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03 Sep 2020, 20:00
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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57% (01:54) correct
44%
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wrong
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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. \(\frac{9}{16}*\pi\)
E. \(\frac{3}{4}*\pi\)
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03 Sep 2020, 20:36
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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
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
Attachments
circle.png [ 60.77 KiB | Viewed 7663 times ]
General Discussion
Jks3000
Joined: 28 Mar 2018
Last visit: 12 Nov 2025
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Location: India
Concentration: Strategy, Technology
Schools: Fuqua '24 McCombs '24 NTU NUS Kellogg '24
GMAT 1: 700 Q49 V36
GMAT 2: 730 Q50 V39
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03 Sep 2020, 20:23
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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
Posted from my mobile device
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
Posted from my mobile device
Attachments
File comment: Pic
IMG_20200904_084931.jpg [ 1.26 MiB | Viewed 7703 times ]
