# Difference between revisions of "1960 AHSME Problems/Problem 37"

## Problem

The base of a triangle is of length $b$, and the altitude is of length $h$. A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:

$\textbf{(A)}\ \frac{bx}{h}(h-x)\qquad \textbf{(B)}\ \frac{hx}{b}(b-x)\qquad \textbf{(C)}\ \frac{bx}{h}(h-2x)\qquad \textbf{(D)}\ x(b-x)\qquad \textbf{(E)}\ x(h-x)$

## Solution

Let $AB=b$, $DE=h$, and $WX = YZ = x$. $[asy] pair A=(0,0),B=(56,0),C=(20,48),D=(20,0),W=(10,0),X=(10,24),Y=(38,24),Z=(38,0); draw(A--B--C--A); draw((10,0)--(10,24)--(38,24)--(38,0)); draw(C--D); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot((20,24)); label("A",A,S); label("B",B,S); label("C",C,N); label("D",D,S); label("W",W,S); label("X",X,NW); label("Y",Y,NE); label("Z",Z,S); label("N",(20,24),NW); [/asy]$ Since $CD$ is perpendicular to $AB$, $ND = WX$. That means $CN = h-x$. The sides of the rectangle are parallel, so $XY \parallel WZ$. That means by AA Similarity, $\triangle CXY \sim \triangle CAB$. Letting $n$ be the length of the base of the rectangle, that means $$\frac{h-x}{n} = \frac{h}{b}$$ $$n = \frac{b(h-x)}{h}$$ Thus, the area of the rectangle is $\frac{bx}{h}(h-x)$, which is answer choice $\boxed{\textbf{(A)}}$.

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