# 1987 AHSME Problems/Problem 21

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

There are two natural ways to inscribe a square in a given isosceles right triangle. If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$. What is the area (in $\text{cm}^2$) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below? $[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (-25,10), W); label("B", (-25,0), W); label("C", (-15,0), E); label("Figure 1", (-20, -5)); label("Figure 2", (5, -5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); [/asy]$ $\textbf{(A)}\ 378 \qquad \textbf{(B)}\ 392 \qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 441 \qquad \textbf{(E)}\ 484$

## Solution

We are given that the area of the inscribed square is $441$, so the side length of that square is $21$. Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$, then the legs of the larger isosceles right triangle ( $BC$ and $AB$) are equal to $42$. $[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); label("S", (25/3,11/6), E); label("S", (11/6,25/3), E); label("S", (5,5), NE); [/asy]$

We now have that $3S=42\sqrt{2}$, so $S=14\sqrt{2}$. But we want the area of the square which is $S^2=(14\sqrt{2})^2= \boxed{\mathrm{(B)}\ 392}$

## See also

 1987 AHSME (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS