Problem:
Circles with centers P,Q, and R, having radii 1,2, and 3, respectively, lie on the same side of line l and are tangent to l at P′,Q′, and R′, respectively, with Q′ between P′ and R′. The circle with center Q is externally tangent to each of the other two circles. What is the area of △PQR?
Answer Choices:
A. 0
B. 32
C. 1
D. 6−2
E. 23 Solution:
Let X be the foot of the perpendicular from P to QQ′, and let Y be the foot of the perpendicular from Q to RR′. By the Pythagorean Theorem,
P′Q′=PX=(2+1)2−(2−1)2=8
and
Q′R′=QY=(3+2)2−(3−2)2=24
The required area can be computed as the sum of the areas of the two smaller trapezoids, PQQ′P′ and QRR′Q′, minus the area of the large trapezoid, PRR′P′ :
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