2) L. H. S. = \(5 x-\frac{2}{y}\)

\(\begin{aligned}
& =5 \times \frac{1}{4}-\frac{2}{6} \\
& =\frac{5}{4}-\frac{1}{3} \\
& =\frac{15-4}{12}
\end{aligned}\)

 

= \(\frac{11}{12}\)
= R. H. S.

∴ \(x=\frac{1}{4}\) & y = 6 satify the equations (1) & (2).

 

2. \(\frac{2}{x}+\frac{3}{y}=2, \frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\)

Solution:

\(\frac{2}{x}+\frac{3}{y}=2\) …(1)

 

\(\frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\)….(2)

From equation (1) \(\frac{3}{y}=2-\frac{2}{x}\)

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

or,(2x-2)y = 3 x

or, \(y=\frac{3 x}{2 x-2}\) ….(3)

Putting the value of x in equation (2),

\(\frac{5}{x}+\frac{10(2 x-2)}{3 x}=\frac{35}{6}\)

Class 9 Mathematics West Bengal Board

or, \(\frac{15+20 x-20}{3 x}=\frac{35}{6}\)

or, \(\frac{20 x-5}{x}=\frac{35}{2}\)

or, 40x-10=35x
or, 40x-35x=10
or, 5x = 10

or, \(x=\frac{10}{5}\)

or, x  = 2

Putting the value of x in equation (3),

or, \(y=\frac{3 \times 2}{2 \times 2-2}\)

or, \(y=\frac{6}{4-2}\)

or, \(y=\frac{6}{2}\)

∴ y = 3, x = 2

From equation (1) L. H.S. = \(\frac{2}{x}+\frac{3}{y}\)

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

= 1 + 1 = 2
= R. H. S.

Again, from equation (2) L. H. S. \(\frac{5}{x}+\frac{10}{y}\)

\(\begin{aligned}
& =\frac{5}{2}+\frac{10}{3} \\
& =\frac{15+20}{6} \\
& =\frac{35}{6} \\
& =5 \frac{5}{6}
\end{aligned}\)

= R. H. S.

3. \(\frac{x+y}{x y}=3, \frac{x-y}{x y}=1\)

Solution: \(\frac{x+y}{x y}=3\) …(1)

\(\frac{x-y}{x y}=1\)   ….(2)

From equation (1) \(\frac{x}{x y}+\frac{y}{x y}=3\)

\(or, \frac{1}{y}+\frac{1}{x}=3
or, \frac{1}{y}=3-\frac{1}{x}
or, \quad \frac{1}{y}=\frac{3 x-1}{x}\)

Class 9 Mathematics West Bengal Board

or, (3x-1)y = x

or, \(y=\frac{x}{3 x-1}\) ….(3)

From equation (1) \(\frac{x}{x y}-\frac{y}{x y}=1\)

or, \(\frac{1}{y}-\frac{1}{x}=1\)  …..(4)

Putting the value \(y=\frac{x}{3 x-1}\) in equation (4)

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

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

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

or, 3x-2=x
or,3x-x=2
or,2x = 2

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

or, X = 1

Putting the value of x in equation (3),

or, \(y=\frac{1}{3 \times 1-1}\)

or, \(y=\frac{1}{3-1}\)

or, \(y=\frac{1}{2}\)

∴ x =1

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

From equation (1) L. H.S. = \(\frac{x+y}{x y}\)

= \(=\frac{1+\frac{1}{2}}{1 \times \frac{1}{2}}\)

\(\begin{aligned}
& \frac{2+1}{2} \\
= & \frac{1}{2} \\
= & \frac{3}{2} \times \frac{2}{1}
\end{aligned}\)

Class 9 Mathematics West Bengal Board

= 3
= R. H. S.

Again, from equation (2) L. H. S. = \(\frac{x-y}{x y}\)

\(\begin{aligned}
& =\frac{1-\frac{1}{2}}{1 \times \frac{1}{2}} \\
& =\frac{\frac{2-1}{2}}{\frac{1}{2}} \\
& =\frac{1}{2} \times \frac{2}{1}
\end{aligned}\)

= 1
= R. H. S.

4.  \(\frac{x+y}{x-y}=\frac{7}{3}, \quad x+y=\frac{7}{10}\)

Solution: \(\frac{x+y}{x-y}=\frac{7}{3}\)   ….(1)

\(x+y=\frac{7}{10}\) ……(2)

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

or, 7x-7y=3x+3y
or, 7x-3x=7y+ 3y
or, 4x = 10y

or, \(x=\frac{10 y}{4}\)

or, \(x=\frac{5 y}{2}\)  …..(3)

Putting the value of y in equation (2),

\(or,\frac{5 y}{2}+y=\frac{7}{10}
or,\quad\frac{5 y+2 y}{2}=\frac{7}{10}
or,\frac{7 y}{1}=\frac{7}{5}
or,\quad y=\frac{7}{5 \times 7}
or, \quad y=\frac{1}{5}\)

Class 9 Math Chapter 5 WBBSE

Putting the value of y in equation (2),

\(or, \quad x+\frac{1}{5}=\frac{7}{10}
or,\quad x=\frac{7}{10}-\frac{1}{5}
or,\quad x=\frac{7-2}{10}
or,\quad x=\frac{5}{10}
or,\quad x=\frac{1}{2}
therefore  x=\frac{1}{2}
y=\frac{1}{5}\)

 

From equation (1) L. H.S.  \(=\frac{x+y}{x-y}\)

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

= \(=\frac{\frac{7}{10}}{\frac{3}{10}}\)

= \(=\frac{7}{3}\)

= R. H. S.

Again, from equation (1) L. H. S. = x + y

\(\begin{aligned}
& =\frac{1}{2}+\frac{1}{5} \\
& =\frac{5+2}{10} \\
& =\frac{7}{10}
\end{aligned}\)

= R. H. S.

Question 5. Let us solve the following equations in two variables by substitution method:

1. 2(x-y) = 3, 5x + 8y = 14

Solution: 2(x-y)=3 ….(1)
5x + 8y = 14  …..(2)

From equation (1) 2x-2y=3
or, 2x = 3 + 2y

or, \(x=\frac{3+2 y}{2}\)  ….(3)

Putting the value of x in equation (2),

\(\frac{5(3+2 y)}{2}+8 y=14\)

Class 9 Math Chapter 5 WBBSE

or, \(\frac{15+10 y+16 y}{2}=14\)

or, 15+26y=28
or, 26y=28-15
or, 26y = 13

\(or, \quad y=\frac{13}{26}
or, \quad y=\frac{1}{2}\)

Putting the value of y in equation (3),

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

 

or, \(x=\frac{3+1}{2}\)

or, \(x=\frac{4}{2}\)

or, x =2

∴ x =2 , y= 1/2

2. \(2 x+\frac{3}{y}=5,5 x-\frac{2}{y}=3\)

Solution:

\(2 x+\frac{3}{y}=5 \) …..(1)

\(5 x-\frac{2}{y}=3\)   …..(2)

Class 9 Math Chapter 5 WBBSE

From equation (1) \(2 x=5-\frac{3}{y}\)

or, \(2 x=\frac{5 y-3}{y}\)

or, \(x=\frac{5 y-3}{2 y}\) ….(3)

Putting the value of x in equation (2),

\(\frac{5(5 y-3)}{2 y}-\frac{2}{y}=3\)

 

or, \(\frac{25 y-15-4}{2 y}=3\)

or, 25y-19 = 6y
or,25y-6y 19
or,19y= 19

or, y = \(\frac{19}{19}\)

or, y = 1

Putting the value of y in equation (3),

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

 

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

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

or, x = 1

∴ x = , y = 1

3. \(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=1\)

Solution: \(\frac{x}{2}+\frac{y}{3}=1\) ….(1)

\(\frac{x}{3}+\frac{y}{2}=1\) ….(2)

 

From equation (1) \(\frac{x}{2}=1-\frac{y}{3}\)

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

or, 3x = 6-2y

or, \(x=\frac{6-2 y}{3}\) …(3)

Putting the value of y in equation (3),

\(\frac{\frac{6-2 y}{3}}{3}+\frac{y}{2}=1\)

Class 9 Math Chapter 5 WBBSE

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

or, \(\frac{12-4 y+9 y}{18}=1\)

or, 12 + 5y = 18
or, 5y = 18-12

or, \(y=\frac{6}{5}\)

Putting the value of y in equation (3),

\(x=\frac{6-2 \times \frac{6}{5}}{3}\)

 

\(or, x=\frac{\frac{30-12}{5}}{3}
or, \quad x=\frac{18}{5} \times \frac{1}{3}
or, \quad x=\frac{6}{5}\)

 

∴ \(x=\frac{6}{5}\)

∴ \(y=\frac{6}{5}\)

4. \(\frac{x}{3}=\frac{y}{4}\), 7x-5y = 2

Solution: \(\frac{x}{3}=\frac{y}{4}\) ….(1)

7x-5y – 2 ….(2)

From equation(1) 4x = 3y

or, \(x=\frac{3 y}{4}\) …..(3)

Putting the value of x in equation. (2),

\(7 \times \frac{3 y}{4}-5 y=2\)

 

or, \(\frac{21 y-20 y}{4}=2\)

or, y = 8

Putting the value of y in equation (3),

\(x=\frac{3 \times 8}{4}\)

Class 9 Math Solution WBBSE In English

∴ x = 6, y = 8

5. \(\frac{2}{x}+\frac{5}{y}=1, \frac{3}{x}+\frac{2}{y}=\frac{19}{20}\)

Solution: \(\frac{2}{x}+\frac{5}{y}=1\) …(1)

\(\frac{3}{x}+\frac{2}{y}=\frac{19}{20}\)……..(2)

 

From equation (1) \(\frac{2}{x}=1-\frac{5}{y}\)

or, \(\frac{2}{x}=\frac{y-5}{y}\)

or, (y-5)x = 2y

or, \(x=\frac{2 y}{y-5}\) ….(3)

Putting the value of x in equation (2),

\(\frac{3}{\frac{2 y}{y-5}}+\frac{2}{y}=\frac{19}{20}\)

 

or, \(\frac{3(y-5)}{2 y}+\frac{2}{y}=\frac{19}{20}\)

or, \(\frac{3 y-15+4}{2 y}=\frac{19}{20}\)

or, \(\frac{3 y-11}{y}=\frac{19}{10}\)

or, 30y-110 = 19y
or,30y-19y=110
or,11y=110

or, \(y=\frac{110}{11}\)

or,y = 10

Putting the value of y in equation (3),

\(x=\frac{2 \times 10}{10-5}\)

Class 9 Math Solution WBBSE In English

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

or,X = 4

∴ X=4
y = 10

6. \(\frac{1}{3}\)(x − y) = \(\frac{1}{3}\) (y-1), \(\frac{1}{3}\)(4x-5y) = x – 7

Solution:  \(\frac{1}{3}\)(x − y) = \(\frac{1}{3}\) (y-1) ….(1)

\(\frac{1}{3}\)(4x-5y) = x – 7 …..(2)

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

or, 4x-4y=3y-3
or, 4x=3y – 3+ 4y
or, 4x = 7y-3

or, \(x=\frac{7 y-3}{4}\)  ….(3)

Putting the value of x in equation (2),

\(\frac{1}{7}\left\{4 \frac{(7 y-3)}{4}-5 y\right\}=\frac{7 y-3}{4}-7\)

Class 9 Math Solution WBBSE In English

or, \(\frac{1}{7}(7 y-3-5 y)=\frac{7 y-3-28}{4}\)

or, \(\frac{2 y-3}{7}=\frac{7 y-31}{4}\)

or, 49y-217 = 8y-12
or, 49y-8y= -12+217
or, 41y=205

or, \(y=\frac{205}{41}\)

or, y = 5

Putting the value of y in equation (3),

\(x=\frac{7 \times 5-3}{4}\)

 

\(or, x=\frac{35-3}{4} or, x=\frac{32}{4}\)

 

or, x =8

∴ x = 8, y =5

7. \(\frac{x}{14}+\frac{y}{18}=1, \frac{x+y}{2}+\frac{3 x-5 y}{4}=2\)

Solution:

\(\frac{x}{14}+\frac{y}{18}=1\) ….(1)

\(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\) ….(2)

Class 9 Math Solution WBBSE In English

From equation, (1) \(\frac{x}{14}+\frac{y}{18}=1\)

or, \(\frac{9 x+7 y}{126}=1[/latex

or, 9x + 7y = 126
or, 9x=126-7y

or, [latex]x=\frac{126-7 y}{9}\)   ….(3)

From equation (1) \(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\)

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

or, 5x – 3y = 8 (4)

From equation (4) \(x=\frac{126-7 y}{9}\)

\(\frac{5(126-7 y)}{9}-3 y=8\)

 

or, \(\frac{630-35 y-27 y}{9}=8\)

or, – 62y + 630 = 72
or,-62y=72-630
or, -62y=-558
or, 62y=558

or, \(y=\frac{558}{62}\)

or, y=9

Putting the value of y in equation (3),

\(x=\frac{126-7 \times 9}{9}\)

Class 9 Math Solution WBBSE In English

\(or, \quad x=\frac{126-63}{9}
or, x=\frac{63}{9}
or, \quad x=7\)

∴ x = 7, y=9

8. p (x+y)= q(xy) =2pq

Solution:
P(x+y)=2pq ….(1)
q (x – y) = 2pq…..(2)

From equation (1) p (x+y)=2pq

or, \(x+y=\frac{2 p q}{p}\)

or, x + y = 2q
or, x = 2q-y ….(3)

Putting the value of x in equation(2)
q(2q – y-y) = 2pq

or, \(2 q-2 y=\frac{2 p q}{q}\)

or,-2y=2p-2q

or, \(y=\frac{2(p-q)}{-2}\)

or,y=q-p

Putting the value of y in equation (3),
x = 2q – (q.-p)
or, x=2q-q+p
or, x=p+q
=x=p+q
y=q-p

 

Class 9 Math Chapter 5 WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.4

Question 1. Let us express the variable x of the equation \(\frac{x}{3}+\frac{y}{2}=8\) in term of variable y.

Solution: \(\frac{x}{3}+\frac{y}{2}=8\)

\(or, \quad \frac{x}{3}=8-\frac{y}{2}
or, \frac{x}{3}=\frac{16-y}{2}
or, x=\frac{3}{2}(16-y)\)

 

Question 2. \(\frac{2}{x}+\frac{7}{y}=1\)=1 Express the value of y in terms of y.

Solution: \(\frac{2}{x}+\frac{7}{y}=1\)

\(or,\quad \frac{7}{y}=1-\frac{2}{x}
or,\quad \frac{7}{y}=\frac{x-2}{x}
or,\quad(x-2) y=7 x
or,\quad y=\frac{7 x}{x-2}\)

Question 3. Let us solve the following equations by comparison method and check whether the solutions satisfy the equations.

1. 2(x − y) = 3, 5x + 8y = 14

Solution: 2(x-y)=3 ….(1)
5x + 8y = 14 ….(2)

From equation (1) 2x-2y=3
or,2x = 3 + 2y

or, \(x=\frac{3+2 y}{2}\) ….(3)

From equation (2) 5x+8y = 14
or, 5x = 14-8y

or, \(x=\frac{3+2 y}{2}\) ….(4)

Comparing the value of x from equations (3) & (4),

\(\frac{3+2 y}{2}=\frac{14-8 y}{5}\)

Class 9 Math Chapter 5 WBBSE

or, 15+10y=28-16y
or, 10y+ 16y=28-15
or, 26y=13

or, \(y=\frac{13}{26}\)

or, \(y=\frac{1}{2}\)

Putting the value of y in equation (3)

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

 

\(\begin{aligned}
& x=\frac{3+1}{2} \\
& x=\frac{4}{2} \\
& x=2 \\
& x=2
\end{aligned}\)

Class 9 Math Chapter 5 WBBSE

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

From equation (1) L. H.S. = 2(x-y)

\(\begin{aligned}
& =2\left(2-\frac{1}{2}\right) \\
& =2\left(\frac{4-1}{2}\right) \\
& =3
\end{aligned}\)

 

= R. H. S.

Again, from equation (1), L. H. S.= 5x + 8y

= \(5 \times 2+8 \times \frac{1}{2}\)

= 10 +4
= 14
= R. H. S.

2.  \(2 x+\frac{3}{y}=5,5 x-\frac{2}{y}=3\)

Solution: \(2 x+\frac{3}{y}=5\) ….(1)

\(5 x-\frac{2}{y}=3\) …(2)

From equation (1) \(\frac{3}{y}=5-2 x\)

or, \(\frac{1}{y}=\frac{5-2 x}{3}\) …..(3)

From equation (2) \(\frac{-2}{y}=3-5 x\)

or, \(\frac{2}{y}=5 x-3\)

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

Comparing the value of \(\frac{1}{y}\) from equations (3)& (4) we get,

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

or, 15x –  9 = 10-4x
or, 15x+4x= 10 +9
or, 19x= 19

or, x = 19/19
or,  = 1

Putting the value of x in equation (3)

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

or, \(\frac{1}{y}=\frac{3}{3}\)

or, \(\frac{1}{y}=1\)

or, y = 1
X = 1

From equation (1)L. H.S. = \(2 x+\frac{3}{y}\)

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

=2+3
= 5
= R. H. S.

From equation (2) L. H. S.= \(5 x-\frac{2}{y}\)

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

=5-2
= 3
= R. H. S.

3. \(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=1\)

Solution:

\(\frac{x}{2}+\frac{y}{3}=1\) …(1)

\(\frac{x}{3}+\frac{y}{2}=1\) …(2)

From equation (1) \(\frac{x}{2}=1-\frac{y}{3}\)

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

or, \(x=\frac{3(2-y)}{2}\) ….(3)

From equation(2) \(\frac{x}{3}=1-\frac{y}{2}\)

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

or, \(x=\frac{3(2-y)}{2}\)….(4)

Comparing the value of x in equations (3) & (4),

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

Class 9 Math Chapter 5 WBBSE

or, 4(3-y) 9 (2-y)
or, 12-4y=18-9y
or, 9y-4y=18-12
or, 5y = 6

or, \(y=\frac{6}{5}\)

Putting the value of y in equation (3) we get,

\(x=\frac{2\left(3-\frac{6}{5}\right)}{3}\)

or, [/latex]x=\frac{2\left(\frac{15-6}{5}\right)}{3}
or, \quad x=2 \times \frac{9}{5} \times \frac{1^{-}}{3}[/latex]

or, \(\begin{aligned}
& x=\frac{6}{5} \\
& =x=\frac{6}{5}
\end{aligned}
y=\frac{6}{5}
\)

Class IX Maths Solutions WBBSE

From equation (1 ) L. H.S. =\(\frac{x}{2}+\frac{y}{3}\)

\(\begin{aligned}
& =\frac{\frac{6}{5}}{2}+\frac{\frac{6}{5}}{3} \\
& =\frac{6}{5} \times \frac{1}{2}+\frac{6}{5} \times \frac{1}{3} \\
& =\frac{3}{5}+\frac{2}{5} \\
& =\frac{5}{5} \\
& =1
\end{aligned}\)

 

= R. H. S.

From equation (2) L. H. S. = \(\frac{x}{3}+\frac{y}{2}\)

\(\begin{aligned}
& =\frac{\frac{6}{5}}{3}+\frac{\frac{6}{5}}{2} \\
& =\frac{6}{5} \times \frac{1}{3}+\frac{6}{5} \times \frac{1}{2} \\
& =\frac{2}{5}+\frac{3}{5} \\
& =\frac{2+3}{5} \\
& =\frac{5}{5}
\end{aligned}\)

Class IX Maths Solutions WBBSE

= 1
= R. H. S.

4. 4x-3y = 18, 4y – 5x = -7

Solution:
4x-3y = 18 …(1)
4y – 5x = -7 ….(2)

From equation (1) 4x = 18+ 3y

or, \(x=\frac{18+3 y}{4}\) ….(3)

From equation (2) -5x=-4y-7.
or, 5x = 4y+ 7

or, \(x=\frac{4 y+7}{5}\) ….(4)

Comparing the value of ‘x’ from equations (3) & (4),

or, \(\frac{18+3 y}{4}=\frac{4 y+7}{5}\)

or,16y+28 = 90+ 15y
or,16y 15y = 90-28
or, y = 62

Putting the value of y, we get

\(x=\frac{18+3 \times 62}{4}\)

 

\(or, \quad x=\frac{18+186}{4}
or, x=\frac{204}{4}
or, \quad x=51\)

x =51,y= 62

From equation (1) L. H.S. = 4x-3y
= 4 x 51-3 x 62
= 204
= 18
=R. H. S.

From equation (2) L. H. S.= 4y-5x
= 4 x 62-5 x 51
=248-255
=-7
= R. H. S.

Class IX Maths Solutions WBBSE Question 4. Let us solve the equations 2x + y = 8 and 2y-3x=-5 by comparison method and justify them by solving graphically.

Solution:

2x + y = 8 ….(1)
2y-3x=-5…(2)

Equation y = 8-2x …(3)

Equation 2y=3x-5

or, \(y=\frac{3 x-5}{2}\) …(4)

From equations (3) & (4),

\(8-2 x=\frac{3 x-5}{2}\)

or, 3x-5=16-4x
or, 3x+4x=16+5
or, 7x=21

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

or, x = 3

Putting this value of x in equation (3),
y=8-2×3

or,y = 2
x = 3

⇒ y = 8-2x

x	0	2	-1
y	8	4	10

 

Class IX Maths Solutions WBBSE

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

 

x	1	3	-7
y	-1	2	-13

 

Question 5. Let us solve the following equations in two variables by comparison method:

1. 3x-2y= 2, 7x + 3y = 43

Solution:
3x-2y=2 …(1)
7x + 3y = 43 …(2)

From equation (1) 3x=2+2y

or, \(x=\frac{2+2 y}{3}\) …(3)

From equation (2) 7x=43-3y

or, \(x=\frac{43-3 y}{7}\) ….(4)

Comparing the value of x from equations (3) & (4)

\(\frac{2+2 y}{3}=\frac{43-3}{7}\)

 

or,14+ 14y = 129-9y
or, 14y+9y=129-14
or, 23y=115

or, \(y=\frac{115}{23}\)

or,y = 5

Putting the value of y in equation (3),

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

 

\(x=\frac{12}{3}\)

X = 4

∴x=4&y=5

2. 2x-3y=8, \(\frac{x+y}{x-y}=\frac{7}{3}\)

Solution: 2x-3y=8 …(1)

\(\frac{x+y}{x-y}=\frac{7}{3}\)….(2)

Class IX Maths Solutions WBBSE

From equation (1) 2x=8+3y

or, \(x=\frac{8+3 y}{2}\) ….(3)

From equation (2) 7x-7y = 3x + 3y
or, 7x-3x=3y+ 7y.
or, 4x = 10y

or, \(x=\frac{10 y}{4}\)

or, \(x=\frac{5 y}{2}\) …(4)

Companing the value of x from equations (3) & (4),

\(\frac{8+3 y}{2}=\frac{5 y}{2}\)

 

or, 8+3y = 5y
or, 8+ 3y = 5y
or, 3y-5y=-8
or,-2y=-8

or, \(y=\frac{-8}{-2}\)

y = 4

Putting the value of y in equation (4),

\(x=\frac{5 \times 4}{2}\)

or, X = 10

∴ X = 10
y = 4

3. \(\frac{1}{3}(x-y)=\frac{1}{4}(y-1), \frac{1}{7}(4 x-5 y)=x-7\)

Solution:

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

 

\(\frac{1}{7}(4 x-5 y)=x-7\) …(2)

Class 9 Mathematics West Bengal Board

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

or,4x-4y=3y-3
or,4x 3y+4y-3

or, \(x=\frac{7 y-3}{4}\) ….(3)

From equation (2) 7(x-7)=4x-5y
or, 7x-49=4x-5y
or, 7x-4x=49-5y
or, 3x=49-5y

or, \(x=\frac{49-5 y}{3}\) ……(4)

Comparing the value of x from equations (3) & (4),

\(\frac{7 y-3}{4}=\frac{49-5 y}{3}\)

 

or, 21y-9-196-20y
or, 21y+20y 196 +9
or, 41y=205

or, \(y=\frac{205}{4}\)

or, y = 5

Putting the value of y in equation (3),

\(x=\frac{7 \times 5-3}{4}\)

Class 9 Mathematics West Bengal Board

\(or, x=\frac{35-3}{4}
or, \quad x=\frac{32}{4}
or, x=8 \)

Required solution x = 8,y = 5

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

Solution:

\(\frac{x+1}{y+1}=\frac{4}{5}\) …….(1)

 

\(\frac{x-5}{y-5}=\frac{1}{2}\) ….(2)

From equation (1) 5x+5=4y+4
or, 5x = 4y+ 4-5
or, 5x= 4y-1

or, \(x=\frac{4 y-1}{5}\) ….(3)

From equation (2) 2x-10-y-5
or, 2x=y-5+10
or, 2x = y +5

or, \(x=\frac{y+5}{2}\) …(4)

Comparing the value of x from equations (2) & (4),

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

or, 8y-2=5y+25
or, 8y – 5y = 25+ 2
or, 3y=27

or, \(y=\frac{27}{3}\)

or,y=9

Putting the value of y in equation (4),

\(\begin{aligned}
&x=\frac{9+5}{2}\\
&x=\frac{14}{2}
\end{aligned}\)

Class 9 Mathematics West Bengal Board

x =7,y =9

5.  x + y = 11, y+2= \(\frac{1}{8}\) (10y+x)

Solution: x + y = 11……(1)

y+2= \(\frac{1}{8}\) (10y+x) …..(2)

From equation (1) x = 11-y ….(3)

From equation (2) \(\frac{(y+2)}{1}=\frac{(10 y+x)}{8}\)

or, 10y + x = 8y + 16
or, x = 8y + 16-10y
or, x = -2y+16 ….(4)

Comparing the value of x from equations (3) & (4),
11 – y = -2y+ 16
or, -y+2y= 16-11
or,y = 5

Putting the value of y in equation (3),
X=11-5
or,X = 6

∴ X = 6
y = 5

6. \(\frac{x}{3}+\frac{y}{4}=1\), 2x + 4y = 11

Solution: \(\frac{x}{3}+\frac{y}{4}=1\)…..(1)

2x + 4y = 11 …(2)

From equation (1) \(\frac{x}{3}+\frac{y}{4}=1\),

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

or, 4x + 3y = 12
or, 4x=12-3y

or, \(x=\frac{12-3 y}{4}\)…(3)

From equation (2) 2x = 11-4y

or, \(x=\frac{11-4 y}{2}\) ….(4)

Comparing the value of x from equations (3) & (4),

or, \(\frac{12-3 y}{4}=\frac{11-4 y}{2}\)

or, 24 – 6y = 44 – 16y
or, 16y – 6y = 44-24
or,10y=20

or, \(y=\frac{20}{10}\)

∴ y = 2

\(x=\frac{12-3 \times 2}{4}\)

Class 9 Mathematics West Bengal Board

or, \(x=\frac{6}{4}\)

∴ \(x=\frac{3}{2}\), y = 2

7. \(x+\frac{2}{y}=7,2 x-\frac{6}{y}=9\)

Solution:

\(x+\frac{2}{y}=7\) …..(1)

\(2 x-\frac{6}{y}=9\) …(2)

From equation (1) \(x=7-\frac{2}{y}\)

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

From equation (2) \(2 x=9+\frac{6}{y}\)

or, \(2 x=\frac{9 y+6}{y}\)

or, \(x=\frac{9 y+6}{2 y}\) …..(4)

Comparing the value of x from equations (3) & (4),

\(\begin{aligned}
& \frac{7 y-2}{y}=\frac{9 y+6}{2 y} \\
& \frac{7 y-2}{1}=\frac{9 y+6}{2}
\end{aligned}\)

 

or, 14y- 4= 9y+6
or,14y-9y=6+4
or, 5y = 10

or, \(y=\frac{10}{5}\),

∴ y = 2

Putting the value of y in equation (3),

\(\begin{aligned}
& x=\frac{7 \times 2-2}{2} \\
& x=\frac{14-2}{2} \\
& x=\frac{12}{2}
\end{aligned}\)

Class 9 Maths WB Board

8. \(\frac{1}{x}+\frac{1}{y}=\frac{5}{6}, \frac{1}{x}-\frac{1}{y}=\frac{1}{6}\)

Solution: \(\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\)…(1)

 

\(\frac{1}{x}-\frac{1}{y}=\frac{1}{6}\) ….(2)

 

From equation (1) \(\frac{1}{x}=\frac{5}{6}-\frac{1}{y}\) …(3)

From equation (2) \(\frac{1}{x}=\frac{1}{6}+\frac{1}{y}\) …(4)

Comparing the value of x from equations (3) & (4),

\(\begin{aligned}
& \frac{5}{6}-\frac{1}{y}=\frac{1}{6}+\frac{1}{y} \\
& \frac{5}{6}-\frac{1}{6}=\frac{1}{y}+\frac{1}{y} \\
& \frac{5-1}{6}=\frac{1+1}{y} \\
& \frac{4}{6}=\frac{2}{y} \\
& 4 y=12 \\
& y=\frac{12}{4} \\
& y=3
\end{aligned}\)

 

Putting the value of y in equation (3).

\(or, \frac{1}{x}=\frac{5-2}{6}
or, \frac{1}{x}=\frac{3}{6}
or, \quad 3 x=6
or, \quad x=\frac{6}{3}
or, \quad x=2
therefore   x=2 \)

Class 9 Maths WB Board

9. \(\frac{x+y}{x y}=2, \frac{x-y}{x y}=1\)

Solution: \(\frac{x+y}{x y}=2\) …(1)

\(\frac{x-y}{x y} = 1\) …(2)

From equation (1) \(\frac{x}{x y}+\frac{y}{x y}=2\)

or,\(\frac{1}{y}+\frac{1}{x}=2\)

or, \(\frac{1}{y}=2-\frac{1}{x}\) ….(3)

From equation (2)=\(\frac{x}{x y}-\frac{y}{x y}=1\)

or, \(\frac{1}{y}-\frac{1}{x}=1\)

or, \(\frac{1}{y}=1+\frac{1}{x}\)

Comparing the value of x from equations (3) & (4),

\(or, \quad 2-\frac{1}{x}=1+\frac{1}{x}
or, \quad 2-1=\frac{1}{x}+\frac{1}{x}
or, \quad 1=\frac{1+1}{x}
or, 1=\frac{2}{x}
or, \quad x=2\)

 

Putting the value of y in equation (3),

\(or, \frac{1}{y}=\frac{4-1}{2}
or, \quad \frac{1}{y}=\frac{3}{2}
or, 3 y=2
or, \quad y=\frac{2}{3}\)

Class 9 Maths WB Board

∴ x =2, \(y=\frac{2}{3}\)

10. \(\frac{x+y}{5}+\frac{x-y}{4}=5, \frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5}\)

Solution: \(\frac{x+y}{5}+\frac{x-y}{4}=5\) …(1)

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

 

From equation (1)=\(\frac{x+y}{5}+\frac{x-y}{4}=5\)

\(or, \quad \frac{4 x+4 y+5 x-5 y}{20}=5
or, \quad 9 x-v=100
or, \quad 9 x=100+y\)

 

or, \(x=\frac{100+y}{9}\) …(3)

From equation (2) \(\frac{x+y}{4}+\frac{x-y}{5}=\frac{29}{5}\)

\(or, \quad \frac{5 x+5 y+4 x-4 y}{20}=\frac{29}{5}
or, \quad 9 x+y=\frac{29}{5} \times 20
or, \quad 9 x+y=116
or, \quad 9 x=116-y\)

\(x=\frac{16-y}{9}\)….(4)

Class 9 Maths WB Board

Comparing the value of x from equations (3) & (4),

\(or, \quad \frac{100+y}{9}=\frac{116-y}{9}
or, \quad 100+y=116-y
or, y+y=116-100
or, 2 y=16
or, y=\frac{16}{2}
or, y=8\)

 

Putting the value of y in equation (3),

\(\begin{aligned}
& x=\frac{100+8}{9} \\
& x=\frac{108}{9}
\end{aligned}\)

 

∴ x = 12
y = 8

11. \(\frac{4}{x}-\frac{y}{2}=-1, \frac{8}{x}+2 y=10\)

Solution: \(\frac{4}{x}-\frac{y}{2}=-1\)….(1)

\(\frac{8}{x}+2y=10\) …(2)

 

From equation (1)\(-\frac{y}{2}=-\frac{4}{x}-1\)

or, \(\frac{y}{2}=\frac{4}{x}+1\)

\(y=2\left(\frac{4}{x}+1\right)\)…(3)

 

From equation (2) 2y = \(=10-\frac{8}{x}\)

or, \(y=\frac{10}{2}-\frac{8}{2 x}\)

or, \(y=5-\frac{4}{x}\) …(4)

Comparing the value of x from equations (3) & (4),

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

 

\(or, \frac{8}{x}+2=5-\frac{4}{x}
or, \frac{8}{x}+\frac{4}{x}=5-2\)

 

\(or, \quad \frac{8+4}{x}=3
or, \quad \frac{12}{x}=3\)

 

or, 3x = 12

\(x=\frac{12}{3}\)

 

or, x = 4

Putting the value of y in equation (4),

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

or, y=5-1

y = 4
∴ x= 4 , y = 4

12. 2-2(3x-y)=10(4-y)- 5x = 4(y-x)

Solution: 2- 2(3x-y) = 4(y-x) …..(1)

10(4-y)-5x = 4(y-x) …(2)

From equation (1) 2-6x+2y= 4y – 4x
or, – 6x+4x=4y-2y-2
or, -2x =-2y-2
or, 2x=2-2y

or, \(x=\frac{2-2 y}{2}\) …(3)

From equation (2) 40-10y-5x = 4y – 4x
or, – 5x + 4x = 4y+ 10y – 40
or, -x=14y-40
or, X = 40-14y

Comparing the value of x from equations (3) & (4),

\(\frac{2-2 y}{2}=40-14 y\)

 

or, \(\frac{2}{2}-\frac{2 y}{2}=40-14y\)

or, 1-y=40-14y
or, -y+14y=40-1
or, 13y=39

or, y = \(\frac{39}{13}\)

y = 3

Putting the value of y in equation (4)
x=40-14 x 3

or, X = 40-42

∴ x=-2,y =3

 

Class IX Maths Solutions WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.3

Question 1. Let us solve the following linear equations in two variables by elimination method and check them graphically:

1. 8x+5y-11=0,3x-4y-10=0

Solution: 8x+5y = 11 …(1)
3x-4y=10 …(2)

To eliminate y we multiply equation (1) by 4 and equation (2) by 5; we get

32x+20y-44 =0…(3)

Adding, eq(3)+(2)

Read and Learn More WBBSE Solutions For Class 9 Maths

\(\begin{array}{r}
32 x+20 y-44=0 \\
15 x-20 y-50=0 \\
\hline 47 x-94=0
\end{array}\)

Class 9 Mathematics West Bengal Board

∴ 47x=94

∴ X = \(\frac{94}{47}\) = 2

Putting, x = 2 in equation (1) we get
8×2+5y-11=0
=>5y=11-16
5y=-5

∴ \(y=-\frac{5}{5}=-1\)

∴ The required solution is x = 2 &y=-1.

 

 

2. 2x + 3y – 7 = 0, 3x + 2y -8=0

Solution: 2x + 3y-7=0 …(1)
3x + 2y -8=0 …(2)

To eliminate x, we multiply equation (1) by 3 & equation (2) by 2, we get,

Class 9 Mathematics West Bengal Board

Subtracting we get

\(\begin{gathered}
6 x+9 y-21=0 \\
6 x+4 y-16=0 \\
(-)(-)(+) \\
5 y-5=0
\end{gathered}\)

∴ 5y = 5
∴ y = 1

Putting, y = 1 in equation (1) we get
2x + 3×1=7=0
2x+3-7=0
∴ 2x = 4

∴ \(x=\frac{4}{2}=2\)

The required solution is x = 2 & y = 1.

 

 

Question 2. To eliminate y, what number is multiplied with the equation 7x-5y + 2 = 0 and then added to the equation 2x + 15y+3=0?

Solution:  To eliminate y, 3 is multiplied with the equation.

Class 9 Mathematics West Bengal Board

Question 3. Let us write the least natural number by which we can multiply both the equations and get the equal co-efficients of x.

Solution: 4x-3y = 16 …(1)
6x+5y= 62….(2)

If we multiply equation (1) by 3 & equation (2) by 2, we will get the equal co- efficient of x.

Question 4. Let us solve the following linear equations in two variables by elimination method:

1. 3x+2y= 6, 2x-3y = 17

Solution: 3x + 2y =-6 …(1)
2x-3y = 17 …(2)

To eliminate y, we multiply equation (1) by 3 and equation (2) by 2; we get

9x+6y=18 …(3)

Adding,eq(3)+(2)

\(\begin{aligned}
& 9 x+6 y=18 \\
& 4 x-6 y=34 \\
& \hline 13 x=52
\end{aligned}\)

Class 9 Mathematics West Bengal Board

∴ \(x=\frac{52}{13}\)

Putting, x = 4 in equation (1) we get
3×4+2y=6
2y=-6
∴ 2y=-6

∴ \(y=-\frac{6}{2}=-3\)

The required solution is x = 4 & y = – 3

2. 2x + 3y = 32, 11y-9x=3

Solution: 2x + 3y = 32 …(1)
– 9x+11y = 3 …(2)

To eliminate x, we multiply equation (1) by 9 & equation (2) by 2, we get

18x+27y=288…(3)

Adding,eq(3)+(2)

\(\begin{array}{r}
18 x+27 y=288 \\
-18 x+22 y=\quad 6 \\
\hline 49 y=294
\end{array}\)

Class 9 Maths WB Board

∴ \(y=\frac{294}{49}=6\)

Putting, y 6 in equation (1) we get
2x + 3 x 6 = 32
or, 2x = 32-18
or, 2x = 14

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

∴ X=7.
The required solution is x = 7 & y = 6

3. x + y = 48, x+4= \(\frac{5}{2}(y+4)\)

Solution: x + y = 48 …(1)

x+4=\(\frac{5}{2}(y+4)\)  …(2)

or, 2x+8=5y + 20
or, 2x-5y=20-8
or, 2x-5y = 12 …(3)

To eliminate y, we multiply equation (1) by 5 and equation (2) by 1, we get

5x+5y =240…(4)

Adding eq(4) +(3)

\(\begin{aligned}
& 5 x+5 y=240 \\
& 2 x-5 y=12 \\
& (-)(+) \quad(-) \\
& \hline 7 x=252
\end{aligned}\)

Class 9 Maths WB Board

Putting x = 36 in equation (1) we get
36+ y = 48
∴ y= 48 – 36 = 12

The required solution is x = 36 & y = 12.

4. \(\frac{x}{2}+\frac{y}{3}=8, \quad \frac{5 x}{4}-3 y=-3\)

Solution: \(\frac{x}{2}+\frac{y}{3}=8\)  ….(1)

\(\frac{5 x}{4}-3 y=-3\) …(2)

To eliminate y, we multiply equation (1) by & equation (2) by 1, we get \(\frac{x}{2}+\frac{y}{3}=8\)

\(\begin{aligned}
& \frac{9 x}{2}+3 y=72 \\
& \frac{5 x}{4}-3 y=-3
\end{aligned}\)

Class 9 Maths WB Board

Adding,

\(\frac{9 x}{2}+\frac{5 x}{4}=69\)

 

⇒ \(\frac{18 x+5 x}{4}=69\)

23x = 69 Χ 4 …(3)

∴ \(x=\frac{69 \times 4}{23}=12\)

Putting, x = 12 in equation (1) we get,

⇒ \(\frac{12}{2}+\frac{y}{3}=8\)

∴ \(\frac{y}{3}=8-6=2\)

The required soluition is x = 12 & y = 6.

5. \(3 x-\frac{2}{y}=5, x+\frac{4}{y}=4\)

Solution: \(3 x-\frac{2}{y}=5\) …(1)

\(x+\frac{4}{y}=\) …(2)

To eliminate y we multiply equation no. (1) by 2 & equation no. (2) by 1, we get

\(6 x-\frac{4}{y}=10\) …(3)

Adding, eq (3) + eq(2)

\(
\begin{aligned}
& 6 x-\frac{4}{y}=10 \\
& x+\frac{4}{y}=4
\end{aligned}
7 x=14\)

Class 9 Math Chapter 5 WBBSE

or, x = 14/7
or, x = 2

Putting, x = 2 in equation no. (2) we get

\(
2+\frac{4}{y}=4
or, \frac{4}{y}=2
or, y=\frac{4}{2}
or, y=2\)

 

The required solution x = 2 & y = 2.

6. \(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=1\)

Solution:
\(\frac{x}{2}+\frac{y}{3}=1\) …(1)

\(\frac{x}{3}+\frac{y}{2}=1\) …(2)

The eliminate x, we multiply equation no. (1) by 1/3 & equation no. (2) by 1/2, we get

\(
\begin{aligned}
& \frac{x}{6}+\frac{y}{9}=\frac{1}{3} \\
& \frac{x}{6}+\frac{y}{4}=\frac{1}{2}
\end{aligned}\)

Class 9 Math Chapter 5 WBBSE

Subtracting \(\frac{y}{9}-\frac{y}{4}=\frac{1}{3}-\frac{1}{2}
\)

 

\(\begin{aligned}
& \Rightarrow \frac{4 y-9 y}{36}=\frac{2-3}{6} \\
& \Rightarrow \frac{-5 y}{36}=\frac{-1}{6} \\
& therefore   y=\frac{1}{6} \times \frac{36}{5}=\frac{6}{5}
\end{aligned}\)

 

Putting y = 6/5 in equation no. (1) we get

\(\begin{aligned}
& \Rightarrow \frac{x}{2}+\frac{6}{5} \times \frac{1}{3}=1 \\
& \Rightarrow \frac{x}{2}=1-\frac{2}{5}=\frac{3}{5} \\
& therefore x=\frac{6}{5}
\end{aligned}\)

 

∴ The required solution, x = 6/5 & y = 6/5

7. \(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2, \frac{x}{14}+\frac{y}{18}=1\)

Solution:

\(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\) …(1)

\(\frac{x}{14}+\frac{y}{18}=1\)….(2)

Class 9 Math Chapter 5 WBBSE

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

 

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

or, 5x-3y = 8 …(3)

\(\frac{x}{14}+\frac{y}{18}=1\)

 

or, \(\frac{9 x+7 y}{126}\) = 1

or, 9x + 7y = 126 ….(4)

To eliminate y, we multiply equation no. (1) by 7 & equation no. (2) by 3; we get

\(\begin{aligned}
& 35 x 121 y=56 \\
& 27 x+21 y=378 \\
& \text { Adding, } 62 x=434
\end{aligned}\)

Class 9 Math Chapter 5 WBBSE

or, \(x=\frac{434}{62}\)

∴ x = 7

Putting x = 7 in equation no. (2), we get
9×7+ 7y = 126
⇒126-63 = 63

or, y= \(\frac{63}{7}\)

∴ y = 9

The required solution, x = 7, & y = 9.

8. \(\frac{x y}{x+y}=\frac{1}{5}, \frac{x y}{x-y}=\frac{1}{9}\)

Solution: \(\frac{x y}{x+y}=\frac{1}{5}\) …(1)

⇒ x+y =5x-y

\(\frac{x y}{x-y}=\frac{1}{9}\) ….(2)

⇒ x + y = 9xy

Adding eq (1)  and eq (2) we get,

x + y = 5xy ….(1)
x – y = 9xy ….(2)

2x=14xy

or, 2 = 14y

\(or, \quad \mathrm{y}=\frac{2}{14}
or, y=\frac{1}{7}\)

 

Again, subtracting  eq (1) and (2)

x + y = 5xy
x – y = 9xy

we get 2y= -4xy
=> -4x=2

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

∴ \(x=-\frac{1}{2}\)

∴ The required solution, x=-1/2 , y = 1/7

9. \(\frac{1}{x-1}+\frac{1}{y-2}=3, \frac{2}{x-1}+\frac{3}{y-2}=5\)

Solution:

\(\frac{1}{x-1}+\frac{1}{y-2}=3\) …(1)

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

To eliminate x, we multiply equation no. (1) by 2 & equation no. (2) by 1, we get

\(
\begin{aligned}
& \frac{2}{x-1}+\frac{2}{y-2}=6 \\
& \frac{2}{x-1}+\frac{3}{y-2}=5 \\
& (-) \quad(-) \quad(-)
\end{aligned}\)

Subtracting, \(\begin{aligned}\frac{2}{y-2}-\frac{3}{y-2}=1
\end{aligned}\)

\(\begin{aligned}
& \Rightarrow \quad \frac{2-3}{y-2}=1 \\
& \Rightarrow \quad \frac{-1}{y-2}=1
\end{aligned} \)

 

∴ y-2=-1
∴ y = 1
Putting, y = 1 in equation (1) we get

\(\begin{aligned}
& \frac{1}{x-1}+\frac{1}{1-2}=3 \\
& \frac{1}{x-1}=3+1 \\
& x-1=\frac{1}{4} \\
& x=\frac{5}{4}
\end{aligned}\)

Class 9 Math Chapter 5 WBBSE

∴ \(x=1 \frac{1}{4}\)

∴ \(x=\frac{5}{4} \& y=1\)

10. \(\frac{14}{x+y}+\frac{3}{x-y}=5, \frac{21}{x+y}-\frac{1}{x-y}=2\)

Solution:

\(\frac{14}{x+y}+\frac{3}{x-y}=5\) …(1)

multiplying equation (1) by 1

\(\frac{21}{x+y}-\frac{1}{x-y}=2\) …(2)

and equation (2) by 3, we get,

\(\frac{63}{x+y}-\frac{3}{x-y}=6\) …(3)

 

eq(3) + eq (1)

\(\begin{aligned}
& \frac{63}{x+y}-\frac{3}{x-y}=6 \\
& \frac{14}{x+y}+\frac{3}{x-y}=5
\end{aligned}\)

Adding, \(\frac{77}{x+y} \quad=11\)

 

or, x + y = 7

or, x – y = 1 …..(5)

or, x+y = 7 ….(4)

Adding eq(5) + eq(4) we get

2x = 8
⇒ x =4 & y=7-4 = 3

∴ x= 4; y =3

11. \(\frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20}, \frac{x+y}{3}-\frac{x-y}{2}+\frac{5}{6}=0\)

Solution: \(\frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20} \ldots \ldots \times(1)\) …(1)

 

\(\frac{x+y}{3}-\frac{x-y}{2}=-\frac{5}{6} \ldots \ldots \times\left(\frac{1}{2}\right)\) …(2)

 

\(\begin{aligned}
& \frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20} \\
& \frac{x+y}{6}-\frac{x-y}{4}=-\frac{5}{12} \\
& -\quad+\quad +\quad \\
& \text { subracting, } \frac{x+y}{5}-\frac{x+y}{6}=\frac{7}{20}+\frac{5}{12} \\
& \end{aligned}\)

 

\(\begin{aligned}
& \Rightarrow \quad \frac{6(x+y)-5(x+y)}{30}=\frac{21+25}{60} \\
& \Rightarrow \quad \frac{x+y}{30}=\frac{46}{60} \\
& \Rightarrow \quad x+y=\frac{46 \times 30}{60} \\
& \Rightarrow \quad y=23
\end{aligned}\)

 

∴Putting, x + y = 23 in equation (1)

\(\begin{aligned}
& \Rightarrow \frac{23}{5}-\frac{x-y}{4}=\frac{7}{20} \\
& \Rightarrow \frac{-(x-y)}{4}=\frac{7}{20}-\frac{23}{5} \\
& \Rightarrow \frac{7-92}{20} \\
& \Rightarrow x-y=-\left(\frac{-85}{20} \times 4\right)=17
\end{aligned}\)

 

\(\begin{aligned}
&  therefore x+y=23 \\
&\frac{x-y=17}{2 x=40} \\
&\end{aligned}\)

 

∴ x = 20 & y = 23-20=3
∴ x 20, y = 3

12. x + y = a + b, ax-by= \(a^2-b^2\)

Solution: x + y = a + b …(1) Χ b we get

ax-by= \(a^2-b^2\) …(2) by 1, we get

Multiplying equation (1) by b & equation (2) by 1, we get

\(b x+b y=a b+b^2\)

 

Adding eq(3)+eq(2)

\(\begin{aligned}
& b x+b y=a b+b^2 \\
& a x-b y=a^2-b^2
\end{aligned}\)

Adding, \(a x+b x=a^2 a b \)

or, (a + b) x = a(a+b)

or, \(x=\frac{a(a+b)}{(a+b)}\)

or x = a

Putting the value of x in equation (1)
a+y=a+b
or, y=a+b-a
or, y=b

∴ Solution is x = a ,y = b

13. \(\frac{x+a}{a}=\frac{y+b}{b}, a x-b y=a^2-b^2\)

Solution: \(\frac{x+a}{a}=\frac{y+b}{b}\)…(1)

\(a x-b y=a^2-b^2\)…(2)

 

From equation (1) \(\frac{x+a}{a}=\frac{y+b}{b}\)
or, bx+ab = ay + ab
or, bx-ay = ab- ab
or, bx – ay = 0 …(3)

Multiplying equation (2) by a & equation (3) by b

\(\begin{aligned}
& a^2 x-a b y=a\left(a^2-b^2\right) \\
& b^2 x-a b y=0 \\
& (-) \quad(+) \\
& \text {subtracting, } a^2 x-b^2 x=a\left(a^2-b^2\right)
\end{aligned}\)

 

or, \(\left(a^2-b^2\right) x=a\left(a^2-b^2\right)\)

or, \(x=\frac{a\left(a^2-b^2\right)}{\left(a^2-b^2\right)}\)

or, x = a

Putting the value of x in equation (3), we get
b.a – ay = 0
or, – ay = – ab

or, \(y=\frac{-a b}{-a}\)

or, y = b

∴ x=a, & y = b

14. ax + by = c, \(a^2 x+b^2 y=c^2\)

Solution:  ax + by = c …(1)

\(a^2 x+b^2 y=c^2\)…(2)

Multiplying equation (1) by a & equation (2) by 1,

\(a^2 x+a b y=a c\)…(3)

 

\(\begin{aligned}
& a^2 x+a b y=a c \\
& a^2 x+b^2 y=c^2 \\
& (-) \quad(-) \quad(-)
\end{aligned}\)

Subtracting, \(aby -b^2 y=a c-c^2\)

 

or, by(ab) = c(a – c)

or, \({y}=\frac{c(a-c)}{b(a-b)}\)

Putting the value of y in equation (1)

\(a x+b \cdot \frac{c(a-c)}{b(a-b)}=c\)

or, \(\quad a x=c-\frac{c(a-c)}{(a-b)}\)

or, \(\quad a x=\frac{a c-b c-a c+c^2}{a-b}\)

or, \(\quad x=\frac{c(c-b)}{a(a-b)}\)

therefore \(x=\frac{c(c-b)}{a(a-b)} \& y=\frac{c(a-c)}{b(a-b)}\)

 

15. ax + by = 1, \(b x+a y=\frac{(a+b)^2}{a^2+b^2}-1\)

Solution: ax + by = 1 ….(1)

\(b x+a y=\frac{(a+b)^2}{a^2+b^2}-1\)…(2)

 

or, \(b x+a y=\frac{a^2+2 a b+b^2-a^2-b^2}{a^2+b^2}\)

or, \(b x+a y=\frac{2 a b}{a^2+b^2}\) …(3)

Multiplying equation (1) by a & equation (3) by b, we get

\(
\begin{aligned}
& a^2 x+a b y=a \\
& b^2 x+a b y=\frac{2 a b^2}{a^2+b^2} \\
& (-) \quad(-) \quad(-)
\end{aligned}\)

Subtracting, \(a^2 x-b^2 x=a-\frac{2 a b^2}{a^2+b^2}\)

\(or, \quad\left(a^2-b^2\right) x=\frac{a^3+a b^2-2 a b^2}{a^2+b^2}\)

 

or, \(\quad\left(a^2-b^2\right) x=\frac{a^3-a b^2}{a^2+b^2}\)

or, \(\quad x=\frac{a\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)}\)

or, \(x=\frac{a}{a^2+b^2}\)

Putting the value of x in equation (1) we get

\(a. \frac{a}{a^2+b^2}+b y=1\)

 

or, \(\quad \frac{a^2}{a^2+b^2}+by=1 \)

or, \(\quad b y=1-\frac{a^2}{a^2+b^2}\)

or, \(\quad b y=\frac{a^2+b^2-a^2}{a^2+b^2}\)

or, \(\quad b y=\frac{b^2}{a^2+b^2}\)

\(or, \quad y=\frac{b^2}{b\left(a^2+b^2\right)}
or, \quad y=\frac{b}{a^2+b^2}
therefore  x=\frac{a}{a^2+b^2} \& y=\frac{b}{a^2+b^2}\)

 

16. (7x-y-6)²+(14x+2y-16)²=0

Solution: 7x-y-6 …(1)
14x+2y-16…..(2)

From equation (1) 7x-y=6 ….(3)
From equation (2) 14x + 2y = 16…(4)

Multiplying the equation (3) by 2 & equation (2) by 1

14x-2y= 12 ….(5)

Adding, eq(5) + eq(4)

\(\begin{aligned}
& 14 x-2 y=12 \\
& 14 x+2 y=16 \\
& \hline 28 x=28
\end{aligned}\)

 

or, \(x=\frac{28}{28}\)

or, x = 1

Putting the value of x in equation (4), we get
14 x 1 + 2y = 16

or, 2y=16-14
or,2y=2

or, \(y=\frac{2}{2}\)

∴ X=1&y= 1

Chapter 18 Area Of Circle Exercise 18

Formulae:

1. Area of a circle = πr² sq.units
2. Area of a semi-circle = \(\frac{1}{2}\) πr² sq.units

3. If the length of the radius of a circle is units r and a sector of the circle makes an angle of degrees then:

1. Length of the arc = \(\frac{\theta}{360}\)  x Circumference of the circle 

= \(\frac{\theta}{360}\) x 2πr

2. Area of the (sector) = \(\frac{\theta}{360}\) x  Area of the circle

= \(\frac{\theta}{360}\) x πr² sq.units

Read and Learn More WBBSE Solutions For Class 9 Maths

Class 9 Math solutions WBBSE Chapter 18

 

 

 

WBBSE Class 9 Math Chapter 18 Question 1

Today the cow of Aminabibi is fastened to a post with a rope of length 2.1 meters in the vacant field. Let us see by calculating how much maximum area the cow can graze.

Solution: ∵ Here radius (r) = 2.1 m

∴ Area = πr² sq. cm

\(=\frac{22}{7} \times(2.1)^2 \text { sq. } \mathrm{cm}\) \(=\frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} . \text { sq. cm }\)

= 13.86 sq. cm.

WBBSE Class 9 Math Chapter 18 Question 2

Suhana will draw a circle whose perimeter will be 35.2 cm. Let us see by calculating what length of radius Suhana takes to draw a circle and what will be the area of that circular field.

Solution: Let the radius = r cm

∴ Circumference of the circle is = 2πr cm

= \(2 \times \frac{22}{7} \mathrm{r} \mathrm{cm}=\frac{44 \mathrm{r}}{7} \mathrm{~cm}\)

By the problem, \(\frac{44 r}{7}=35.2\)

or, \(r=\frac{352 \times 7}{10 \times 44}\)

or, r = 5.6

∴ Radius of the circle drawn by Suhana is 5.6 cm.
Area of the circle = πr²

= \(\frac{22}{7}\) x (5.6)² sq. cm

= \(\frac{22}{7} \times \frac{56}{10} \times \frac{56}{10}\) sq. cm

= 98.56 sq. cm

∴ 5.6 cm, 98.56 sq. cm

Question 3. Grandmother of Rekha has made a circular table cover whose area is 5544 sq. cm. She wants to paste coloring tape surrounding this circular cover of the table, let us see by calculating how much long coloring tape she will buy.

Solution: Let the radius = r cm

∴ Area = πr² sq. cm = \(\frac{22}{7} r^2\) sq.cm

By the problem, \(\frac{22}{7} r^2\) = 5544

= \(\mathrm{r}^2=\frac{5544 \times 7}{22}\)

or, r² = 252 x 7
or, r²= 1764

or, r = √1764 = 42
∴ Radius of the table = 42 cm

∴ Circumference of the table = 2πr cm

= 2 x \(\frac{22}{7}\) x 42 cm = 264 cm

∴ Length of the tape = 264 cm.

WBBSE Class 9 Math Chapter 18 Question 4

The cost of fencing of our village playground with railing is Rs. 924 at the rate of Rs. 21 per meter. Let us write by calculating how much sq. meter canvas will be bought for covering the field.

Solution: The cost of fencing of our village playground with railing is Rs. 924 at the rate of Rs. 21 per meter.

∴ Circumference of the field = 924 = 21 m = 44 m.

Let the radius of the field = r m.

∴ Circumference of the field = 2πr m

= \(2 \times \frac{22}{7} \times r m\)

= \(\frac{44 r}{7}\)

By the problem, \(\frac{44 r}{7}\) = 44

or, \(r=\frac{44 \times 7}{44}=7\)

∴ Radius of the field = 7 m

∴ Area of the ground = r² sq. cm

=\(\frac{22}{7} \times 7 \times 7\) sq.cm

= 154 sq. cm

Area of Circle Class 9 wbbse Question 5.
Faruk will draw a circle whose area will be 616 sq. meters. Let us see by calculating what length of radius Faruk will take to draw a circle and what perimeter he will get.

Solution: Let the radius = rm.

∴ Area of the circle = πr² sq. cm = \(\frac{22}{7} r^2\) sq.cm

By the problem, \(\frac{22}{7} r^2\) = 616

or, sq. cm \(r^2=\frac{616 \times 7}{22}\)

or, r² = 196
or, r = √196
or, r = 14

∴ Length of the radius of the circle drawn by Farukh = 14 cm.

Circumference of the circle = 2πr

=  \(2 \times \frac{22}{7} \times 14 \mathrm{~cm}\) = 88 cm

∴ 14 cm, 88 cm

Question 6. Palash and Piyali have drawn two circles, the ratio of whose lengths of radii is 4 5. Let us write by calculating the ratio of areas of the two circular fields drawn by them.

Solution: Let the radii of two circles are 4r & 5r units respectively.

∴ Ratio of areas of the two circles

= π(4г)²: (5г)²
= 16r²: 25r²
= 16:25

Area of Circle Class 9 wbbse Question 7.
Sumit and Reba have taken two copper wires of having the same length. Sumit bent the wire in the form of a rectangular shape whose length and breadth are 48 cm and 40 cm. But Reba bent the copper wire with the same length in the form of a circle. Let us see by calculating which will cover the maximum place between the rectangle drawn by Sumit and the circle drawn by Reba.

Solution: Area of the rectangle made by Sumit = 48 x 40 sq. cm = 1920 sq. cm

∴ Perimeter of the rectangle = 2(L + B)
= 2(48 +40) cm
= 2=88 cm
= 176 cm

∴ Circumference of the circle drawn by Reba is 176 cm.

Let the radius of the circle drawn by Reba is r cm.
∴ Circumference of the circle = 2πr cm

\(\begin{aligned}
& =2 \times \frac{22}{7} \times r \mathrm{~cm} \\
& =\frac{44 r}{7} \mathrm{~cm}
\end{aligned}\)

By the problem, \(\frac{44 r}{7}=176\)

or, \(r=\frac{176 \times 7}{44}=28 \mathrm{~cm}\) = 28 cm

∴ Area of the circle =πr² sq. cm

\(\begin{aligned}
& =\frac{22}{7}(28)^2 \text { vsq. cm } \\
& =\frac{22}{7} \times 28 \times 28 \text { sq. } \mathrm{cm}
\end{aligned}\)

= 2464 sq. cm

Area of Circle Class 9 wbbse Question 8.

At the center of the rectangular field of Pioneer athletic club, there is a circular pool whose length of the radius is 14 meters. The length and breadth of the rectangular field are 60 meters and 42 meters respectively. Let us see by calculating how much cost it will take for planting grass of the remaining place of rectangular field except the pool at the rate of Rs. 75 per square meter.

Solution:

 

 

Area of the rectangular field = 60 x 42 sq. cm = 2520 sq. cm

Area of the circular pool = πr² sq. cm

\(\begin{aligned}
& =\frac{22}{7} \times(14)^2 \text { sq. cm } \\
& =\frac{22}{7} \times 14 \times 14 \text { sq. cm }
\end{aligned}\) = 616 sq. cm

 

The remaining area of the field = (2520-616) sq. cm = 1904 sq. cm

∴ Cost for planting grass at the rate of Rs. 75 per sq. cm
= Rs. 1904 x 75
= Rs. 142800

 

Ganit Prakash Class 9 Solutions Question 9.

A 7 meter wide path runs outside a circular park of Etalgacha Friends association club along its perimeter. Let us write by calculating the area of the path if the perimeter of the circular park is 352 meters; let us write by calculating how much the cost for concreting the path at the rate of Rs. 20 per square meter.

Solution: Let the radius of the park = r m

∴ Perimeter of the park = 2πr m

\(\begin{aligned}
& =2 \times \frac{22}{7} \times r \mathrm{~m} \\
& =\frac{44 r}{7} \mathrm{~m}
\end{aligned}\)

 

By the problem, \(\frac{44 r}{7}=352\)

or, \(r=\frac{352 \times 7}{44}\)

or, r = 8 x7 = 56 m

∴ Now radius of park with the path of width 7 m = (56+7) m = 63 m

∴ Area of the path = π {(63)²-(56)²} sq. cm

= \(\frac{22}{7}\) (63+56) (63-56) sq. cm

= \(\frac{22}{7}\) x 119 x 7 sq. cm

= 2618 sq. cm

Now cost for concreting the path at the rate of Rs.20 per sq. cm
= Rs. 2618 x 20
= Rs. 52360

Question 10. Anwarbibi has spent Rs. 2664 at the rate of Rs. 18.50 per meter for fencing of her semi-circular land. Let us write by calculating how much cost it will take if she makes the semi-circular land plow at the rate of Rs. 32 per sq. meter.

Solution:
∴ Perimeter of the land = \(\frac{2664}{18.50} \mathrm{~m}\) = 144m

Let the radius of the land = r m.

∴ Perimeter of the land = (πr +2r) m
= г(π+2) m

= \(r\left(\frac{22}{7}+2\right) m\)

= \(r\left(\frac{22+14}{7}\right) m\)

= \(\frac{36 r}{7} \cdot m\)

By the problem, \(\frac{36 r}{7}=144\)

or, \(r=\frac{144 \times 7}{36}=28\)

∴ Radius of the land = 28 m

Area of the land \(\frac{1}{2} \pi r^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times 28 \times 28 \text { sq. cm }\)

= 1232 sq. cm

∴ Cost of plowing at 32 Rs. per sq.m = Rs. 1232 x 32
= Rs. 39424

Ganit Prakash Class 9 Solutions Question 11.

The time my friend Rajat took today running with uniform speed to round one of a circular field of school is 30 seconds less than that when he ran diametrically with same speed. Let us write by calculating the area of the field of the school if his speed is 90 meters/second.

Solution: Let the radius of the circular field = r m.

∴ Diameter of the field = 2r m.
∴ Perimeter of the field = 2πr m.

∵ Rajat covers up 90 m distance in 60 sec.

∴ Rajat covers up 1 m distance in \(\frac{60}{90}\) sec.

∴ Rajat covers up 2πr m distance in \(\frac{2}{3} \times 2 \pi r\) sec.

∴ Rajat covers up 2r m distance in \(\frac{2}{3} \times 2 r \mathrm{sec}\)

By the problem,

\(\frac{2}{3} \times 2 \pi r-\frac{2}{3} \times 2 r=30\)

or, \(\)

or, \(\frac{2}{3} \times 2 r(\pi-1)=30\)

or, \(\frac{4 r}{3}\left(\frac{22}{7}-1\right)=30\)

or, \(\frac{4 r}{3}\left(\frac{22-7}{7}\right)=30\)

or, \(\frac{4 r}{3} \times \frac{15}{7}=30\)

or, \(4 r \times \frac{5}{7}=30\)

or, \(r=\frac{30 \times 7}{5 \times 4}=\frac{21}{2}\)

∴ Area of school’s field= π r² sq. cm

= \(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \text { sq. cm }\)

= \(\frac{693}{2} \text { sq. cm }\)

= 346.5 sq. cm

Question 12. An equally wide path runs outside the circular field of Bakultala. The length of the outer circumference exceeds the inner circumference by 132 meters. If the area of the path is 14190 sq. meters, let us write by calculating the area of the circular path.

Solution:

 

 

Let the inner radius = r m
& the outer radius = Rm

∴ Circumference of the small circle = 2πr m
& circumference of the big circle = 2πR m

2πR – 2πг = 132
or, 2π(R-r) 132

or, \(2 \times \frac{22}{7}(R-r)=132\)

or, R-r= \(\frac{132 \times 7}{2 \times 22}\)

or, R-r=21 ….()

According to the second condition,

πR² – πr² = 14190
or,  π(R² -r²) = 14190

or,\(\frac{22}{7}\) (R+r)(R-r) =14190

or, \(\frac{22}{7}\) (R+r).21 = 14190 [∵R-r=21]

or, R+r= \(\frac{14190}{22 \times 3}=215\)

or, R+r=215……… (2)

Subtracting, eq (2)-(1)

R+r=215
R-r=21
(-) (+) (-)

∵ 2r = 194

or, \(r=\frac{194}{2}\)

or, r = 97

∴ Radius of the field = 97 m

∴ Area of the field = πr² sq. cm

= \(\frac{22}{7} \times(97)^2 \text { sq. cm }\)

= \(\frac{22}{7} \times 97 \times 97 \text { sq. cm }\)

= \(29571 \frac{1}{7} \text { sq. cm }\)

Ganit Prakash Class 9 Solutions Question 13.

Let us write by calculating the area of the shaded regions in the pictures below:

1. ABCD is a square. The length of the radius of the circle is 7 cm.

 

 

2. The length of the radius of each circle is 3.5 cm. The centers of the four circles are A, B, and D respectively.

 

 

Solution: Length of the radius (r) = 7 cm

∴ Area of the circle = πr² sq. cm

= \(\begin{aligned}
& =\frac{22}{7} \times(7)^2 \text { sq. } \mathrm{cm} \\
& =\frac{22}{7} \times 7 \times 7 \text { sq. } \mathrm{cm}
\end{aligned}\)

= 154 sq. cm

Length of the diagonal = 2r
= 2 x 7 cm
= 14 cm

∴ Area of square = \(\frac{1}{2}(diagonal)2\) Χ (diagonal)²

\(\begin{aligned}
& =\frac{1}{2} \times(14)^2 \text { sq. cm } \\
& =\frac{1}{2} \times 196 \text { sq. cm }
\end{aligned}\)

= 98 sq. cm

∴ Area of shaded region (154-98) sq. cm
= 56 sq. cm

2. Radius of each circle = 3.5 cm

∴ Área of each circle = \(\frac{22}{7} \times(3.5)^2\) sq. cm

= \(\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \text { sq. } \mathrm{cm}\)

= 38.5 sq. cm

∴ Total area of 4 circles = 4 x 38.5 sq. cm = 154 sq. cm

White portion of the pictures in the 4 circles

= \(4 \times \frac{90}{360} \times \frac{22}{7} \times(3.5)^2 \text { sq. cm }\)

= \(4 \times \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 \text { sq. cm }\)

= 38.5 sq. cm

∴ Area of shaded region picture = (154-38.5) sq. cm
= 115.5 sq. cm

Question 14. Dinesh has made a pie chart of the students of their class who want to play some games. He has taken the length of the radius of the circle of 3.5 cm, let us write by calculating the perimeter and area of each sector of circles.

Solution: Let in Dinesh’s class there are 100 students where 50 play cricket, 30 play football and 20 play badminton.

Name of Game	Percentage	Fraction	Angle (part of 360°)
Cricket	50	\(\frac{50}{100}=\frac{1}{2}\)	\(\frac{1}{2} \times 360^{\circ}=180^{\circ}\)
Football	30	\(\frac{30}{100}=\frac{3}{10}\)	\(\frac{3}{10} \times 360^{\circ}=108^{\circ}\)
Badminton	20	\(\frac{20}{100}=\frac{1}{5}\)	\(\frac{1}{5} \times 360^{\circ}=72^{\circ}\)

 

Dinesh made a pie chart of the list.

 

 

1. Length of the chord = \(\overline{\mathrm{AB}}=\frac{180}{360} \times 2 \pi r\)

=\(\frac{1}{2} \times 2 \times \frac{22}{7} \times 3.5 \mathrm{~cm}\)

= 11 cm

The perimeter of AOB (Sector)

= \(\overline{\mathrm{AB}}\) Length of + 2 x radius

= (11 + 2 x 3.5) cm = 18 cm

Area of sector AOB = \(\frac{180}{360} \times \pi r^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times(3.5)^2 \text { sq.cm }\) = 19.25 sq.cm

2. Length of the chord \(\overline{\mathrm{AC}}=\frac{108}{360} \times 2 \times \frac{22}{7} \times 3.5 \mathrm{~cm}\)

= \(\frac{3}{10} \times 2 \times \frac{22}{7} \times \frac{35}{10} \mathrm{~cm}\)

= 6.6 cm

Perimeter of the sector AOC = \(\overline{\mathrm{AC}}\) length of 2 x length of radius
= (6.6+2 x 3.5) cm
= 13.6 cm

Area of the sector AOC = \(\frac{108}{360} \times \frac{22}{7} \times(3.5)^2 \text { sq.cm }\)

= \(\frac{3}{10} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \text { sq.cm }\)

= 11.55 sq. cm

3. Length of the sector \(\overline{\mathrm{BC}}=\frac{72}{360} \times 2 \times \frac{22}{7} \times 3.5 \mathrm{~cm}\)

= \(\frac{1}{5} \times 2 \times \frac{22}{7} \times \frac{35}{10} \mathrm{~cm}\)

= 4.4 cm

The perimeter of the sector BOC = \(\overline{\mathrm{BC}}\) length of + 2 x length of the radius
= (4.4+7) cm
= 11.4 cm

Area of the sector BOC = \(\frac{72}{360} \times \pi r^2\)

\(\begin{aligned}
& =\frac{1}{5} \times \frac{22}{7} \times(3.5)^2 \text { sq. cm } \\
& =\frac{1}{5} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \text { sq. cm }
\end{aligned}\)

= 7.7 sq. cm

Wbbse Maths Solution Class 9 

Question 15. Nitu has drawn a square ABCD whose length of each side is 12 cm. My sister has drawn four circular arcs with a length of radius 6 cm centering A, B, and D like pictures beside and she has designed some portions. Let us write by calculating the area of the shaded region.

Solution:

 

 

Length of each side of the square = 12 cm.

∴ Area of square ABCD = (12)² sq. cm
= 144 sq. cm

Length of the arc of radius 6 cm

= \(\frac{90}{360} \times \text { Circumference of the circle }\)

\(\begin{aligned}
& =\frac{1}{4} \times 2 \times \frac{22}{7} \times 6 \mathrm{~cm} \\
& =\frac{66}{7} \mathrm{~cm}
\end{aligned}\)

Area of the sector of radius 6 cm = \(\frac{90}{360} \times \pi r^2\)

= \(\frac{1}{4} \times \frac{22}{7} \times(6)^2 \text { sq. cm }\)

∴ Area of the 4 sectors

= \(\begin{aligned}
& =4 \times \frac{1}{4} \times \frac{2}{7} \times 36 \text { sq. cm } \\
& =\frac{792}{7} \text { sq. cm }
\end{aligned}\)

Area of the 4 sectors = \(4 \times \frac{66}{7} \mathrm{~cm}\)

\(\begin{aligned}
& =\frac{264}{7} \mathrm{~cm} \\
& =37 \frac{5}{7} \mathrm{~cm}
\end{aligned}\)

 

Area of shaded region= \(\left(144-\frac{792}{7}\right) \text { sq. cm }\)

= \(\left(\frac{1008=792}{7}\right) \text { sq. } \mathrm{cm}\)

\(\begin{aligned}
& =\frac{216}{7} \text { sq. } \mathrm{cm} \\
& =30 \frac{6}{7} \text { sq. } \mathrm{cm}
\end{aligned}\)

 

∴ \(37 \frac{5}{7} \cdot \mathrm{cm} ; 30 \frac{6}{7} \mathrm{sq} . \mathrm{cm}\)

 

Question 16. The area of a circular field is 154 sq cm. Let us write by calculating the perimeter and area of circumfering the circular field with a square.

Solution:

 

 

 

 

Let the radius of the circular field = r cm

∴ Area of the field = r² sq. cm

= \(\frac{22}{7} r^2 \text { sq. } \mathrm{cm}\)

By the problem,

\(\frac{22}{7} r^2=154\)

or, \(r^2=\frac{154 \times 7}{22}\)

or, r² = 49

or, r = √49
or, r = 7

∴ Diameter of the field = 2 x 7 cm
= 14 cm

∴ Length of one side of the square outside the circle = 14 cm.
∴ Perimeter of the square = 4 x 14 cm = 56 cm

∴ And the area of the square = (14)² sq. cm = 196 sq. cm

Length of the diagonal of the internal square = 14 cm
= Diameter of the circle

∴ Length of one side of the inner square= \(\frac{14}{\sqrt{2}}\)

= \(\begin{aligned}
& =\frac{14 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \mathrm{~cm} \\
& =\frac{14 \sqrt{2}}{2} \mathrm{~cm}
\end{aligned}\)

= 7√2 cm

Perimeter of the inner square = 4×7√2 cm = 28√2 cm

Area of the internal square = \(\frac{1}{2}\) x (Diagonal)²

= \(\frac{1}{2}\) Χ (14)² 

= \(\frac{1}{2}\)  x 196 cm = 98 cm

Question 17. Let us write the perimeter and area of the circular region shaded sector below. 

Solution:

1.

 

 

length of AB = \(\sqrt{(12)^2+(12)^2} \mathrm{~cm}\)

= \(\begin{aligned}
& =\sqrt{144+144} \mathrm{~cm} \\
& =\sqrt{288} \mathrm{~cm} \\
& =\sqrt{2 \times 12 \times 12} \mathrm{~cm} \\
& =12 \sqrt{2} \mathrm{~cm}
\end{aligned}\)

= 12 x 1.414 cm
= 16.97 cm (approx).

Length of the arc \(\widehat{\mathrm{AB}}=\frac{90}{360}\) x Circumference of the circle

\(\begin{aligned}
& =\frac{1}{4} \times 2 \times \frac{22}{7} \times 12 \mathrm{~cm} \\
& =\frac{132}{7} \mathrm{~cm}
\end{aligned}\)

= 18.86 cm (approx).

∴ Perimeter of the shaded region = (16.97 +18.86) cm
= 35.83 cm (approx).

Area of AOB (Sector) = \(\frac{90}{360} \times \pi r^2 \text { sq. cm }\)

\(\begin{aligned}
& =\frac{1}{4} \times \frac{22}{7} \times(12)^2 \text { sq. cm } \\
& =\frac{1}{4} \times \frac{22}{7} \times 12 \times 12 \text { sq. cm } \\
& =\frac{792}{7} \text { sq. } \mathrm{cm}
\end{aligned}\)

 

Area of the right-angled ΔAOB = \(\frac{1}{2}\) x 12 x 12 sq. cm = 72 sq. cm

∴ Area of the shaded region = \(\left(\frac{792}{7}-72\right) \text { sq. cm }\)

\(\begin{aligned}
& =\frac{792-504}{7} \text { sq. cm } \\
& =\frac{288}{7} \text { sq. cm } \\
& =41 \frac{1}{7} \text { sq. } \mathrm{cm}
\end{aligned}\)

 

2.

 

Length of AC = 42 cm ( ∵ ABC is an equilateral triangle

Length of arc \(\widehat{\mathrm{AB}}=\frac{60}{360}\) x Circumference of the circle

\(\begin{aligned}
& =\frac{1}{6} \times 2 \pi r \\
& =\frac{1}{6} \times 2 \times \frac{22}{7} \times 42 \mathrm{~cm} \\
& =44 \mathrm{~cm}
\end{aligned}\)

 

∴ Perimeter of the shaded region = (42 + 44) cm = 86 cm

Area of sector ABC = \(\frac{60}{360} \times \pi r^2\)

= \(\frac{1}{6} \times \frac{22}{7} \times 42 \times 42 \text { sq. } \mathrm{cm}\)

= 924 sq. cm

Area of equilateral triangle ABC = \(\frac{\sqrt{3}}{4} \times(\text { side })^2\)

\(\begin{aligned}
& =\frac{\sqrt{3}}{4} \times(42)^2 \text { sq. cin } \\
& =\frac{\sqrt{3}}{4} \times 42 \times 42 \text { sq. cm }
\end{aligned}\)

 

= 441√3 sq. cm
= 441 x 1.732 sq. cm
= 763.812 sq. cm

∴ Area of shaded region = (924-763.812) sq. cm
= 160.188 sq. cm
= 160.19 sq. cm (approx)

Wbbse Maths Solution Class 9 

Question 18. Buying a bangle from the fair Nila wears in her hand. The bangle contains 269.5 sq. cm. metal. If the length of the outer diameter of the bangle is 28 cm, let us write by calculating the length of the inner diameter.

Solution: Length of the radius (R) = \(\frac{28}{2}\) = 14cm
& let the radius of the inner r cm.

∴ Area of the metal of the bangle = \(\pi\left(R^2-r^2\right)\)

= \(\frac{22}{7}\left\{(14)^2-r^2\right\} \text { sq. cm }\)

= \(=\frac{22}{7}\left(196-r^2\right) \text { sq. cm }\)

According to the problem,

\(\frac{22}{7}\left(196-r^2\right)=269.5\)

or, \(196-r^2=\frac{2695 \times 7}{10 \times 22}\)

or, 196-r² = 85.75
or, -r² = 85.75 -196

or, -r² = -110.25
or, r² = 110.25

or, r= √110.25 = 10.5
∴ Length of the inner diamter
= 2 x 10.5 cm 21 cm.

Question 19. Protul has drawn an equilateral triangle ABC (picture is given beside), whose length of each side is 10 cm. Sumita has drawn three circular arcs centering A, B,C with a length of radius 5 cm and has colored some portion at the middle. Let us write by calculating the area of the colored portion.

Solution:

 

 

Area of the equilateral triangle ABC = \(A B C=\frac{\sqrt{3}}{4} \times(\text { side })^2\)

= \(\frac{\sqrt{3}}{4} \times(10)^2 \text { sq. cm }\)

= \(\frac{\sqrt{3}}{4} \times 100 \text { sq. cm }\)

= 25√3 sq. cm
= 25 x 1.732 sq. cm
= 43.3 sq. cm

Area of each sector = \(\frac{60}{360} \times \pi r^2\)

= \(\frac{1}{6} \times \frac{22}{7} \times(5)^2 \text { sq. cm }\)

∴ Total area of the three sectors = \(3 \times \frac{1}{6} \times \frac{22}{7} \times 25 \text { sq. cm }\)

=\(\frac{11 \times 25}{7} \text {.sq. cm }\)

= 39.28 sq. cm
∴ Area of the coloured portion = (43.3-39.28) sq. cm = 4.02 sq. cm

Question 20. Rabeya drew an equilateral triangle with side of 21 cm on a big piece of paper. Drawing a circle inscribing that triangle the colored circular region, I write by calculating the area of the colored region.

Solution:

 

 

Length of each side of the equilateral triangle = 21 cm.

Length of the radius of the incircle = \(\frac{1}{3}\) Χ height of the equilateral triangle

= \(\begin{aligned}
& =\frac{1}{3} \times \frac{\sqrt{3}}{2} \times 21 \mathrm{~cm} \\
& =\frac{7 \sqrt{3}}{2} \mathrm{~cm}
\end{aligned}\)

Area of incircle = πr²

= \(\frac{22}{7}\left(\frac{7 \sqrt{3}}{2}\right)^2 \text { sq. cm }\)

= \(\frac{22}{7} \times \frac{7 \sqrt{3} \times 7 \sqrt{3}}{2 \times 2} \text { sq. } \mathrm{cm}\)

= 115.5 sq. cm

Question 21. The area of circumscribing an equilateral triangle is 462 sq. cm. Let us write by calculating the length of each side of this triangle.

Solution:

 

Area of the circumcircle of the equilateral triangle = 462 sq. cm
Let the radius of the circumcircle = r cm

∴ Area of the circumcircle = πr² sq. cm

By the problem, \(\frac{22}{7} r^2=462\)

or, \(r^2=\frac{462 \times 7}{22}\)

or, r² = 21 x 7

or, r = \(\sqrt{21 \times 7}\) = 7√3

∴ Radius of the circumcircle = \(\frac{2}{3}\) x height of the triangle

or, 7√3 = \(\frac{2}{3} \times \frac{\sqrt{3}}{2}\) x length of the triangle

∴ Length of each side of the triangle = \(\frac{7 \sqrt{3} \times 3}{\sqrt{3}} \mathrm{~cm}\) = 21 cm

∴ Length of each side of the equilateral triangle = 21 cm.

 

Wbbse Maths Solution Class 9 

Question 22. The perimeter of a triangle is 32 cm and the area of its inscribing circle is 38.5 sq. cm. Let us write by calculating the area of this triangle.

Solution:

 

 

Let the radius of the incircle of the equilateral triangle = r cm.

∴ Area of the circle = πr² sq. cm = \(\frac{22}{7} r^2 \text { sq. cm }\)

According to the problem,

\(\frac{22}{7} r^2\) = 38.5

or, \(r^2=\frac{385 \times 7}{10 \times 22}\)

or, \(r^2=\frac{49}{4}\)

or, \(r=\sqrt{\frac{49}{4}}=\frac{7}{2}\)

Area of ΔABC = Area of (ΔAOB + area of ΔBOC+ Area of ΔCOA)

\(\begin{aligned}
& =\frac{1}{2} \cdot A B \cdot r+\frac{1}{2} \cdot B C \cdot r \cdot+\frac{1}{2} C A \cdot r \\
& =\frac{1}{2} r(A B+B C+C A) \\
& =\frac{1}{2} \times \frac{7}{2} \times 32 \text { sq. cm }
\end{aligned}\)

= 56 sq. cm.

Question 23. Let us write by calculating the length of the radius of the incircle and circumcircle of a triangle, whose sides are 20 cm, 15 cm, and 25 cm. Let us calculate the area of the incircle and the circumcircle.

Solution:

 

Let ABC is a triangle whose sides AB = 15 cm, BC = 25 cm, and AC = 20 cm, and the incentre is O and in radius is r cm.

∵ (15)²+(20)² =(25)²

∴ Area of ΔABC = \(\frac{1}{2}\) x 15 x 20 sq. cm
= 150 sq. cm

∵ ΔABC = ΔAOB + ΔBOC+ΔCOA

\(\text { or, } 150=\frac{1}{2} \cdot 15 \cdot r+\frac{1}{2} \cdot 25 \cdot r+\frac{1}{2} \cdot 20 \cdot r\)

 

\(\text { or, } 150=\frac{1}{2} r(15+25+20)\)

 

\(\text { or, } r=\frac{150 \times 2}{60}\)

or, r = 5
∴ Radius of the incirle = 5 cm

Area of the incircle = πr²

\(\begin{aligned}
& =\frac{22}{7} \times(5)^2 \text { sq. cm } \\
& =\frac{22}{7} \times 25 \text { sq. cm } \\
& =\frac{550}{7} \text { sq. } \mathrm{cm} \\
& =78 \frac{4}{7} \text { sq. } \mathrm{cm}
\end{aligned}\)

 

∵ The hypotenuse of a right-angled triangle is the diameter of its circumcircle.

∴ Radius of the circumcircle = \(\frac{25}{2} \mathrm{~cm}\) = 12.5 cm

∴ Area of circumcircle = \(\frac{22}{7} \times \frac{25}{2} \times \frac{25}{2} \text { sq. cm }\)

\(\begin{aligned}
& =\frac{6875}{14} \text { sq. cm } \\
& =491 \frac{1}{14} \text { sq. } \mathrm{cm}
\end{aligned}\)

 

∴ The inradius and circumradius of the triangle are respectively, 5 cm and 12.5 cm and the areas of incircle and circumcircle are respectively \(78 \frac{4}{7}\) sq. cm and \(491 \frac{1}{14} \text { sq. cm. }\)

Question 24. Jaya drew an incircle of a square. That circle is also circumscribed an equilateral triangle, of which the length of each side is 4√3 cm. Let us write by calculating the length of the diagonal of the square.

Solution:

 

Length of each side of the equilateral triangle = 4√3 cm.

∴ Height of the triangle = \(\frac{\sqrt{3}}{2}\) x side

= \(\frac{\sqrt{3}}{2} \times 4 \sqrt{3} \mathrm{~cm}\) = 6 cm

∴ The radius of the circumcircle = \(\frac{2}{3}\) x height of the triangle

= \(\frac{2}{3} \times 6 \mathrm{~cm}\)

= 4 cm

∴ Diameter of the circumcircle
= 2 x 4 cm
= 8 cm

∴ Length of the diagonal of the square = √2 × side
= √2×8 cm
= 8√2 cm.

Question 25. Sumit cut a wire into two equal parts. One part he bent in the form of a square and another part he bent in the form of a circle. If the area of the circle exceeds that of the square by 33 sq. cm, let us write by calculating the original length of the wire.

Solution: Let the length of the wire = x cm

∴ Each part of the wire = \(\frac{x}{2}\)

∴ Perimeter of square= \(\frac{x}{2}\)

∴ Length of each side of the square= \(\frac{x}{8}\)

∴ Area of the square =\(\left(\frac{x}{8}\right)^2 \text { sq. cm }\)

= \(\frac{x^2}{64} \text { sq. cm }\)

Again, the perimeter of the circle = \(\frac{x}{2}\)

oг, 2πr = \(\frac{x}{2} \mathrm{~cm}\)

or, \(r=\frac{x}{4 \pi}\)

Area of the circle = πr²

\(\begin{aligned}
& =\pi\left(\frac{x}{4 \pi}\right)^2 \text { sq. cm } \\
& =\pi \cdot \frac{x^2}{16 \pi^2} \text { sq. } \mathrm{cm} \\
& =\frac{x^2}{16 \pi} \text { sq. cm }
\end{aligned}\)

 

According to the problem, \(\frac{x^2}{16 \pi}-\frac{x^2}{64}=33\)

or, \(\frac{x^2}{16}\left(\frac{1}{\pi}-\frac{1}{4}\right)=33\)

or, \(\frac{x^2}{16}\left(\frac{1}{\frac{22}{7}}-\frac{1}{4}\right)=33\)

or, \(\frac{x^2}{16}\left(\frac{7}{22}-\frac{1}{4}\right)=33\)

or, \(\frac{x^2}{16}\left(\frac{14-11}{44}\right)=33\)

or, \(\frac{x^2}{16} \times \frac{3}{44}=33\)

or, \(x^2=\frac{33 \times 44 \times 16}{3}\)

or, x² = 11 x 44 x 16

or, x = \(\sqrt{11 \times 11 \times 4 \times 16}\)

or, x = 11 x 2 x 4 = 88
∴ Length of the wire =  88 cm.

Chapter 18 Area Of Circle Multiple choice questions

 

1. If the area of a circular field is x sq unit, the perimeter is y unit, and the length of diameter is z unit then the value of \(\frac{x}{y z}\) is

1. \(\frac{1}{2}\)

2. \(\frac{1}{4}\)

3. 1

4. \(\frac{1}{8}\)

Solution: Let the radius = r unit

∴ \(\frac{x}{y z}=\frac{\pi r^2}{2 \pi r \cdot 2 r}=\frac{1}{4}\)

∴ 2. \(\frac{1}{4}\)

2. The ratio of areas of two squares circumscribed and inscribed by a circle is

1. 4: 1
2. 1:4
3. 2:1
4. 1:2

Solution: Let the radius = r unit
∴ Diameter = 2r unit

∴ Length of one side of the outer square = 2r unit
& diagonal of the inner square = 2r unit

∴ One side of the inner square= \(\frac{2 r}{\sqrt{2}} \text { unit }\)

∴ The ratio of areas of the two squares = \((2 r)^2:\left(\frac{2 r}{\sqrt{2}}\right)^2\)

\(\begin{aligned}
& =4 r^2: \frac{4 r^2}{2} \\
& =1: \frac{1}{2}
\end{aligned}\)

= 2:1

∴ 3. 2:1

Wbbse class 9 Maths  Question 3.

The numerical value of the perimeter of a square and area of a circular field is equal. The length of the diagonal of the square circumscribed by a circle is

1. 4 unit
2. 2 unit
3. 4√2 unit
4. 2√2 unit

Solution: Let the radius = r unit
∴ Perimeter = 2πr unit
∴ & Area = πr² unit

According to the problem,
πr² = 2πr
or, r = 2

∴ Length of one side of the outer square Diameter of the circle
= 2r unit

∴ Length of the diagonal of the square = √2 × side
= √2x2r
=√2x2x2 unit
= 4√2 unit

3. 4√√2 unit

Question 4. The ratio of the areas of an equilateral triangle and its inscribing circle is

1. 4: 1
2. 1:4
3. 2:1
4. 1:2

∴ 1. 4:1

Question 5. The inner diameter and external diameter of an iron ring plate are 20 cm and 22 cm. The quantity of iron plates in the ring is

1. 22 sq. cm
2. 44 sq. cm
3. 66 sq. cm
4. 88 sq. cm

Solution: Area of iron ring = π(R² – r²)

= \(\frac{22}{7}\left\{(11)^2-(10)^2\right\} \text { sq. cm }\)

= \(\frac{22}{7}(11+10)(11-10) \text { sq. cm }\)

= \(\frac{22}{7} \times 21 \times 1 \text { sq. cm }\)

= 66 sq. cm

∴ 3. 66 sq. cm

Class 9 Maths Chapter 18 Area Of Circle Short Question and Answers

 

Question 1. If the length of the radius of a circular field was increased by 10%, let us write by calculating what percent it increased the area of a circular field.

Solution: Let the radius of the circular field = r unit.
∴ Its area = r² sq unit

Now, radius after increasing 10 %= \(\frac{110 r}{100} \text { unit }=\frac{11 r}{10} \text { sq unit }\)

∴ New area = \(=\pi\left(\frac{11 r}{10}\right)^2 \text { sq unit }\)

= \(\pi \cdot \frac{121 r^2}{100} \text { sq unit }\)

∴ Increase in area

\(\begin{aligned}
& =\pi \cdot \frac{121 r^2}{100}-\pi r^2 \text { sq unit } \\
& =\pi r^2\left(\frac{121}{100}-1\right) \text { sq unit } \\
& =\pi r^2\left(\frac{121-100}{100}\right) \text { sq unit } \\
& =\frac{21 \pi r^2}{100} \text { sq unit }
\end{aligned}\)

 

∴ Percentage increase in area

= \(\frac{21 \pi r^2}{\frac{100}{\pi r^2}} \times 100\)

= 21%

Wbbse class 9 Maths  Question 2.

If the perimeter of a circular field was decreased by 50%, let us write by calculating what percent it decreases the area of the circular field.

Solution: Let the radius of the circular field = r unit.

∴ Permeter = 2πr unit
& Area = πr² sq. unit

Now, new radius after decreasing 50%= \(\frac{50}{100} \times 2 \pi r=\pi r \text { unit }\)

∴ Perimeter = \(=\frac{r}{2} \text { unit }\)

New area = \(\pi\left(\frac{r}{2}\right)^2 \text { sq. unit } \frac{\pi r^2}{4} \text { sq. unit }\)

Decrease in the area = \(\left(\pi r^2-\frac{\pi r^2}{4}\right) \text { sq. unit }\)

\(\begin{aligned}
& =\frac{4 \pi r^2-\pi r^2}{4} \text { sq. unit } \\
& =\frac{3 \pi r^2}{4} \text { sq. unit }
\end{aligned}\)

Percentage decrease in area = \(\frac{3 \pi r^2}{4 \times \pi r^2} \times 100=75 \%\)

3. The length of radius of a circular field is r meter. If the area of the other circle is x times of the first circle, let us see by calculating how much length is of the radius of the other circle.

Solution: Radius first circular field = r m
∴ Let the radius of the other circular field = πr² sq.m

Let the radius of the second circle be R m.
∴ Area of the second circle = πR² sq.m

By the problem,

πR² = x.πr²
or, R² = xr²

or, R= \(R=\sqrt{x r^2}\)

or, R = r√x
∴ Radius of the other circle = r√x m

Wbbse class 9 Maths  Question 4. Let us calculate the area of a circle which circumscribes a triangle, of which sides are 3 cm, 4 cm, and 5 cm.

Solution: ∵ (3)2 + (4)2 = (5)2
∴ The triangle whose sides are 3 cm, 4 cm & 5 cm is a right-angled triangle.

∴ The hypotenuse of a right-angled triangle is the diameter of its circumcircle.
∴ Length of the diameter of the circumcircle = 5 cm.

∴ Radius of the circle = \(\frac{5}{2} \mathrm{~cm}\)

∴ Area of the circumcircle = \(\pi\left(\frac{5}{2}\right)^2 \text { sq. cm }\)

\(\begin{aligned}
& =\frac{22}{7} \times \frac{25}{4} \text { sq. cm } \\
& =\frac{275}{14} \text { sq. } \mathrm{cm}
\end{aligned}\)

 

= \(19 \frac{9}{14} \text { sq. cm }\)

5. Three circular plates were cut off from a tin plate with equal width. If the ratio of lengths of the diameter of three circles is 3:5: 7, let us see by calculating the ratio of their weights.

Solution: Let the diameters of the 3 circles = 3r, 5r & 7r unit.

∴ Radius of 3 circles = \(\frac{3 r}{2} \text { unit } \frac{5 r}{2} \text { unit, and } \frac{7 r}{2} \text { unit }\)

∴ Ratio of areas of 3 circles = \(\pi\left(\frac{3 r}{2}\right)^2: \pi\left(\frac{5 r}{2}\right)^2: \pi\left(\frac{7 r}{2}\right)^2\)

\(=\frac{9 r^2}{4}: \frac{25 r^2}{4}: \frac{49 r^2}{4}\)

=9:25:49

∴ Ratio of weights of three circular rings is equal to the ratio of their areas.
∴ Ratio of weights of three circular rings 9:25:49

Class IX Maths Solutions WBBSE Chapter 11 Statistics Exercise 11.1

Question 1. I have written the number of children belonging to each of 40 families in our locality below

1	2	6	5	1	5	1	3	2	6
2	3	4	2	0	4	4	3	2	2
0	0	1	2	2	4	3	2	1	0
5	1	2	4	3	4	1	6	2	2

 

I prepare a frequency distribution table of the above-given data whose classes are 0-2, 2-4,…….., etc.

Solution: Frequency distribution table

Class Interval	Class Size	Length of Class	Tally Mark	Frequency
0-2	0-2	2	I	11
2-4	2-4	2	II	17
4-6	4-6	2	llll	9
6-8	8-Jun	2	III	3
			Total	40

 

Read and Learn More WBBSE Solutions For Class 9 Maths

Question 2. Given below are the marks obtained by 40 students in a test of school:

34	27	45	21	30	40	11	47	1	15
3	40	12	47	48	18	30	24	25	28
32	31	25	22	27	41	12	13	2	44
43	7	9	49	13	19	32	39	24	3

 

I construct a frequency distribution table of these marks by taking classes 1-10, 11-20, ………, 41-50.

Solution: Frequency distribution table

Class	Tally mark	Frequency
1-10	I	6
11-20	III	8
21-30	I	11
31-40	II	7
41-50	III	8
	Total	40

 

Question 3. There are many oranges in a basket. From this basket, by aimlessly taking 40 oranges, I wrote below their weights (gm):

45, 35, 30, 55, 70, 100, 80, 110, 80, 75, 85, 70, 75, 85, 90, 75, 90, 30, 55, 45, 40, 65, 60, 50, 40, 100, 65, 60, 40, 100, 75, 110, 30, 45, 84, 70, 80, 95, 85, 70.

Now, I construct a frequency distribution table and a less-than-type cumulative frequency distribution table for the above-given data.

Solution: Maximum weight = 110 gm
Minimum weight = 30 gm
∴ Width = 110-30=80 gm
Size of class = 10

Class	Tally mark	Frequency	Cumulative Frequency
30-40	llll	4	4
40-50	I	6	4 + 6=10
50-60	III	3	10 + 3=13
60-70	llll	4	13 + 4=17
70-80	III	8	17 + 8 = 25
80-90	II	7	25 + 7 = 32
90-100	III	3	32 + 3 = 35
100-120	III	3	35 + 3 = 38
110-120	II	2	38 + 2 = 40
	Total	40

 

Note: If the same value occurs in two classes put the value in a higher class.

Class IX Maths Solutions WBBSE

Question 4. Mitali and Mohidul wrote below the amount of money of electricity bills of the 45 houses of their village for this month:

116, 127, 100, 82, 80, 101, 91, 65, 95, 89, 75, 92, 129, 78, 87, 101, 65, 52, 59, 65, 95, 108, 115, 121, 128, 63, 76, 130, 116, 108, 118, 61, 129, 127, 91, 130, 125, 101, 116, 105, 92, 75, 98, 65, 110.

I construct a frequency distribution table for the above data.

Solution: Frequency distribution table

Class	Tally Mark	Frequency
50-60	II	2
60-70	I	6
70-80	llll	4
80-90	llll	4
90-100	II	7
100-110	II	7
110-120	I	6
120-130	II	7
130-140	II	2
	Total frequency	45

 

Question 5. Maria has written the ages of 300 patients of a hospital in the table given below:

Ages (years)	10-20	20-30	30-40	40-50	50-60	60-70
The number of patients	80	40	50	70	40	20

 

I construct a more than type cumulative frequency distribution table for the above data.

Solution: More than type cumulative frequency distribution label

Age (years) Class	No. of patients Frequency	Cumulative frequency
10-20	80	220 + 80 = 300
20-30	40	180 + 40 = 220
30-40	50	130 + 50 = 180
40-50	70	60 + 70 = 130
50-60	40	20 + 40 = 60
60-70	20	20
Total	300

 

Question 6. Let us observe the following cumulative frequency distribution table and construct a frequency distribution table:

Classes	Below 10	Below 20	Below 30	Below 40	Below 50	Below 60
The number of students	17	22	29	37	50	60

 

Solution: Frequency distribution table

Class no. of students	Frequency	Cumulative frequency
10	17	17
10-20	22-17 = 5	22
20-30	29-22 = 7	29
30-40	37-29 =8	37
40-50	50-37 = 13	50
50-60	60-50 = 10	60

 

Question 7. Let us observe the following cumulative frequency distribution table and construct the frequency distribution table :

Marks obtained	The number of students
More than 60	0
More than 50	16
More than 40	40
More than 30	75
More than 20	87
More than 10	92
More than 0	100

 

Class IX Maths Solutions WBBSE

Solution: Frequency distribution table

Marks	Class	The no. of students	Frequency
More than 60	More than 60	0	0
More than 50	50-60	16	16-0 = 16
More than 40	40-50	40	40-16 = 24
More than 30	30-40	75	75-40 = 35
More than 20	20-30	87	87-75 = 12
More than 10	10-20	92	92-87 = 5
More than 0	0-10	100	100-92 = 8

 

Question 8.

1. Which one of the following is a graphical (pictorial) representation of statistical data?

1. Line graph
2. Raw data
3. Cumulative frequency
4. Frequency.

Solution: 1. Line graph

2. The range of the data 12, 25, 15, 18, 17, 20, 22, 26, 6, 16, 11, 8, 19, 10, 30, 20, 32 is

1. 10
2. 15
3. 18
4. 26

Solution: Greatest value = 32
Minimum value = 6
Range 326 = 26

∴ 4. 26

3. The class size of classes 1-5, 6-10 is

1. 4
2. 5
3. 4.5
4. 5.5

Solution: Class size=5-110-6=4
∴ 1. 4

Class 9 Maths WB Board

4. In a frequency distribution table, the mid-points of the classes are 15, 20, 25, 30 ……. respectively. The class having a mid-point as 20 is

1. 12.5-17.5
2. 17.5-22.5
3. 18.5 21.5
4. 19.5 20.5

Solution:  20-15-25-20-30-25=5
∴ Class size = 5
Class size of 1st class = 17.5-12.5 = 5

Mid value = \(\frac{12.5+17.5}{2}=\frac{30.0}{2}=15 \neq 20\)

In the class, 17.5-22.5, class size = 22.5 17.5 = 5

Mid value = \(\frac{12.5+17.5}{2}\) = 20

The mid value of the last two classes is 20, but the class length is not 5.

∴ 2. 17.5 22.5

5. In a frequency distribution table if the mid-point of a class is 10 and the class size of each class is 6; the lower limit of the class is

1. 6
2. 7
3. 8
4. 12

Solution: Let upper limit of = x and lower limit = y.

According to the problem, 1st conditions: \(\frac{x+y}{2}=10\)

or, x + y = 20 …..(1)

According to the problem, 2nd conditions:
or, x – y = 6 ….(2)

Subtracting (2) from (1), 2y = 14

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

∴ Lower limit of the class
∴ 2. 7

Question 9.

1. In a continuous frequency distribution table if the mid-point of a class is m and the upper-class boundary is u, then let us find out the lower-class- boundary.

Solution: Mid value of the class = m
Upper boundary of the class = u
∴ Mid value = m
∴ Lower class boundary= 2 x mid value =2 x m-u = 2m – u

2. In a continuous frequency distribution table, if the mid-point of a class is 42 and class size is 10, then let us write the upper and lower limits of the class.

Solution: Let the upper-class boundary be x and the lower boundary be y. According to the problem, 1st condition

According to the problem, 1st condition \(\frac{x+y}{2}=42\)

or, x + y = 84 ……(1)

According to the problem, 2nd condition 2x = 94

or, \(x=\frac{94}{2}=47\)

Putting the value of x in equation (1), 47 + y = 84
⇒ y= 84-47 = 37
∴ Upper limit = 47
Lower limit 37

Class 9 Maths WBBSE

3. Let us write the frequency density of the first class of the frequency distribution table.

Class- limit	70-74	75-79	80-84	85-89
Frequency	3	4	5	8

Solution: 1st class = 70-74
Class size the 1st class = 74-70 = 4
Frequency of the 1st class = 3

Frequency dencing of the 1st class \(=\frac{\text { Frequency }}{\text { Classsize }}\)

\(=\frac{3}{4}\)= 0.75

4. Let us write the frequency density of the last class

Class- limit	70-74	75-79	80-84	85-89
Frequency	3	4	5	8

 

Solution: Last class = 85-89
Frequency = 8
Total frequency=3+4+5+8=20

∴ Frequency of the class \(y=\frac{\text { Frequency of the class }}{\text { Total frequency }}\)

\(\begin{aligned}
& =\frac{8}{20} \\
& =\frac{2}{5}=0.4
\end{aligned}\)

 

5. Let us write from the following examples which one indicates attribute and which one indicates variable.

1. Population of the family.
2. Daily temperature.
3. Educational value.
4. Monthly income.
5. Grade obtained in Madhyamik Examination.

Solution:
1. Population of the family – Variable
2. Daily temperature – Variable
3. Educational – Attribute
4. Monthly income – Variable
5. Grade obtained in – Attribute

Class 9 Maths WBBSE Chapter 11 Statistics Exercise 11.2

 

Question 1. I construct the frequency polygon for the following marks obtained by 75 learners of Pritha’s school:

Marks obtained	30	40	50	60	70	80
Number of students	12	18	21	15	6	3

 

In the graph paper, taking suitable measures along horizontal and vertical lines, the points (20, 0), (30, 12), (40, 18), (50, 21), (60, 15), (70, 6), (80, 3) and (90, 0) are plotted on the graph paper and then I draw the frequency polygon by adding them.

Solution: Along x-axis one side of the smallest square = 1 mark & along y-axis two sides of the smallest square 1 student. And putting the points and joining then ABCDEFGH frequency polygon is obtained.

 

 

Question 2. I draw the frequency polygon for the following frequency distribution table:

Classes	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	4	10	24	12	20	8

 

Solution:

Class	Mid value	Frequency
0-5	2.5	4
5-10	7.5	10
10-15	12.5	24
15-20	17.5	12
20-25	22.5	20
25-30	27.5	8

 

 

1. Class along x-axis & frequency along y-axis.
2. Taking the points of mid-value & frequency points are plotted, B(2.5, 4) C(7.5, 10), D(12.5, 24), E(17.5, 12), F(22.5, 20), G(27.5, 8)
3. Joining A, B, C, D, E, F, G, H with a scale we get, ABCDEFGH a frequency polygon.

Question 3. I write below in tabular form the daily profit of the 50 shops of the village of Bakultala:

Daily profit (Rs.)	0-50	50-100	100-150	150-200	200-250
Number of shops	8	15	10	12	5

 

 

Solution: Along x-axis – Daily profit
Along y-axis No. of shops

One side of the smallest square along x-axis = Rs. 10 (Profit); two sides of the smallest square along y-axis = 1 shop.

I draw the histogram for the above data.

Question 4. By measuring, Mita wrote the heights of her 75 friends of their school in the table given below : I draw the histogram of the data collected by Mita.

Height (cm.)	136-142	142-148	148-154	154-160	160-166
Number of friends	12	18	26	14	5

 

Solution: Along x-axis – height (in cm) &
along y-axis No. of friends

unit:- 5 sides of the smallest square along x-axis = 6 cm & 1 side of the smallest square along y-axis = 1 friend. Putting the value, I draw the histogram for the given data

 

 

Question 5. In our locality, by collecting the number of Hindi-speaking people between the  ages of 10 years to 45 years, I write them in the table given below:

Age (in years)	10-15	16-21	22-27	28-33	34-39	40-45
Number of Hindi-speaking people	8	14	10	20	6	12

 

Solution: First make a frequency distribution table

Class (year)	Class boundary	Length of class	Frequency
10-15	9.5 – 15.5	6	8
16-21	15.5 – 21.5	6	14
22-27	21.5-27.5	6	10
28-33	27.5 – 33.5	0.6	20
34-39	33.5 – 39.5	6	6
40-45	39.5 – 45.5	6	12

 

Along the x-axis – Age (year) & Along the y-axis – No. of Hindi-speaking people Unit 5 sides of the smallest square along the x-axis & 2 sides of the smallest square along the y-axis = 1 people

Putting the value, I draw the histogram.

 

 

Question 6. I draw the histogram of the frequency distribution table given below:

Class	1-10	11-20	21-30	31-40	41-50	51-60
Frequency	8	3	6	12	2	7

 

Solution: Frequency distribution table

Class	Class-boundary	Length of Class	Frequency
1-10	0.5 – 10.5	10	8
11-20	10.5-20.5	10	3
21-30	20.5 – 30.5	10	6
31-40	30.5 – 40.5	10	12
41-50	40.5 – 50.5	10	2
51-60	50.5 – 60.5	10	7

 

Along x-axis-class boundary
& Along y axis – Frequency along x-axis
unit one side of the smallest square = 2 units
& Four side of the smallest square along y-axis = 1 unit
Putting the value, Histogram is obtained

 

 

Question 7. By drawing the histogram, I draw the frequency polygon of the frequency distribution table given below:

Amount of subscriptions (Rs.)	20	25	30	35	40	45	50
Number of Members	20	26	16	10	418	6

 

Solution: Frequency distribution table

Class (Rs.)	Mid value	Frequency (no. of members)
17.5-22.5	20	20
22.5-27.5	25	26
27.5-32.5	30	16
32.5-37.5	35	10
37.5-42.5	40	4
42.5-47.5	45	18
47.5-52.5	50	6

 

Along x-axis Amount of Subscription (Rs.)
& Along y-axis Number of members (Frequency)

Unit 2 sides of the smallest square along the x-axis = Re. 1; 2 sides of the smallest square along y-axis = 1 member.

Now I draw the histogram.

Now for the drawing of frequency polygon just before the first class interval I take class interval 12.5 17.5 and just after class interval, I take class interval 52.5 –
57.5. The frequencies of these two class intervals are 0.

(30,16), (35,10), (40,4), (45,18), (50-6) & (55,0) successively with straight lines, I have drawn the frequency polygon ABCDEFGHI.

Class 9 Math Solution WBBSE In English

 

 

Question 8. I draw the histogram for the following frequency distribution table:

Number of children	0	1	2	3	4	5
Number of families		85	50	25	15	5

 

Hints: At first, by the exclusive class method the statistical data will be constructed as a frequency distribution table with class boundaries given below:

Number of children	0-1	1-2	2-3	3-4	4-5	5-6
Number of families	120	85	50	25	15	5

 

Solution: Frequency distribution table

Class (No. of children)	Frequency (No. of families)
0-1	120
1-2	85
2-3	50
3-4	25
4-5	15
5-6	5

 

Along x-axis – No of children & Along y-axis – Number of families

Unit: 5 sides of the smallest square along x-axis = 1 child. 2 sides of the smallest square along y-axis = 5 families

Putting the values Histogram is obtained.

 

Question 9. I have written the ages of 32 teachers of Primary Schools in the village of Virsingha in a table given below:

Ages (years)	25-31	31-37	37-43	43-49	49-55
Number of teachers	10	13	5	3	1

 

Now, we have to draw a histogram and frequency polygon with the given data graphically:

Solution: Frequency distribution table

Class Age (years)	Class boundary	Mid value	Length	Frequency(No. of teachers)
25-31	25-31	28	6	10
31-37	31 -37	34	6	13
37-43	37-43	40	6	5
43-49	43-49	-46	6	3
49-55	49-55	52	6	1

 

Along x-axis Age (years) & Along y-axis – The number of teachers

Unit: 1 side of the smallest square along x – axis = 1 year and 5 sides of the smallest square along the y-axis = 1 teacher.

Now I draw the histogram.

Class 9 Math Solution WBBSE In English

Now for drawing frequency polygon just before 1st class interval, I take a class interval of 19-25, and just after the last class interval, I take another class interval of 55- 61. The frequencies of these two class intervals are 0 (Zero).

Then by joining the points (22,0), (28,10), (34,13), (40,5), (46,3), (52,1), and (58,0) successively with straight lines, frequency polygon ABCDEFG is obtained.

Question 10. I draw the frequency polygon for the following frequency distribution table:

Class	75-80	80-85 I	85-90	90-100	100-105
Frequency	12	18	22	10	8

Solution: Frequency distribution table

Class	Mid value	Frequency
75-80	77.5	12
80-85	82.5	18
85-90	87.5	22
90-100	95	10
100-105	102.5	8

 

Along x-axis Class & along y-axis Frequency

Unit: 2 sides of the smallest square along x-axis = 1 unit and 2 sides of the smallest square along the y-axis= 1 unit.

Now for drawing frequency polygon just before 1st class interval I take a class interval of 70-75 and just after the last class interval, I take another class interval of 105- 110. The frequencies of the two class intervals are 0 (Zero).

Then by doing the points (72.5,0) & (107.5,0) successively with a straight line the freguency polygon ABCDEFG is obtained.

Question 11. I draw the frequency polygon for the following frequency distribution table.

Class	1-10	11-20	21-30	31-40	41-50
Frequency	8	3	6	12	4

 

Solution: Frequency distribution table

Class	Class boundary	Mid Value	Frequency
1-10	0.5-10.5	5.5	8
11-20	10.5-20.5	15.5	3
21-30	20.5-30.5	25.5	6
31-40	30.5 – 40.5	35.5	12
41-50	40.5 – 50.5	45.5	4

 

Along x-axis – Class & along y-axis – Frequency

Unit: 1 side of the smallest square along x-axis = 1 unit and 4 sides of the smallest square along y-axis = 1 unit

Now for drawing frequency polygon just before 1st class interval I take a class interval of -10,-0 and just after the last class interval, I take another class interval 51- 60.

The frequencies of these two class intervals are 0 (Zero). Then by joining the points (-4.5,0), (5.5,8), (15.5,3), (25.5,6), (35.5,12), (45.5,4), and (55.5,0) successively with straight lines we obtained ABCDEFG, frequency polygon.

 

 

Question 12. A special drive will be taken for women’s literacy in total in our village. For this reason, we have collected the following data:

Age	10-15	15-20	20-25	25-30	30-35
Number of illiterates	40	90	too	60	160

 

Solution: To draw the frequency polygon

Frequency distribution table

Class Age (years)	Mid value	Frequency (No. of illiterates)
10-15	12.5	40
15-20	17.5	90
20-25	22.5	100
25-30	27.5	60
30-35	32.5	160

 

Along x-axis-Class (Age) & along y-axis is the Number of illiterates

Unit: 2 sides of the smallest square along x axis = 1 year & 1 side of the smallest square along the y-axis = 5 number of illiterates.

Now for drawing frequency polygon just before 1st class interval I take a class interval 5-10 and just after the last class interval, I take another ass interval of 35-40.

The frequencies of these two class intervals are 0 (Zero). Then by joining the points (7.5,0), (12.5,40), (17.5,90), (22.5,100), (27.5,60), (32.5,160), and (37.5,0) straight lines we get the required frequency polygon ABCDEFG.

 

Question 13. I have written in the following the frequency of the number of goals given by the teams in our Kolkata football league in the previous month. I draw the frequency polygon for the representation of the data.

Scores	0	1	2	3	4	5	6
Frequency	15	20	12	8	6	3	1

 

Solution: Frequency distribution table

Score	Frequency
0	15
1	20
2	12
3	8
4	6
5	3
6	1

 

Along x-axis-score (goal) & along y-axis – Frequency

Unit 10 sides of the smallest square along x-axis = 1 unit and 2 sides of the smallest square along y-axis = 1 unit.

Now for drawing frequency polygon just before 1st class interval I take a class interval of -1,-0 and just after the last class interval, I take another class interval of 7-0.

The frequencies of these two class intervals are 0 (Zero). Then by joining the points (1,0), (0,15), (1,20), (2,12), (3,0), (4,6) and (5,3), (6,1) & (7,0) successively with straight lines we obtained frequency polygon ABCDEFGHI,

 

 

Question 14.  Let us discuss

1. Each of the area of each of the rectangle of a histogram is proportional to

1. The mid-point of that class
2. The class size of that class
3. The frequency of that class
4. The cumulative frequency of that class

Solution: 3. The frequency of that class

2. A frequency polygon is drawn

1. Upper limit of the class
2. Lower limit of the class
3. Mid-value of the class
4. Any value of the class

Solution 3. Mid-value of the class

3. To draw a histogram, the class

1. Along y-axis
2. Along x-axis
3. Along x-axis and y-axis both
4. In between x-axis and y-axis

Solution: 2. Along x-axis

4. In the case of drawing a histogram,

1. Frequency
2. Class boundary
3. Range.
4. Class size.

Solution: 4. Class size.

5. A histogram is the graphical representation of grouped data whose class- boundary and frequency are taken respectively,
1. Along the vertical axis and horizontal axis,
2. Only along the vertical axis,
3. Only along horizontal axis,
4. Along the horizontal axis and vertical axis.

Solution: 4. Along the horizontal axis and vertical axis.

 

 

 

Chapter 17 Theorems On Concurrence Exercise 17

Question 1. The bisectors of ∠B and ∠C of ABC intersect each other at point I. Let us prove that \(\angle B I C=90^{\circ}+\frac{\angle B A C}{2}\)

Solution:

 

 

∵ Bl and Cl respectively are the bisectors of ∠ABC and ∠ACB.

∴ ∠IBC = \(\frac{1}{2}\) ∠ABC and ∠ICB = \(\frac{1}{2}\) ∠ACB

In ΔBIC, ∠BIC+∠IBC + ∠ICB = 180°

or, ∠BIC+ \(\frac{1}{2}\) ∠ABC + \(\frac{1}{2}\) ∠ACB=180°

or, ∠BIC+ \(\frac{1}{2}\) + (∠ABC + ∠ACB) = 180°

or, ∠BIC + \(\frac{1}{2}\) (∠ABC+∠ACB+ ∠BAC) = 180° + \(\frac{1}{2}\) ∠BAC

Adding \(\frac{1}{2}\) ∠BAC on both sides)

or, ∠BIC+ \(\frac{1}{2}\)  x 180° = 180° + \(\frac{1}{2}\) ∠BAC

or, ∠BIC = 90° + \(\frac{\angle B A C}{2}\) Proved

Read and Learn More WBBSE Solutions For Class 9 Maths

Question 2. If the lengths of the three medians of a triangle are equal, let us prove that the triangle is an equilateral triangle.

Solution: Let ABC is a triangle whose three medians AD, BE and CF cut each other at O and AD = BE = CF.
To prove ABC is an equilateral triangle

 

 

Proof: Centroid divides the median in the ratio of 2: 1.

∴ AO = \(\frac{2}{3}\), BO = \(\frac{2}{3}\) BE and CO = \(\frac{2}{3}\) CF

∵ AD = BE = CF (∵ Lengths of 3 medians of a triangle are equal.)

∴ \(\frac{2}{3}\) AD = \(\frac{2}{3}\) BE = \(\frac{2}{3}\) CF

∴ AO = BO=CO
∴ AD – AO = BE – BO = CF-CO
or, OD = OE = OF

In ΔAOF and ΔCOD,
OF = OD, AO = CO
∠AOF = VOA ∠COD

∴ ΔAOF ≅ ΔCOD (S-A-S congruency)
∴ AF = CD
or, 2AF 2CD

∴ AB = BC   ……(1)

Similarly, from ABOD and AAOE, it can be proved that AE = BD.
or, 2AE = 2BD

∴ AC = BC ….(2)

∴ In ΔABC AB = BC = CA
∴ ΔABC is an equilateral triangle. Proved

Question 3. Let us prove that in an equilateral triangle, circumcentre, incentre, centroid, and orthocentre will coincide.

Solution:

 

 

 

ABC is an equilateral triangle.
To prove cireum center, incentre, other center & centroid will coincide.

Proof: From A, B & C perpendiculars are drawn on BC, CA & AB. The perpendiculars AD, BE & CF meet at G

In ΔABD and ΔACD,
1. AB = AC
2. ∠ABD = ∠ACD
3. ∠ADB = ∠ADC

∴ ΔABD ≅ ΔACD
∴ ∠BAD = ∠CAD

∴ AD, ∠BAC is the bisector of ∠BAC. Similarly, BE & CF are the bisectors of ∠ABC & ∠ACB respectively. G is the centroid of the triangle. BD
∴ AD is the median

Similarly, BE & CF are two medians.
∴ G is the centroid.

Three medians are perpendicular on respective sides. C is the circumcentre of the triangle. (Proved)

Question 4. AD, BE and CF are three medians of a triangle ABC. Let us prove that the centroid of ABC and DEF are the same point.

Solution:

 

 

To prove the centroid of ΔABC and ΔDEF are the same point.

Proof: Three medians AD, BE & CF meet at G.
∴ G is the centroid of ΔABC

In ΕABC, F & E are the midpoints of AB & AC respectively.
∴ EF II BC, i.e., EF II BD.

Similarly, DE II BF
∴ BDEF is a parallelogram.

Its diagonals BE & DF intersect each other at N.
∴ N is the mid-point of DF.

Similarly, in parallelogram AFDE, M is the midpoint of EF. The two medians EN & DM meet each other at G.
∴ G is the centroid of ADEF.
∴The centroid of ΔABC & ΔDEF is the same point.

Question 5. Let us prove that the two medians of a triangle are together greater than the third median.

Solution:

 

 

 

In ΔABC, AD, BE & CF are the medians
To prove, BE+ CF > AD

Produce AD to H such that GD = DH
B, H and C, H are joined.

Proof: In □BHCG, BD = CD (∵ AD is median)
GD = DH (By construction)
BC & GH bisect each other at D.

∴ BHCG is a parallelogram.
∴ BH = CG

Now, ΔBGH in BG + BH > GH
or, BG+CG > 2GD ( ∵ BH = CG and GD = DH)

or, \(\frac{2}{3}\) BE +\(\frac{2}{3}\) CF > 2. \(\frac{1}{3}\) AD

or, \(\frac{2}{3}\) BE+ \(\frac{2}{3}\) CF > \(\frac{2}{3}\) AD

or, BE+ CF > AD Proved

Question 6. AD, BE and CF are the three medians of ΔABC. Let us prove that

1. 4(AD + BE + CF) > 3(AB+ BC + CA);
2. 3(AB+ BC + CA) > 2(AD + BE + CF).

Solution: In triangle ABC, 3 medians AD, BE and CF cut each other at G. Prove that

1. 4(AD+ BE + CF) > 3(AB + BC + CA);
2. 3(AB+ BC + CA) > 2(AD + BE + CF)

 

 

Proof: ∵Medians of a triangle cut each other in a ratio-2: of 1 at the centroid.

∴ AG = \(\frac{2}{3}\) AD, BG = \(\frac{2}{3}\) BE & CG = \(\frac{2}{3}\) CF

In ΔABG, AG + BG > AB

In ΔBCG, BG + CG > BC

In ΔACG, AG + CG > CA

∴ 2 (AG + BG+CG) > AB + BC + CA

or, \(2\left(\frac{2}{3} A D+\frac{2}{3} B E+\frac{2}{3} C F\right)>A B+B C+C A\)

or, \(\frac{4}{3}\)(AD+BE+CF) > AB + BC + CA

or, 4(AD + BE + CF) > 3 (AB+ BC + CA)

2.  In ΔABD, AB + BD > AD

or, AB + \(\frac{1}{2}\) BC > AD ( ∵ D, is the midpoint of BC)

In ΔBCE, BC + CE > BE

or, BC+ \(\frac{1}{2}\) CA> BE ( ∵ E, is the midpoint of CA)

In ΔCAF, CA + AF > CF

or, CA + \(\frac{1}{2}\) AB > CF (∵ F is the midpoint of AB)

∴ AB + \(\frac{1}{2}\) BC+BC+ \(\frac{1}{2}\) CA + CA + \(\frac{1}{2}\) AB > AD+ BE + CF

or, \(\frac{3}{2}\) (AB+BC+ CA)> (AD + BE+CF)

or, 3(AB+ BC + CA) > 2(AD + BE + CF)

Question 7. Three medians AD, BE and CF of ΔABC intersect each other at point G. If the area of AABC is 36 sq cm, let us calculate

1. Area of ΔAGB;
2. Area of ΔCGE;
3. Area of quadrilateral ΔDGF.

 

 

Solution: AD is the median of ΔABC
∴ ΔABD = ΔACD
∴ ΔGBD = ΔGCD  [GD is the median of AGBC]

ΔABD- ΔGBD = ΔACD- ΔGCD
∴ ΔAGB = ΔAGC

Similarly,
ΔAGB = ΔBGC

∴ ΔAGB = ΔBGC = ΔAGC = \(\frac{1}{3}\) (ΔAGB+ΔBGC + ΔAGC)

= \(\frac{1}{3}\) ΔABC

∴ ΔAGB = \(\frac{1}{3}\) x 36 sq. cm = 12 sq. cm

2. ΔCGE = \(\frac{1}{2}\) ΔAGC

= \(\frac{1}{2}\).\(\frac{1}{3}\) ΔABC

= \(\frac{1}{6}\)ΔΑΒC

= \(\frac{1}{6}\) x 36 sq. cm = 6 cm

3. Area of quadrilateral BDGF

= \(\frac{1}{6}\) x area of ΔABC + \(\frac{1}{6}\) x area of ΔABC

= \(\frac{1}{3}\) x 36 sq. cm + \(\frac{1}{6}\) x 36 sq. cm

= (3+5) sq. cm = 12 sq. cm

Question 8. AD, BE and CF are the medians of ΔABC. If \(\frac{2}{3}\) AD = BC, then let us prove that the angle between two medians is 90°.

Solution:

 

 

Produce GD to point H, such that GD = DH. As G is the centroid of ΓABC,
∴ AG: GD = 2:1

∴ GD = \(\frac{1}{3}\) AD = DH (∵ GD = DH)

∴ GD = GD + DH

= \(\frac{1}{3}\) AD + \(\frac{1}{3}\) AD

= AD \(\frac{2}{3}\) = BC (given)

∵ AD is the median.
∴ BD = DC & GD = DH

∴ In quadrilateral BHCG, diagonals bisect each other at D.
∴ BHCG is a parallelagram.

∵ The other diagonals bisect equally  ∴BC= GH
∴ BHCG is a rectangle.

∴ ∠BGC = 90° = ∠EGF

Question 9. P and Q are the mid-points of sides BC and CD respectively of a parallelogram ABCD; the diagonals AP and AQ cut BD at the points K and L. Let us prove that, BK = KL = LD.

Solution:

 

 

Diagonal AC bisects BD at O.
Two medians of AADC, DO & AQ intersect at L.

∴ DL: LO = 2:1

∴ DL = \(\frac{2}{3}\) DO and LO = \(\frac{1}{3}\) DO

KO = \(\frac{1}{3}\) BO = \(\frac{1}{3}\) DO and BK = \(\frac{2}{3}\) BO = \(\frac{2}{3}\) DO

KL = LO+KO = \(\frac{1}{3}\) DO + \(\frac{1}{3}\) DO

= \(\frac{\mathrm{DO}+\mathrm{DO}}{3}=\frac{2}{3} \mathrm{DC}\)

∴ BK = KL = LD = \(\frac{2}{3}\) DO Proved

Question 10. Multiple choice questions

1. O is the circumcentre of ABC; if ∠BOC = 80°, the measure of ∠BAC is

1. 40°
2. 160°
3. 130°
4. 110°

 

 

 

Solution: ∠BOC = 2 ∠BAC

∴ ∠BAC = \(\frac{1}{2}\) ∠BOC

= \(\frac{1}{2}\) × 80° = 40°

Solution: 1. 40°

2. O is the orthocentre of ABC; if ∠BAC = 40°, the measure of ∠BOC is

1.  80°
2. 140°
3. 110°
4. 40°

 

 

Solution: Let in ΔABC AD, BE and CF on sides BC, CA, and AB are perpendicular; the orthocentre is O.

In □AFOE,

∠FOE = 360°-(∠OFA + ∠OEA + ∠FAE)
= 360° (90° + 90° + 40°) = 360° – 220° = 140°

∴ ∠BOC = VOA  ∠FOE = 140°

∴ 2. 140°

3. O is the orthocentre of ABC; if ∠BAC 40° then

1. 80°
2. 110°
3. 140°
4. 40°

 

 

Solution: ΔABC
∠A+∠B+∠C = 180°
or, 40°+ ∠B+∠C = 180°
or, ∠B+∠C= 180° – 40° = 140°

In ΔBOC,
∠BOC+∠OBC+∠OCB = 180°

or, ∠BOC + \(\frac{1}{2}\) ∠B+ \(\frac{1}{2}\) ∠C=180°

or, ∠BOC + \(\frac{1}{2}\)(∠B + ∠C) = 180°

or, ∠BOC+ \(\frac{1}{2}\) x 140° = 180°

or, ∠BOC+70° = 180°
or, ∠BOC = 180° – 70° = 110°

∴ 2.110°

4. G is the centroid of triangle ABC; if the area of GBC is 20 sq cm, then the area of ABC is

1.  24 sq. cm
2. 6 sq. cm
3.  36 sq. cm
4. none of them

 

 

Solution: In a triangle, ABC G is the centroid.
∴ ΔAGB = ΔGBC = ΔCGA

∴ \(\frac{1}{3}\) ΔABC = ΔGBC

or, ΔABC = 12 sq. cm

or, ΔABC = 3 x 12 sq. cm 36 sq. cm

∴ 3. 36 sq. cm

5. If the length of the circumradius of a right-angled triangle is 5 cm, then the length of its hypotenuse is

1. 2.5 cm
2. 5 cm
3. 10 cm
4. none of this

 

 

Solution: In a right-angled triangle, the circumcentre lies at the mid-point of the hypotenuse.

∵ Length of circumradius = 5 cm
∴ Lengths of hypotenuse 2 x 5 cm = 10 cm.

∴ 3. 10 cm

Question 11. Short answer type questions:

1. If the lengths of the sides of a triangle are 6 cm, 8 cm, and 10 cm, then let us write where the circumcentre of this triangle lies.

Solution: ∵ (10)² = (6)² + (8)²

∴ The triangle whose three sides are 6 cm, 8 cm, and 10 cm is a right-angled triangle. And we know that the circumcentre is at the midpoint of the hypotenuse.
∴ Circumcentre lies at the midpoint of the side of 10 cm.

2. AD is the median and G is the centroid of an equilateral triangle. If the length of side is 3√3 cm, then let us write the length of AG.

Solution: ∵ Median of an equilateral triangle = Height of the triangle

∴ AD = \(\frac{\sqrt{3}}{2}\) x side 3√3 = \(\frac{9}{2}\)

∴ AG = \(\frac{2}{3}\) AD= \(\frac{2}{3}\) x \(\frac{9}{2}\) cm = 3 cm.

3. Let us write how many points are equidistant from the sides of a triangle.

Solution: One point is equidistance from the sides of the triangle.

4. DEF is a pedal triangle of an equilateral triangle ABC. Let us write the measure of ∠FDA.

Solution: The triangle formed by the foot of perpendiculars from the vertices to the opposite sides in a triangle is called Pedal Triangle.
‘O’ is the orthocentre of ΔABC.

 

 

D, E, and F are the foot of perpendiculars of AD, BE & CF.
The triangle joining the points D, E, and F is a pedal triangle.
Δ DEF is a pedal triangle.

∵ ΔABC is an equilateral triangle.
∴ DA is the bisector of ∠FDE

∴ ∠FDA = \(\frac{1}{2}\) x 60° = 30°

5. ABC is an isosceles triangle in which ∠ABC = ∠ACB and median AD =  BC. If AB = √2 cm, let us write the length of the circumradius of this triangle.

Solution: ABC is an isosceles triangle whose ∠ABC = ∠ACB.
∴ AB AC = √2 cm

 

 

∵ AD is median
∴BD = CD

∵ AD = \(\frac{1}{2}\) BC

∴ BD = CD = AD

The height of an isosceles triangle is the median of that triangle.
∴ AD ⊥ BC

∴ In rt. angled ΔADB,
AD²+ BD² = AB²

or, AD² + AD² =   (√2)² [∵ BD = AD]

or, 2 AD²=2

or, AD² = \(\frac{2}{2}\) = 1

or, AD = √1
or, AD = 1

∴ Circumradius of the triangle = 1 cm.

 

 

Chapter 16 Circumference Of Circle

Wbbse Class 9 Maths Chapter 16 Definitions:

1. Circle: When a curved line rotates around a point keeping a fixed distance, then the figure formed by its rotation is called a circle.
2. Circumference: The curved line which forms the circle is called the circumference.
3. Centre of a circle: The point inside a circle from which the distance between any point on the circumference is equal, is called the center of the circle.

4. Radius: The distance between the center and the circumference is called the radius.
5. Diameter: The straight line passing through the center which touches the circumference on both sides is called the diameter.
6. Arc: A part of the circumference is called arc.

7. Chord: The straight line joining any two points on the circumference is called a chord. The longest chord is the diameter.
8. Cyclic Quadrilateral: The quadrilateral whose vertices lie on the circumference of a circle is called a cyclic quadrilateral.
9. Segment: A figure formed by an arc and a chord of a circle is called a segment.

Read and Learn More WBBSE Solutions For Class 9 Maths

Wbbse Class 9 Maths Chapter 16 Memorable Facts:

1. Infinite number of circles can be drawn from a point.
2. Infinite number of circles can be drawn from two points.
3. Only one circle can be drawn from three non-collinear points.

4. It is not possible to draw a circle from more than three points. If it is, then those points are called the same circle points.
5. There are an infinite number of diameters in a circle and all diameters are equal in length.
6. There are an infinite number of radii in a circle and all radii are equal in length.”

7. The part of a circle that is enclosed on one side by diameter and on the other by circumference is called a semi-circle.
8. Angle in a semi-circle is a right angle.
9. The greatest chord of a circle is its diameter.
10. All angles of the same segment of a circle are equal.

Wbbse Class 9 Maths Chapter 16  Formulae:

1. Circumference of a circle = 2πr=πd units
2. Perimeter of a semicircle = (x+2) units
3. Circumference of a semi-circle = πr units

 

Chapter 16 Circumference Of Circle Exercise 16

Question1. Let us calculate the perimeter of each of the following Images

 

 

 

 

Solution: DE = (CE-CD) = (8-5) m = 31m

In right-angled ΔADE

\(\begin{aligned}
& A E=\sqrt{A D^2+D E^2} \\
& =\sqrt{(4)^2+(3)^2} \mathrm{~m} \\
& =\sqrt{16+9} \mathrm{~m} \\
& =\sqrt{25} \mathrm{~m} \\
& =5 \mathrm{~m}
\end{aligned}\)

 

 

 

The perimeter of the semi-circle = πr²

\(\begin{aligned}
& =\frac{22}{7} \times \frac{4}{2} \mathrm{~m} \\
& =\frac{44}{7} \mathrm{~m} \\
& =6 \frac{2}{7} \mathrm{~m}
\end{aligned}\)

 

∴ Perimeter of 1st Image = (5+5+8+\(6 \frac{2}{7}\))m = \(24 \frac{2}{7}\) m

2. Perimeter of a semi-circle = πr

= \(\frac{22}{7} \times 7\) cm = 22 cm

∴ Perimeter of 2nd Image = (14+14 +14 +22) cm = 64 cm.

Question 2. Let us calculate how long the wire will be taken to make a circular ring of radius 35 meters.

Solution: Circular ring of radius = 35 m
∴ Circumference = 2πr

= \(2 \times \frac{22}{7} \times 35 \mathrm{~m}\)= 220 m

∴ 220 m is the required length of the wire.

Wbbse Class 9 Maths Chapter 16  Question 3.

The radius of a wheel of a train is 35 meters. If It makes 450 revolutions per minute, let us calculate the velocity of the train per hour.

Solution: The radius of the wheel of the train is = 0.35 m.
∴ Circumference of wheel = 2πr

= \(2 \times \frac{22}{7} \times \frac{35}{100} \mathrm{~m}\) = 2.2 m

∵ The wheel of the train moves in 1 minute = 450 revolution
∴ The wheel moves in 60 minutes = 60 x 450 revolutions.
∴ Wheel moves in 1 hr = 27000 revolution

∵ Wheel moves in 1 revolution = 2.2 m
∴ Wheel moves in 27000 revolution = 27000 × 2.2 m = 59400 m = 59.4 km

∴ Speed = 59.4 km/hr.

Question 4. The radius of the circular field of the village Amadpur is 280 meters. Chaltall wants to go around the field by walking with a speed 5.5 km/hour, let us calculate how long time will be taken by Chaltall to complete one revolution.

Solution: Radius of a circular field = 280 m.

∴ Circumference = 2πr = \(2 \times \frac{22}{7} \times \frac{35}{100} \mathrm{~m}\) = 1760 m

5.5 km 5.5 x 1000 m = 5500 m
1 hr = 60 minute = 60 x 60 sec = 3600 sec

∵ Chaitali goes 5500 m in 3600 sec

∴ She goes 1 m in \(\frac{3600}{5500}\) sec

∴ She goes 1760 m in \(\frac{3600}{5500} \times 1760\) sec

∴ She goes 1760 m in 1152 sec = 19 minutes 12 sec.

Wbbse Class 9 Maths Chapter 16 Question 5.

Tathagata bent a copper wire in the form of a rectangle whose length is 18 cm and breadth is 15 cm. I made a circle by bending this copper wire. Let us calculate the length of the radius of the circular copper wire.

Solution: Perimeter of the rectangle = 2(18+15) cm = 66 cm

If the wire is bent in form of a circle, its circumference is 66 cm.
Let the radius of the circle be r cm.

∴ Circumference of the circle = 2πr cm

\(\begin{aligned}
& =2 \times \frac{22}{7} r \mathrm{~cm} \\
& =\frac{44 r}{7} \mathrm{~cm}
\end{aligned}\)

 

 

 

As for questions,

 

 

 

\(
\frac{44 r}{7}=66
or, r=\frac{66 \times 7}{44}
or, r=\frac{21}{2}=10.5\)

∴ Radius = 10.5 cm.

Question 6. The perimeter of a semi-circular field is 108 meters. Let us calculate the diameter of the field.

Solution: Let the radius of the semi-circular field be r m.

∴ Perimeter of semi-circle = (πr+2r) m

\(\begin{aligned}
& =r(\pi+2) m \\
& =r\left(\frac{22}{7}+2\right) m \\
& =r\left(\frac{22+14}{7}\right) m \\
& =\frac{36 r}{7} m
\end{aligned}\)

As for question,

\(
\frac{36 r}{7}=108
or, r=\frac{108 \times 7}{36}\)

or, r = 21

∴ Diameter of the field = 2r m = 2 × 21 m = 42 m

Question 7. The difference between the circumference and the diameter of a wheel is 75 cm, let us calculate the length of the radius of this wheel.

Solution: Let the radius of the wheel be r cm.
∴ Circumference of the wheel 2πr cm
Diameter of the wheel = 2r cm

∴ Difference between the circumference and diameter = (2πr-2г) cm

\(\begin{aligned}
& =2 r(\pi-1) \mathrm{cm} \\
& =2 r\left(\frac{22}{7}-1\right) \mathrm{cm} \\
& =2 r\left(\frac{22-7}{7}\right) \mathrm{cm} \\
& =2 r \times \frac{15}{7} \mathrm{~cm}
\end{aligned}\)

 

As for question,

\(
2 r \times \frac{15}{7}=75
or, r=\frac{7-\times 7}{2 \times 15}
or, r=\frac{35}{2}\)

or, r = 17.5
∴ Radius = 17.5 cm.

Question 8. In a race, Puja and Jakir start to compete from the same point and same time on a circular track of a length of diameter 56 meter. When Puja finishes the race at the competition by 10 revolutions, Jakir is one revolution behind. Let us calculate how many meters is the length of the race and by how many meters Puja beats Jakir.

Solution: Radius of track = \(\frac{56}{2} \mathrm{~m}\) = 28 m

∴ Circumference = 2πr = \(2 \times \frac{22}{7} \times 28 \mathrm{~m}\) = 176 m

∴ Length of the race = 10 x 176 m = 1760 m
& Puja beats Jakir by 176 m.

Ganit Prakash Class 9 Solutions Question 9.

The perimeter of the borewell of our village is 440 cm. There is an equally wide stone parapet around this borewell. If the perimeter of the borewell with parapet is 616 cm, let us write by calculating how much is the width of the stone parapet.

Solution: Let the radius of the (borewell) be r₁ cm and the radius including the stone parapet be r₂ cm.

∴ Circumference of the borewell = 2πr₁ cm
∴ 2πr₁ = 440  ……(1)

∴ Circumference of the borewell including the stone parapet = 2πr₂ cm
∴ 2πr₂ =616   ……(2)

 

 

Subtracting (1) from (2),
2πr₂-2πr₁ = 616-440
or, 2π(r₂-r₁) =176

\(or, 2 \times \frac{22}{7}\left(r_2-r_1\right)=176
or, r_2-r_1=\frac{176 \times 7}{2 \times 22}\)

 

or, r₂ -r₁ = 28
∴ Width of stone parapet is 28 cm.

Question 10. Niyamat chacha of the village attaches the motor’s wheel with a machine’s wheel with a belt. The length of diameter of the motor wheel is 14 cm and the machine wheel is 94.5 cm. If the motor’s wheel revolves 27 times in a second then let us calculate how many times the machine’s wheel will revolve in an hour.

Solution: Radius of the wheel of motor (r) = \(\frac{14}{2}\) = 7cm

∴ Circumference of motor’s wheel = 2πr

= \(2 \times \frac{22}{7} \times 7 \mathrm{~cm}\) = 44cm

∴ In 1 revolution motor wheel moves 44 cm.
∵ In 27 revolutions it moves = 44 x 27 cm
∴ In 1 hour it moves = 3600 x 27 x 44 cm

Radius of the motor wheel = \(\frac{94.5}{2} \mathrm{~cm}\) = 47.25 cm

Circumference of the motor wheel = 2πr

= \(2 \times \frac{22}{7} \times \frac{4725}{100} \mathrm{~cm}\) = 297 cm

∴ The motor wheel can traverse a 297 cm distance in 1 revolution.

∴ The motor wheel can traverse 1 cm distance in \(\frac{1}{297}\) revolution

∴ The motor wheel can traverse a 3600×27 x 44 cm distance in \(\frac{1 \times 3600 \times 27 \times 44}{297}\) revolution.

∴ The motor wheel can traverse 3600 x 27 x 44 cm distance in 14400 revolution.
∴ In an hour the motor wheel will revolve 14400 times.

Question 11. The lengths of hour’s hand and minute’s hand are 8.4 cm and 14 cm respectively of our club clock. Let us calculate how much distance will be covered by each hand in a day.

Hints: Hour’s hand goes in 12 hours = \(2 \times \frac{22}{7} \times 8.4 \mathrm{~cm}\)

Minute’s hand goes in one hours = \(2 \times \frac{22}{7} \times 14 \mathrm{~cm}\)

Solution: Radius of hour hand (r) = 8.4 cm.

∴Hour’s hand will traverse distance in 12 hours = \(2 \times \frac{22}{7} \times 8.4 \mathrm{~cm}[/latex = 52.8 cm

∴ Hour’s hand will traverse distance in 1 hours = [latex]\frac{52.8}{12} \mathrm{~cm}\)

∴ Hour hand will move in a day (= 24 hours) twice = \(\frac{52.8}{12} \times 24 \mathrm{~cm}\)

 

 

∵ Circumference = \(2 \times \frac{22}{7} \times 14 \mathrm{~cm}\) = 88 cm.

∴ The minute hand will moves in a day (= 24 here) 24 x 88 cm = 2112 cm.
∴ In a day the hour’s hand and minute’s hand will move 105.6 cm and 2112 cm respectively.

Question 12. The ratio of diameters of two circles which are drawn by me and my friend Mihir is □:□ It was found by calculating that the ratio of perimeters of two circles is □:□

Solution: If the ratio of diameter = x:y of two circles then the ratio of their perimeter (circumference) is also x:y.

Ganit Prakash Class 9 Solutions Question 13.

The time that Rahim takes to cover up by running a circular field is 40 seconds less when he goes from one end to another end diametrically. The velocity of Rahim is 90 meters per minute. Let us calculate the length of the diameter of the field.

Solution: Let the radius of the circular field be r m.

∴ Circumference of the circular field = 2πr m.
Diameter of the circular field = 2r m.

∵ Rahim can travel a 1m distance in \(\frac{60}{90}\)

∴ Rahim can travel 2πr m distance in \(\frac{2}{3} \times 2 \pi r \text { sec }\)

∴ Rahim can travel 2 r m distance in \(\frac{2}{3} \times 2 r\)

By the problem,

\(
\frac{2}{3} \times 2 \pi r-\frac{2}{3} \times 2 r=40
or, \frac{2}{3} \times 2 r(\pi-1)=40
or, \frac{2}{3} \times 2 r\left(\frac{22}{7}-1\right)=40\)

 

\(or, \frac{2}{3} \times 2 r\left(\frac{22-7}{7}\right)=40
or, \frac{2}{3} \times 2 r \times \frac{15}{7}=40
or, r=\frac{40 \times 7 \times 3}{2 \times 2 \times 15}\)

or, r = 14
∴ Diameter of the field = 2r m = 2 x 14 m = 28 m.

Question 14. The ratio of perimeters of two circles is 2 : 3 and the difference of their length of radii is 2 cm. Let us calculate the lengths of diameters of the two circles.

Solution: The radius of the smaller circle = r cm
∴ Radius of the larger circle = (r + 2) cm
∴ Radius of the smaller circle circle = 2πr cm and circumference of the larger circle = 2π(r+2) cm

B.T.P.,
2πг: 2π(г+2)=2:3

\(or,$\frac{2 \pi r}{2 \pi(r+2)}=\frac{2}{3}
or, \frac{r}{r+2}=\frac{2}{3}\)

or, 3r = 2r+4

or, 3r – 2r=4
or, r = 4

∴ Diameter of the first circle = 2r cm
= 2 x 4 cm = 8 cm.

∴ Diameter of the second circle = 2(r + 2) cm
= 2(4+2) cm = 12 cm.

Ganit Prakash Class 9 Solutions Question 15.

The four maximum-sized circular plates are cut out of a brass plate of square size. Having an area of 196 sq. cm; let us calculate the circumference of each circular plate.

Solution: The area of the square plate is 196 sq. cm

∴ Length of each side = √196 cm = 14.cm

Diameter of each circle = \(=\frac{14}{2} \mathrm{~cm}\) = 7 cm

∴ The radius of the maximum-sized circular plate cut out of a square plate of side 7 cm will be 7 cm.

 

 

∴ Radius of each circular plate (r) = latex]\frac{7}{\dot{2}} \mathrm{~cm}[/latex]

∴ Circumference of each circular plate p = 2πr cm

= \(2 \times \frac{22}{7} \times \frac{7}{2} \mathrm{~cm}\) = 22 cm

Question 16. The time that Nashifer takes to cover up a circular field from one end to other end is 45 seconds less when he goes diametrically. Let us calculate the length of the diameter of this field.

Solution:

 

Let the radius of the circular field r m
∴ Semi-circumference of the circular field = r m
Diameter of the circular field = 2r m

∴ Nashifer covers up 80 m distance in 60 sec

∴ Nashifer covers up 1 m distance in \(\frac{60}{80} \mathrm{sec}\)

∴ Nashifer covers up r m distance in \(\frac{3}{4} \times \pi r \sec\)

∴ Nashifer covers up 2r m distance in \(\frac{3}{4} \times 2 r\)

B.T.P.,

\(
\frac{3}{4} \times \pi r-\frac{3}{4} \times 2 r=45
or, \frac{3}{4} r(\pi-2)=45\) \(\text { or, } \frac{3}{4} r\left(\frac{22}{7}-2\right)=45\) \(\text { or, } \frac{3}{4} r\left(\frac{22-14}{7}\right)=45\) \(\text { or, , } \frac{3}{4} r \times \frac{8}{7}=45\) \(\text { or, } r=\frac{45 \times 7 \times 4}{8 \times 3}\) \(\text { or, } r=\frac{105}{2}\)

 

Diameter of the field = 2r m

\(\begin{aligned}
& =2 \times \frac{105}{2} \mathrm{~m} \\
& =105 \mathrm{~m}
\end{aligned}\)

Ganit Prakash Class 9 Solutions Question 17.

Mohim takes 46 seconds and 44 seconds respectively to go around along the outer and the inner edges of a circular path with 7 meters 5 dcm width by a cycle. Let us calculate the diameter of the circle along the inner edge of the path.

Solution: Let the inner radius of the circular path = r m
Width of the path = 7 m 5 dcm = 7.5 m

∴ The other radius of the circular path = (r+7.5) m
∴ Inner circumference = 2πr m
And other circumference = 2(r+7.5) m

To cover up 2π(r+7.5) m distance it takes 46 seconds

∴ To cover up 1 m distance it takes \(\frac{46}{2 \pi(r+7.5)} \text { seconds }\)

∴ To cover up 2πr m distance it takes \(\frac{46 \times 2 \pi r}{2 \pi(r+7.5)} \text { seconds }\)

B.T.P..

 

 

 

\(\begin{aligned}
& \frac{46 \times 2 \pi r}{2 \pi(r+7.5)}=44 \\
& \text { or, } \frac{46 r}{r+7.5}=44
\end{aligned}\)

or, 46r = 44r + 330
or, 46r- 44r330
or, 2r=330

∴ Diameter of the circle along the inner edge of the path = is 330 m. Ans.

Class 9 Ganit Prakash Solutions

Question 18. The ratio of time taken by a cyclist to go around the outer and inner circumference of a circular path is 20: 19; if the path is 15 meters wide, let us calculate the length of the diameter of the inner circle.

Solution: Width of the path = 5 m.
Let the inner radius of a circular path be r m and
the speed of the cycle is v m/second.

∴ The outer radius of the circular path = (r + 5) m
∴ The inner and outer circumferences of the path is 2r m and 2π(r+5) m respectively.

 

 

∴ Cycle covers up v m distance in 1 sec

∴ Cycle covers up 1 m distance in \(\frac{1}{v} \sec\)

∴ Cycle covers up 2πr m distance in \(\frac{2 \pi r}{v} \sec\)

∴ Cycle covers up, 2π(r+5) m distance in \(\frac{2 \pi(r+5)}{v} \sec\)

B. T. P.,

\(\frac{2 \pi\left(r+5\right)}{v}: \frac{2 \pi r}{v}=20: 19\) \(\text { or, } \frac{\frac{2 \pi(r+5)}{v}}{\frac{2 \pi r}{v}}=\frac{20}{19}\) \(\text { or, } \frac{2 \pi(r+5)}{v} \times \frac{v}{2 \pi r}=\frac{20}{19}\) \(\text { or, } \frac{r+5}{r}=\frac{20}{19}\)

or, 20r = 19r+ 95
or, 20r – 19r = 95
or, r = 95

∴Length of the inner diameter of the path = 2r m
= 2 x 95 m 190 m.

Class 9 Maths Chapter 16 Circumference Of Circle Multiple Choice Questions

 

1. The ratio of the velocity of the hour’s hand and minute’s hand at a clock is

1. 1: 12
2. 12:1
3. 1:24
4. 24: 1

Solution: Let the radius of the clock be r unit.
∴ Circumference of the clock = 2πr unit.

∵ The hour’s hand in 12 hours covers up 2πr unit distance.

∴ The hour’s hand in 1 hours covers up \(\frac{2 \pi r}{12}\)

∵ The minute’s hand in 1 hour covers up 2πr unit distance.

∴ Rotio of the velocity of hour’s hand and minute’s hand = \(\frac{2 \pi r}{12}: 2 \pi r\)

\(\begin{aligned}
& =\frac{1}{12}: 1 \\
& =\frac{1}{12} \times 12: 1 \times 12 \\
& =1: 12
\end{aligned}\)

∴ 1. 1: 12

Class 9 Ganit Prakash Solutions

2. Soma takes \(\frac{\pi x}{100}\) minutes to go one complete round of a circular path. Soma will take how much time for going around the park diametrically?

1. \(\frac{x}{200}\) minute

2. \(\frac{x}{100}\) minute

3. \(\frac{\pi}{100}\) minute

4. \(\frac{\pi}{200}\) minute

Solution: Let the radius of the circular park be r m.
∴ Circumference of the circular park = 2πr m and
Diameter of the circular park = 2r m

∴ Soma will take to cover up 2πr m distance \(\frac{\pi x}{100}\) minute

∴ Soma will take to cover up 1 m distance \(\frac{\pi \mathrm{x}}{100.2 \pi \mathrm{r}}\) minute

∴ Soma will take to cover up 2r m distance \(\frac{\pi x .2 r}{100.2 \pi r}\) minute

∴ Soma will take to cover up 2r m distance \(\frac{x}{100}\) minut

∴ Soma will take time for going round the park diametrically \(\frac{x}{100}\) minute

∴ 2. \(\frac{x}{100}\) minute

3.  A circle is inscribed by a square. The length of a side of the square is 10 cm. The length of diameter of the circle is

1. 10 cm
2. 5 cm
3. 20 cm
4. 10√2 cm

 

 

Solution: Let ABCD is a square in which
AB = BC= CD = DA = 10 cm

∵ A circle is inscribed by the square.
∴ Length of the diameter of the circle Length of each side of the square

∴ 1. 10 cm

Class 9 Ganit Prakash Solutions

4. The minute’s hand of a clock is 7 cm. How much length does the minute’s hand move in 15 minutes?

1. 5√2 cm
2. 10√2 cm
3. 5 cm
4. 10 cm

 

 

Solution: Length of each side of the square 5 cm.

∴ Length of diagonal of the square = √2x side
= √2×5 cm =5√2 cm

∵ A square is inscribed in a circle.
∴ Diameter of the circle = 5√2 cm

Solution: 1. 5√2 cm

Wbbse Class 9 Ex 16

5. A circular ring is 5 cm wide. The difference between the outer and inner radii is

1. 5 cm
2. 2.5 cm
3. 10 cm
4. none of these.

 

 

Solution: Width of the circular ring = 5 cm.
Let the inner radius of the circular ring be r cm.

∴ The outer radius of the ring = (r+ 5) cm.
∴ Difference between outer and inner radius of the ring = {(r+5)-r} cm = (r+5-r) cm = 5 cm

Answer: 1. 5 cm

Class 9 Maths Chapter 16 Circumference Of Circle Short Answer Type Questions

1. Perimeter of a semi-circle is 36 cm. What is the length of the diameter?

Solution: Let the radius of the semi-circle be r cm.
∴ Perimeter of the semi-circle = (r+2r) cm
Diameter of the semi-circle = 2r cm.

 

 

B. T. P., πг+2r=36
or, r(x+2)=36

\(\text { or, } r\left(\frac{22}{7}+2\right)=36\) \(\text { or, } r\left(\frac{22+14}{7}\right)=36\) \(or, r \times \frac{36}{7}=36
or, r=\frac{36 \times 7}{36}\)

or, r = 7
∴ Length of the diameter of the semi-circle = 2r cm = 2 x 7 cm = 14 cm.

Class 9 Ganit Prakash Solutions 

2. The length of the minute’s hand is 7 cm. How much length will a minute’s hand go to rotate 90°?

Solution: Length of the minute’s hand of the watch = 7 cm

∴ Circumference of the clock = 2πr

\(2 \times \frac{22}{7} \times 7 \mathrm{~cm}\) = 44 cm

 

 

∵ To rotate 360° degrees the minute’s hand will cover 44 cm distance.

∴ To rotate 1° degrees the minute’s hand will cover \(\frac{44}{360} \mathrm{~cm}\) distance

∴ To rotate 90° degrees the minute’s hand will cover \(\frac{44}{360} \times 90\) distance.

∴ To rotate 90° degrees the minute’s hand will cover \(\frac{44}{360} \times 90\) cm distance.

 Wbbse Class 9 Ex 16 

3. What is the ratio of radii of the inscribed and circumscribed circles of a square?

Solution: Let each side of the square = a unit.

∴ Diameter of the inner circle = a unit.

∴ Radius of the inner circle = \(\frac{a}{2}\) unit.

 

 

Diameter of the circumcircle = Length of the diagonal of the square = √2a unit.

∴ Radices of the circumcircle = \(\frac{\sqrt{2} a}{2} \text { unit. }\)

∴ Ratio of radii of inscribed circle and circumscribed circle = \(=\frac{a}{2}: \frac{\sqrt{2} a}{2}=1: \sqrt{2}\)

4. The minute’s hand of a clock is 7 cm. How much longer does the minute’s hand move in 15 minutes?

Solution: Radius of the circle = 7 cm.

∴ Circumference of the clock = 2πr

= \(2 \times \frac{22}{7} \times 7 \mathrm{~cm}=44 \mathrm{~cm}\)

∴ The minute hand moves in 60 minutes = 44 cm.

∴ It will move in 15 minute =\(\frac{44}{60}\)

∴ The minute’s hand in 15 minutes will cover = \(\frac{44}{60} \times 15 \mathrm{~cm}\) = 11 cm.

Wbbse Class 9 Ex 16

5. What is the ratio of the side of a square and the perimeter of a circle when the length of the diameter of the circle is equal to the length of the side of the square?

Solution: Let the radius of the circle = r cm.
∴ One side of the square = 2r unit.

 

 

∴ Circumference 2πr cm.
∴ Length of one side of the square = 2r cm
= 4 × side
= 4 x 2r cm = 8r cm.

∴ Ratio of the side of the square & the circumference
= 2πг: 8r
= π: 4

\(\begin{aligned}
& =\frac{22}{7}: 4 \\
& =\frac{22}{7} \times 7: 4 \times 7=22: 28
\end{aligned}\)

∴ 11:14

Chapter 14 Construction Of A Triangle Exercise 14

Question 1. Pritam drew a quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm, and ∠ABC = 60°. I drew a triangle with an equal area of that quadrilateral.

Solution:

 

Construction: First, a straight line BX is drawn. From BX, BC is cut off equal to 6cm. On point B equal to a 60° angle ∠CBY is drawn. From BY equal to AB 5 is cut off Taking A as the center and taking a radius equal to 3 cm a radius is drawn. Again, taking C as a center and taking a radius equal to 4 cm, an arc is drawn.

Read and Learn More WBBSE Solutions For Class 9 Maths

Both arcs intersect each other at the D point. A, D, and C, D are joined. Consequently, ABCD, a quadrilateral whose sides AB = 5 cm, BC = 6 cm, CD = 4 cm, AD = 3 cm, and ∠ABC 60°, is formed.

Diagonal AC is drawn D From point parallel to AC a st. line is drawn which cuts BX at point E. A, and E is joined. ABE is the required triangle whose area is equal to quadrilateral ABCD.

Question 2. Sahana drew a quadrilateral ABCD of which AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 4.2 cm and diagonal AC = 6 cm. I draw a triangle with an equal area of the quadrilateral.

Solution:

 

 

Equal in area to quadrilateral ABCD, triangle ABE is drawn.

Question 3. Sahana drew a rectangle ABCD, of which AB = 4 cm and BC = 6 cm. I draw a triangle with an equal area of that rectangle.

Solution:

 

 

Equal in the area to rectangle ABCD, triangle ADE is drawn.

Question 4. I draw a quadrilateral ABCD of which BC = 6 cm, AB = 4 cm, CD = 3 cm, ∠ABC 60°, ∠BCD = 55°. I draw a triangle with an equal area of that quadrilateral of which one side is alongside AB and another side is alongside BC.

Solution:

 

 

Equal in area to quadrilateral ABCD, triangle ABE is drawn.

Question 5. I draw a square with side of 5 cm. I draw a parallelogram of which one angle is 60°, equal area to the square.

Solution:

 

 

Equal in area to ABCD a square of side 5 cm ΔABE is drawn and equal in area to ΔABE a parallelogram CEGF is drawn whose angle ∠FCE = 60°.

Question 6. I draw a square with a side 6 cm and I draw a triangle with an equal area to that square.

Solution:

 

 

A square of side 6 cm, ABCD is drawn and equal in area to this square, a triangle ABE is drawn.

Question 7. I draw a quadrilateral ABCD, of which AD and BC are perpendicular on side AB and AB = 5 cm, AD = 7 cm, and BC = 4 cm. I draw a triangle with an equal area of that quadrilateral of which one angle is 30°.

Hints: I draw a ΔABQ with equal area of quadrilateral ABCD. Taking BQ as base of ΔABC I draw another triangle with same base and between two parallels of which one angle is 30°.

Solution:

 

 

Construction: From AY straight line a 5 cm long segment, AB is cut off. On A and B. points respectively perpendiculars AX and BM are drawn. From AX a 7 cm long segment AD and from BM a 4 cm long segment BC are cut. C and D are joined. Consequently, quadrilateral ABCD is formed.

Equal in area to □ABCD ΔADP is drawn. From point D parallel to AP a straight line DN is drawn. On point A, ∠PAF = 30° is drawn. AF cuts DN at point E.

Joining E, P, ΔAPE is drawn whose area is equal to □ABCD and one angle is 30°.

Proof: ΔAPD = ΔBCD

Again, ΔAPD and ΔAPE lie on same base AP and in between parallel lines AP and DE.

∴ ΔAPD = ΔAPE
∴ ΔAPE = □ABCD (Both are equal to ΔAPD)

Question 8. I draw any pentagon ABCDE and draw a triangle equal in area to it, of which one vertex is C.

Solution:

 

 

ABCDE, a pentagon is drawn. AC and CE diagonals are drawn. B From the point parallel to CA a straight line is drawn which cuts extended EA at point P.

Again, from point D parallel to CE a straight line is drawn which cuts extended AE at point Q C, P and C, Q are joined. ΔCPQ is the required triangle equal in area to pentagon ABCDE.