AMC8 2003

AMC8 2003 · Q21

AMC8 2003 · Q21. It mainly tests Triangles (properties), Area & perimeter.

The area of trapezoid ABCD is 164 cm$^2$. The altitude is 8 cm, AB is 10 cm, and CD is 17 cm. What is BC, in centimeters? [figure]

梯形ABCD的面积是164 cm$^2$。高是8 cm，AB是10 cm，CD是17 cm。BC的长是多少厘米？[figure]

(A) 9 9

(B) 10 10

(D) 15 15

(E) 20 20

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Label the feet of the altitudes from $B$ and $C$ as $E$ and $F$ respectively. Considering right triangles $AEB$ and $DFC$, $AE=\sqrt{10^2-8^2}=\sqrt{36}=6\text{ cm}$, and $FD=\sqrt{17^2-8^2}=\sqrt{225}=15\text{ cm}$. So the area of $\triangle AEB$ is $\frac12(6)(8)=24\text{ cm}^2$, and the area of $\triangle DFC$ is $\left(\frac12\right)(15)(8)=60\text{ cm}^2$. Rectangle $BCFE$ has area $164-(24+60)=80\text{ cm}^2$. Because $BE=CF=8\text{ cm}$, it follows that $BC=10\text{ cm}$.

（B）将从 $B$ 和 $C$ 作出的高的垂足分别标为 $E$ 和 $F$。考虑直角三角形 $AEB$ 和 $DFC$，$AE=\sqrt{10^2-8^2}=\sqrt{36}=6\text{ cm}$，且 $FD=\sqrt{17^2-8^2}=\sqrt{225}=15\text{ cm}$。因此，$\triangle AEB$ 的面积为 $\frac12(6)(8)=24\text{ cm}^2$，$\triangle DFC$ 的面积为 $\left(\frac12\right)(15)(8)=60\text{ cm}^2$。矩形 $BCFE$ 的面积为 $164-(24+60)=80\text{ cm}^2$。因为 $BE=CF=8\text{ cm}$，所以 $BC=10\text{ cm}$。

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