2023 AMC 12B 第 11 题

Question

一个等腰梯形的两腰长为

1

，且一条底边是另一条底边的两倍。这个等腰梯形的最大面积是多少？

What is the maximum area of an isosceles trapezoid that has legs of length

1

and one base twice as long as the other?

\dfrac{5}{4}

\dfrac{8}{7}

\dfrac{5\sqrt{2}}{4}

\dfrac{3}{2}

\dfrac{3\sqrt{3}}{4}

作出两条高；每条腰的水平跨度为

\frac{\text{长底}-\text{短底}}{2}

Drop the two heights; each leg spans a horizontal distance of

\frac{\text{long base}-\text{short base}}{2}

设短底为

a

，面积为

\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}

；最大化它的平方

With shorter base

a,

the area is

\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}};

maximize its square

设两条底边为

a

和

2a

。每条腰的水平偏移为

\tfrac{a}{2}

，所以高为

\sqrt{1-\tfrac{a^2}{4}}

，面积为

A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}

。于是

A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right)

，当

a^2=2

时最大。此时高为

\tfrac{1}{\sqrt2}

，且

A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}

。

因此，正确答案是 D。

Let the bases be

a

and

2a.

Each leg has horizontal offset

\tfrac{a}{2},

so the height is

\sqrt{1-\tfrac{a^2}{4}}

and the area is

A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}.

Then

A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right),

maximized when

a^2=2.

There the height is

\tfrac{1}{\sqrt2}

and

A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}.

Thus, the correct answer is D.

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