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?

$19\frac14$

$20\frac14$

$21 \frac34$

$22\frac12$

$23\frac34$

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 $\frac{B}{H}=\frac{3}{H-3}$ , so $B=\frac{9}{2}$ . The area is $\frac12\cdot\frac92\cdot9=\frac{81}{4}=20\frac14$ .

Thus, the answer is B .

2021 AMC 10B Fall Problem 13

Solution:

Problem 13 in Other Years