2021 AMC 10B Fall Problem 13

Question

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?

19\frac14

20\frac14

21 \frac34

22\frac12

23\frac34

Use similar triangles to relate widths at different heights

The triangle widths are

2

and

3

at heights separated by

2

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},

so

B=\frac{9}{2}.

The area is

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

Thus, the answer is B .

Problem 13 in other years