2021 Fall AMC 10B Problems/Problem 13

Problem

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?

(The official problem came with a daigram...)

$(\textbf{A})\: 19\frac14\qquad(\textbf{B}) \: 20\frac14\qquad(\textbf{C}) \: 21 \frac34\qquad(\textbf{D}) \: 22\frac12\qquad(\textbf{E}) \: 23\frac34$

Solution

By similarity, the height is $3+\frac31\cdot2=9$ and the base is $\frac92\cdot1=4.5$ . Thus the area is $\frac{9\cdot4.5}2=20.25=20\frac14$ , or $\boxed{(\textbf{B})}$ .

~Hefei417, or 陆畅 Sunny from China

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2021 Fall AMC 10B Problems/Problem 13

Problem

Solution

See Also