2021 Fall AMC 10B Problem 13

Problem 13 of 25IntermediateGeometry

A square with side length

3

is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length

2

has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?

Show solution

Solution

Let the isosceles triangle have height

H

and base

B.

By similarity, horizontal widths in the triangle are proportional to distance from the top vertex. The top side of the larger square has width

3

and is

H-3

units from the top. The top side of the smaller square has width

2

and is

H-5

units from the top. Hence

\frac{3}{H-3}=\frac{2}{H-5}.

Solving gives

H=9.

Also

\frac{B}{H}=\frac{3}{H-3},

B=\frac{9}{2}.

The area is

\frac12\cdot\frac92\cdot9=\frac{81}{4}=20\frac14.

Thus, the answer is B .