# Difference between revisions of "1953 AHSME Problems/Problem 48"

## Problem

If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is:

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5}$

## Solution

$[asy] draw((0,0)--(1,3)--(4,3)--(5,0)--cycle); draw((0,0)--(4,3)); draw((4,3)--(4,0)); draw((3.8,0)--(3.8,0.2)--(4,0.2)); label("A",(0,0),W); label("B",(1,3),NW); label("C",(4,3),NE); label("D",(5,0),E); label("E",(4,0),S); label("1",(2,1.5),NW); [/asy]$

Let $a$ be the length of the smaller base of isosceles trapezoid $ABCD$, and $1$ be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is $\frac a1=a$. Let point $E$ be the foot of the altitude from $C$ to $\overline{AD}$.

Since the larger base of the isosceles trapezoid equals a diagonal, $AC=AD=1$. Since the smaller base equals the altitude, $BC=CE=a$. Since the trapezoid is isosceles, $DE=\frac{1-a}{2}$, so $AE = 1-\frac{1-a}{2} = \frac{a+1}{2}$. Using the Pythagorean Theorem on right triangle $ACE$, we have $$a^2+\left(\frac{a+1}{2}\right)^2=1$$ Multiplying both sides by $4$ gives $$4a^2+(a+1)^2=4$$ Expanding the squared binomial and rearranging gives $$5a^2+2a-3=0$$ This can be factored into $(5a-3)(a+1)=0$. Since a must be positive, $5a+3=0$, so $a=\frac 35$. The ratio of the smaller base to the larger base is $\boxed{\textbf{(D) } \frac 35}$.