1988 AMC 12 Problem 24

Question

An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is

16,

and one of the base angles is

\arcsin(0.8).

Find the area of the trapezoid.

72

75

80

90

not uniquely determined

For a circumscribed quadrilateral, the sums of opposite side lengths are equal

Use

\sin\alpha=\frac{4}{5}

and

\cos\alpha=\frac{3}{5}

to relate the leg, height, and difference of the bases

Let the shorter base be

b

and each leg be

\ell.

Tangency gives

16+b=2\ell.

If the base angle is

\alpha,

then

\sin\alpha=\frac{4}{5},\ \cos\alpha=\frac{3}{5},

and

16-b=2\ell\cos\alpha=(\frac{6}{5})\ell.

Solving gives

b=4,\ \ell=10,

and height

\ell\sin\alpha=8.

The area is

\frac12(16+4)(8)=80.

Thus the correct answer is C.

Problem 24 in other years