2023 AMC 12B Problems/Problem 11

Problem

What is the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other?

$\textbf{(A) }\frac 54 \qquad \textbf{(B) } \frac 87 \qquad \textbf{(C)} \frac{5\sqrt2}4 \qquad \textbf{(D) } \frac 32 \qquad \textbf{(E) } \frac{3\sqrt3}4$

Solution 1

Denote by $x$ the length of the shorter base. Thus, the height of the trapezoid is $\begin{align*} \sqrt{1^2 - \left( \frac{x}{2} \right)^2} . \end{align*}$

Thus, the area of the trapezoid is $\begin{align*} \frac{1}{2} \left( x + 2 x \right) \sqrt{1^2 - \left( \frac{x}{2} \right)^2} & = \frac{3}{4} \sqrt{x^2 \left( 4 - x^2 \right)} \\ & \leq \frac{3}{4} \frac{x^2 + \left( 4 - x^2 \right)}{2} \\ & = \boxed{\textbf{(D) } \frac{3}{2}} , \end{align*}$

where the inequality follows from the AM-GM inequality and it is binding if and only if $x^2 = 4 - x^2$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2

~cantalon

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 12 Problems and Solutions

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2023 AMC 12B Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

See Also