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Suppose you are given a square with an inscribed circle if the area of the square is 100 meters squared what is the area of the circle?

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It reeks of homework. Could someone retag this homework please? –  Neil Apr 23 '12 at 14:17
@Neil I have been told that the protocol on this site is that, regardless of the reek, the homework tag should be added only by the original poster, not by others. –  Dilip Sarwate Apr 23 '12 at 17:45
@DilipSarwate is that so, I really did not know. Thanks, I guess :) –  Bidit Acharya Apr 23 '12 at 18:01
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2 Answers


Diameter of the circle is also the length of a side of the square, $10$ units.

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Would I do pir^2 with radious of five? –  El Cholo Grande Apr 23 '12 at 14:08
Exactly. BTW, this site allows you to use TeX mark up for the Math. So, you'd write the area of the circle $\pi r^2$ as $\pi r^2$. Regards, –  user21436 Apr 23 '12 at 14:09
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Note that area of a square is given by $s^2$ where $s$ is the length of one of its sides (whichever side it may be, it doesn't matter for they are all equal.

So if $s^2=100 \implies s=\sqrt{100}=10 \text{ meters}$. The side of your square is $ 10$ meters. A circle inscribed in it would look something like this


Now Note that in the diagram, Since $\bar{DC}=10\text{ meters} \implies \bar{XY}= \bar{DC}=10 \text{m}$.

So we now have that the diameter $XY$ of the circle at $O$ is $10\text{ meters}$.(I made a mistake while making the diagram, it should have been $10 \text{m}$ not $10 \text{cm}$, I hope this won't be much of a problem)
Figuring out the area of a circle with a given diameter is an elementary geometry problem. The area of the circle hence is $\pi\cdot\big(\frac12\cdot10\text{m}\big)^2=(25\pi)\;\text m^2$

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Why the down-vote? If it is because of the figure, I have mentioned about that in the answer. And it would be great if you didn't sulk behind in anonymity and tell me how I can improve the answer. –  Bidit Acharya Apr 24 '12 at 4:01
I gave this answer a +1 for filling in what my post lacked. –  user21436 Apr 24 '12 at 7:12
@KannappanSampath ahaha, thanks mate :) –  Bidit Acharya Apr 24 '12 at 11:28
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