soumya_joy wrote:On a semicircle with diameter AD, chord BC is parallel to the diameter. Further, each of
the chords AB and CD has length 2, while AD has length 8. What is the length of BC?
A. 6.0 B. 6.5 C. 7.0 D. 7.5
In any circle, an inscribed angle that intercepts the diameter is a RIGHT ANGLE.
Thus, ∠ACD is a right angle, and ∆ACD is a right triangle.
In any right triangle, a height drawn through the right angle forms SIMILAR triangles.
Thus, in ∆ACD, height CE forms the following similar triangles:
∆ACE, ∆CED, and ∆ACD.
As shown in the figure above, each of these triangles has the same combination of angles: x-y-90.
Corresponding sides of similar triangles are always in the SAME RATIO.
Thus, in ∆ACD and ∆CED:
(leg opposite angle x)/hypotenuse = (leg opposite angle x)/hypotenuse
2/8 = DE/2
DE = 1/2.
The same reasoning can be used to determine that AF = 1/2.
Thus, FE = AD - AF - DE = 8 - 1/2 - 1/2 = 7.
Since BC || AD, quadrilateral BCEF is a rectangle, implying that BC=FE.
Thus, BC=FE=7.
The correct answer is
C.
Last edited by
GMATGuruNY on Sat Nov 24, 2012 4:36 pm, edited 1 time in total.
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