Question

在梯形 $ABCD$ 中，边 $AB$ 和 $CD$ 相等。 $ABCD$ 的周长是

In trapezoid $ABCD,$ the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is

$27$

$30$

$32$

$34$

$48$

从 $B$ 作一条高

Drop an altitude from $B$

两侧直角三角形的水平直角边都是 $4$

The two side right triangles have horizontal leg $4$

设从 $B$ 到 $\overline{AD}$ 的高交 $\overline{AD}$ 于 $H$ 。

于是我们有 $AH = \dfrac{16 - 8}{2} = 4\text{，}$ 因为 $AB = CD$ 。

于是 $AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\text{。}$

周长为 $8 + 16 + 2 \cdot 5 = 34\text{。}$

所以正确答案是 D。

Let $H$ be where the altitude from $B$ to $\overline{AD}$ intersects $\overline{AD}.$

Then we have that $AH = \dfrac{16 - 8}{2} = 4,$ since $AB = CD.$

We then have that $AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5.$

Then the perimeter is $8 + 16 + 2 \cdot 5 = 34.$

Thus, D is the correct answer.