BTGmoderatorDC wrote:
If arc PQR above is a semicircle, what is the length of diameter PR ?
(1) a = 4
(2) b = 1
An INSCRIBED ANGLE is formed by two chords.
Thus, angle PQR is an inscribed angle.
An inscribed angle that intercepts the diameter is a RIGHT ANGLE.
Thus, angle PQR is a right angle, implying that triangle PQR is a RIGHT TRIANGLE.
In the figure above, PS is a height drawn through right angle PQR.
A height drawn through the right angle of a triangle forms SIMILAR TRIANGLES.
Proof:
If angle QPR = x and angle PQS = y, then x+y = 90.
Since angle PQR = 90, angle SQR = 90-y = x.
Since angle QSR = 90, angle SQR = x, and x+y=90, angle QRP = y.
Thus, all three triangles -- PQS, QRS and PQR -- have the SAME COMBINATION OF ANGLES, as shown in the figure above:
x - y - 90.
Triangles that have the same combination of angles are SIMILAR.
The legs of similar triangles are in the SAME RATIO.
Thus, in all 3 triangles:
(leg opposite x) : (leg opposite y) = (leg opposite x) : (leg opposite y).
In triangle PQS, (leg opposite x) : (leg opposite y) = 2/a.
In triangle QRS, (leg opposite x) : (leg opposite y) = b/2.
Since the two ratios are equal, we get:
2/a = b/2
ab = 4.
Statement 1: a=4
Since ab=4, b=1, implying that PR = 4+1 = 5.
SUFFICIENT.
Statement 2: b=1
Since ab=4, a=4, implying that PR = 4+1 = 5.
SUFFICIENT.
The correct answer is
D.
Problems that test the same concept:
https://www.beatthegmat.com/inscribed-tr ... 74152.html
https://www.beatthegmat.com/length-of-th ... 71979.html
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