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Explanation:

O is the centre of inscribed circle.
Consider one Quarter circle, perpendicular to this tangent will pass through the centre of Quarter circle to Point A.
For other Quarter circle, we do similar construction, the perpendicular to this tangent will pass through the centre of Quarter circle to Point B.
Intersection of these line must contains center O.

As sqauare side is 2, means radius of Quarter circle is 2.

Let inscribed cicle radius is r. So distance between 2 centre is 1-r.

Draw a perpendicular from radius of inscribed circle. Which will bisect side of Square at point E.

Using Pythagoras theorem on Triangle AOE.

1^2 + r^2 = (2-r)^2
4r=4-1
r=3/4

Area of circle: πr^2
= π(3/4)^2
= 9π/16

IMO-D

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unlike the similar one posted recently, this can be answered in under 2 mins :)


Answer : D

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Two quarter circles are drawn in a square and a circle is inscribed between the quarter circles. If the side of the square is 2 cm, what is the area of the circle?


A. \(\frac{1}{9}*\pi\)

B. \(\frac{9}{64}*\pi\)

C. \(\frac{3}{8}*\pi\)

D. \(\frac{9}{16}*\pi\) --> correct

E. \(\frac{3}{4}*\pi\)

Solution:
radius of the small circle = r
for triangle AOE

1^2+r^2 = (2-r)^2
=> 2*(2-2r) =1
=> r=3/4
so the area of the small circle= \(\pi*(\frac{3}{4})^2\)=\(\frac{9}{16}*\pi\)


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Two quarter circles are drawn in a square and a circle is inscribed between the quarter circles. If the side of the square is 2 cm, what is the area of the circle?
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IMO D the explanation is as the picture
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