Question

四边形 $ABCD$ 是梯形， $AD = 15$ 、 $AB = 50$ 、 $BC = 20$ ，高为 $12$ 。这个梯形的面积是多少？

Quadrilateral $ABCD$ is a trapezoid, $AD = 15,$ $AB = 50,$ $BC = 20,$ and the altitude is $12.$ What is the area of the trapezoid?

$600$

$650$

$700$

$750$

$800$

可以作如下高，更容易求面积。

用勾股定理可得且 $DE = \sqrt{15^2 - 12^2} = 9$ $FC = \sqrt{20^2 - 12^2} = 16.$

还知道所以 $EF = AB = 50,$ $DC = DE + EF + FC = 75.$

因此 $ABCD$ 的面积为 $\dfrac{1}{2} \cdot (DC + 50) \cdot 12 = 6 \cdot 125$ $= 750.$

所以正确答案是 D。

We can drop the following altitudes to more easily find the area.

We can use the Pythagorean Theorem to get that $DE = \sqrt{15^2 - 12^2} = 9$ and $FC = \sqrt{20^2 - 12^2} = 16.$

We also know that $EF = AB = 50,$ so $DC = DE + EF + FC = 75.$

Then the area of $ABCD$ is $\dfrac{1}{2} \cdot (DC + 50) \cdot 12 = 6 \cdot 125$ $= 750.$

Thus, D is the correct answer.