AMC12 2023 B

AMC12 2023 B · Q11

AMC12 2023 B · Q11. It mainly tests Triangles (properties), Area & perimeter.

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

具有腿长为$1$且一条底边是另一条底边两倍长的等腰梯形的最大面积是多少？

(A) \frac 54 \frac 54

(B) \frac 87 \frac 87

(D) \frac 32 \frac 32

(E) \frac{3\sqrt3}4 \frac{3\sqrt3}4

Answer

Correct choice: (D)

正确答案：（D）

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 $[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}.$

设梯形为$ABCD$，$AD = BC = 1$，$AB = x$，$CD = 2x$。将$AD$和$BC$延长相交于点$E$。则$\triangle ABE \sim \triangle DCE$，边长比为$1:2$，且$AE = BE = 1$。因此，$[DCE] = 4 \cdot [ABE]$，$[ABCD] = [DCE] - [ABE] = \frac{3}{4} \cdot [DCE]$。问题归结为最大化$[DCE]$的面积，这是一个腿长为$2$的等腰三角形。分析正弦面积公式，显然当$\angle DEC = 90^{\circ}$时最大，因此$[DCE] = 2$，$[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}$。

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