Question

13. A circular pond is surrounded by a 2 m wide circular path. If outer circumference of circular
path is 44 m, find the inner circumference of the circular path. Also, find area of the path.
​

Answer 1

Answer:

Let R be radius of outer circle and r be radius of inner circle.

It has been given that 44m is the circumference of outer circle,

⇒2πR=44m

⇒2×

7

22

×R=44

⇒R=

2×

7

22

44

=

2×22

7×44

=7m

It is given that circular path is 2 m wide.

So, radius of inner circle r=R−2=7−2=5 m

So, inner circumference of path =2πr=2×

7

22

×5=31.42 m

Therefore,

Area of the path = area of outer circle - area of inner circle

⇒π(R

2

−r

2

)

=

7

22

(7

2

−5

2

)=

7

22

×24

=75.43m

2

Answer 2

⠀⠀⠀⠀⠀⠀⠀${\huge{\underbrace{\rm{Question}}}}$

A circular pond is surrounded by a 2 m wide circular path. If outer circumference of circular path is 44 m, find the inner circumference of the circular path. Also, find area of the path.

⠀⠀⠀⠀⠀⠀⠀⠀${\huge{\underbrace{\rm{Answer}}}}$

Given:

⠀⠀

The circular pond is surrounded by a 2 m wide circular path.

The outer circumference of circular path is 44 m

⠀⠀

To find:

⠀⠀

• find the inner circumference of the circular path.

• Also, find area of the path.

⠀⠀

Solution:

⠀⠀

Let the inner radius of the path is ' r '

As the circular path is 2 m wide.

So, the outer radius of the path is (r + 2)

.°. The outer circumference = 2π (r + 2)

⠀⠀

According to the question,

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies 2π(r+2)=44}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies 2×\dfrac{22}{7}(r+2)=44}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies (r+2)=\dfrac{7}{22}×\dfrac{1}{2}×44}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies (r+2)=7}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies r=7-2}$

⠀⠀

⠀⠀⠀⠀⠀⠀$\boxed{\bf{\pink{:⟹\:r\:=\:5}}}$

⠀⠀

Now,

⠀⠀

The inner circumference of the path is

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies 2πr}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies 2×\dfrac{22}{7}×5\:m}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies \dfrac{220}{7}\:m}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies 31.43\:m}$

⠀⠀

Therefore,

⠀⠀

$\dashrightarrow\:\:\underline{\boxed{\sf the\:inner\:circumference\:of\:the\:path=31.43\:m}}$

Now,

Area of the circular path is

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies π(r+2)^{2}-πr^{2}}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies \dfrac{22}{7}×[(5+2)^{2}-5^{2}]m^{2}}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies \dfrac{22}{7}×[49-25]\:m^{2}}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies \dfrac{22}{7}×24\:m^{2}}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies \dfrac{24×22}{7}\:m^{2}}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies {\cancel{\dfrac{528}{7}}}\:m^{2}}$

⠀⠀

⠀⠀⠀⠀⠀$\sf{:\implies 75.43\:m^{2}......(approx)}$

⠀⠀

Hence,

⠀⠀

$\dashrightarrow\:\:\underline{\boxed{\sf area\:of\:the\:circular\:path\:=\:75.43\:m^{2}.....(approx)}}$

⠀⠀

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