A circular swimming pool is surrounded by a wall with a height of 2m.

Question

A circular swimming pool is surrounded by a walk with a width of 2m. The area of the walk is 21% of the area of the swimming pool. What is the area of the swimming pool?

A.  400π   m^2  
B.  300π   m^2  
C.  250π   m^2  
D.  200π   m^2  
E.  100π   m^2

yashikaaggarwal · Answer

Sir, kindly check the height of wall. Are you sure it's not 1m?
Because the last value I m getting for r(h+r)=42
And if h is 2, r will be of some imaginary value. 
Kindly revert.

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SiddharthR · Answer

This question seems incomplete or isn't worded properly. Can someone help

rsrighosh · Answer

Can anybody please explain how answer is 400 pi sq m

MathRevolution · Answer

There were mistakes in words. They are fixed now.
The words "wall" and "height" should be fixed with "walk" and "width", respectively.

Sorry about the mistakes.

Happy Studying !!!

yashikaaggarwal · Answer

Let the radius of walk with the swimming pool in between = r+2
Area of whole area ( swimming pool+ walk) = π(r+2)^2

Given: area of walk is 21% of area of swimming pool.

Area of walk = π(r+2)^2 - πr^2
Therefore,
π(r+2)^2 - πr^2 = 21/100(πr^2)
π{(r+2)^2 - r^2}=π{21/100(r^2)}
(r^2+4+4r)- r^2={21/100(r^2)
4+4r=21/100(r^2)
400+400r=21(r^2)
21(r^2)-400r-400=0
21(r^2)-420r+20r-400=0
21r(r-20)+20(r-20)=0
(r-20)(21r+20)=0
R=20 and -20/21(not possible)
R=20

Area of swimming pool is πr^2
π(20)^2 => 400π

Answer is A

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MathRevolution · Answer

=>

Assume the radius of the swimming pool is  r .

Then the area of the wall is  2πr·2 = 4πr.

Since we have  4πr = (\frac{21}{100}) πr^2,  we have  21r^2 – 400r + 400 = 0  or  (21r + 20)(r - 20) = 0.

Thus, we have  r = 20.

Then, the area of the swimming pool is  πr^2 = 400π.

Therefore, A is the answer.
Answer: A

bumpbot · Answer

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A circular swimming pool is surrounded by a wall with a height of 2m.

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A circular swimming pool is surrounded by a wall with a height of 2m.

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