1988 AHSME Problems/Problem 24

Problem

An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16$ , and one of the base angles is $\arcsin(.8)$ . Find the area of the trapezoid.

$\textbf{(A)}\ 72\qquad \textbf{(B)}\ 75\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 90\qquad \textbf{(E)}\ \text{not uniquely determined}$

Solution

Let the trapezium have diagonal legs of length $x$ and a shorter base of length $y$ . Drop altitudes from the endpoints of the shorter base to the longer base to form two right-angled triangles, which are congruent since the trapezium is isosceles. Thus using the base angle of $\arcsin(0.8)$ gives the vertical side of these triangles as $0.8x$ and the horizontal side as $0.6x$ . Now notice that the sides of the trapezium can be seen as being made up of tangents to the circle, and thus using the fact that "the tangents from a point to a circle are equal in length" gives $2y + 0.6x + 0.6x = 2x$ . Also, using the given length of the longer base tells us that $y + 0.6x + 0.6x = 16$ . Solving these equations simultaneously gives $x=10$ and $y=4$ , so the height of the trapezium is $0.8 \times 10 = 8$ . Thus the area is $\frac{1}{2}(4+16)(8) = 80$ , which is $\boxed{\text{C}}$ .

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1988 AHSME Problems/Problem 24

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