HOT Competition 3 Sep/8PM: Two quarter circles are drawn in a square a

Question

https://gmatclub.com/forum/download/file.php?id=57266 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 E. \frac{3}{4}*\pi Untitled.png

Rogermoris · Accepted Answer

ON= OM=r, AB=2

AM = 1 (due to symmetry)

By inspection, it can be observed that the circle with center O touches the circle with center A, internally at N. If two circles touch each other internally, then the distance between their centers is the difference of their radii.

AN=AB= 2

OA = Distance between the centers of circles = AN-ON= 2-r

From the right angled triangle AOM,

OA²= AM²+OM²

(2-r)² = (1)²+r²

4+r²-4r = 1+r²

1-r = 1/4

r = 3/4

Area of circle=  pi*r^2  =  pi*\frac{9}{16}   OPTION D

Jks3000 · Answer

Draw 2 tangent to the inscribed circle at a point where quarter circle touches it. And then draw a perpendicular to those 2 tangents. These perpendicular lines will pass through the center of the circle. So, the point where these 2 perpendicular lines will intersect will be the center of circle.

Let r be the radius of inscribed circle.

So, Applying Pythagoras Theorem, (Refer to image below)

(2-r)^2 = ( 1^2 ) + r^2

Solving we get, r = 3/4

Therefore, Area = π(9/16)

Answer is  Option D

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rajatchopra1994 · Answer

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

Posted from my mobile device

petrichor · Answer

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

Answer : D Posted from my mobile device

hiranmay · Answer

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= $\pi*(\frac{3}{4})^2$=$\frac{9}{16}*\pi$

Posted from my mobile device

jhavyom · Answer

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?

NeoNguyen1989 · Answer

IMO D the explanation is as the picture

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HOT Competition 3 Sep/8PM: Two quarter circles are drawn in a square a

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HOT Competition 3 Sep/8PM: Two quarter circles are drawn in a square a

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