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25 Aug 2021, 08:00
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E
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Difficulty:
95%
(hard)
Question Stats:
42% (01:56) correct
58%
(02:17)
wrong
based on 137
sessions
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Time
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An yellow semicircle and its tangent blue quarter-circle are drawn in a square as shown above. If AB is tangent to both yellow semicircle and blue quarter-circle, what is the length of AB ?
A. 10
B. 12
C. 14
D. 16
E. 18
M37-24
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M37-24
Attachment:
Untitled14.png
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25 Aug 2021, 08:30
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Itried to explain in the photo below.
My answer is E
My answer is E
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22 Nov 2021, 05:06
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Official Solution:
An yellow semicircle and its tangent blue quarter-circle are drawn in a square as shown above. If AB is tangent to both yellow semicircle and blue quarter-circle, what is the length of AB ?
A. \(10\)
B. \(12\)
C. \(14\)
D. \(16\)
E. \(18\)
Join the centers of the circles and the tangency points (recall that the radius of a circle is perpendicular to the tangent line). This would mean that CD is a straight line. Since CD equals to the radius of the semicircle plus the radius of the quarter circle, then \(CD = 6 + 12= 18\)
Draw line segment AE perpendicular to FD.
In a right triangle BGD, \(\angle BDG+\angle GBD =90°\).
In a right triangle CFD, \(\angle BDG+\angle DCF=90°\). So, \(\angle GBD=\angle DCF\).
In a right triangle ABE, \(\angle BAE+\angle GBD=90°\). So, \(\angle BAE=\angle BDG\).
So, two right tringles ABE and DCF are congruent: all angles are equal AND \(AE = DF = side \ of \ the \ square\). Therefore, the hypotenuses of these triangles must be equal: \(AB = CD = 18\)
Answer: E
An yellow semicircle and its tangent blue quarter-circle are drawn in a square as shown above. If AB is tangent to both yellow semicircle and blue quarter-circle, what is the length of AB ?
A. \(10\)
B. \(12\)
C. \(14\)
D. \(16\)
E. \(18\)
Join the centers of the circles and the tangency points (recall that the radius of a circle is perpendicular to the tangent line). This would mean that CD is a straight line. Since CD equals to the radius of the semicircle plus the radius of the quarter circle, then \(CD = 6 + 12= 18\)
Draw line segment AE perpendicular to FD.
In a right triangle BGD, \(\angle BDG+\angle GBD =90°\).
In a right triangle CFD, \(\angle BDG+\angle DCF=90°\). So, \(\angle GBD=\angle DCF\).
In a right triangle ABE, \(\angle BAE+\angle GBD=90°\). So, \(\angle BAE=\angle BDG\).
So, two right tringles ABE and DCF are congruent: all angles are equal AND \(AE = DF = side \ of \ the \ square\). Therefore, the hypotenuses of these triangles must be equal: \(AB = CD = 18\)
Answer: E
