Track / Stage 3 / 103 of 260 #103 of 2000

Problem 103

AMC 10/12, early questions

Geometry Difficulty 3.4 Find the answer

What is the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other?
$\textbf{(A) }\frac 54 \qquad \textbf{(B) } \frac 87 \qquad \textbf{(C)} \frac{5\sqrt2}4 \qquad \textbf{(D) } \frac 32 \qquad \textbf{(E) } \frac{3\sqrt3}4$

Your answer

Official solution

Let the trapezoid be $ABCD$ with $AD = BC = 1, \; AB = x, CD = 2x$ . Extend $AD$ and $BC$ to meet at point $E$ . Then, notice $\triangle ABE \sim \triangle DCE$ with side length ratio $1:2$ and $AE = BE = 1$ . Thus, $[DCE] = 4 \cdot [ABE]$ and $[ABCD] = [DCE] - [ABE] = \frac{3}{4} \cdot [DCE]$ .
The problem reduces to maximizing the area of $[DCE]$ , an isosceles triangle with legs of length $2$ . Analyzing the sine area formula, this is clearly maximized when $\angle DEC = 90^{\circ}$ , so $[DCE] = 2$ and $[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}.$
-PIDay

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.