AMC8 2017

AMC8 2017 · Q18

AMC8 2017 · Q18. It mainly tests Pythagorean theorem, Area & perimeter.

In the non-convex quadrilateral ABCD shown below, $\angle BCD$ is a right angle, AB = 12, BC = 4, CD = 3, and AD = 13. What is the area of quadrilateral ABCD?

如下所示的非凸四边形ABCD中，$\angle BCD$为直角，AB = 12，BC = 4，CD = 3，AD = 13。四边形ABCD的面积是多少？

(A) 12 12

(B) 24 24

(D) 30 30

(E) 36 36

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): In right triangle BCD, $3^2 + 4^2 = 5^2$, so $BD = 5$. In $\triangle ABD$, $13^2 = 12^2 + 5^2$, so $\triangle ABD$ is a right triangle with right angle $\angle ABD$. The area of $\triangle ABD$ is $\frac{1}{2} \cdot 5 \cdot 12 = 30$. The area of $\triangle BCD$ is $\frac{1}{2} \cdot 3 \cdot 4 = 6$. So the area of the quadrilateral is $30 - 6 = 24$.

答案 (B)：在直角三角形 BCD 中，$3^2+4^2=5^2$，所以 $BD=5$。在 $\triangle ABD$ 中，$13^2=12^2+5^2$，因此 $\triangle ABD$ 为直角三角形，且直角为 $\angle ABD$。 $\triangle ABD$ 的面积是 $\frac{1}{2} \cdot 5 \cdot 12 = 30$。 $\triangle BCD$ 的面积是 $\frac{1}{2} \cdot 3 \cdot 4 = 6$。因此该四边形的面积为 $30 - 6 = 24$。

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