AMC10 2009 A

AMC10 2009 A · Q10

AMC10 2009 A · Q10. It mainly tests Pythagorean theorem.

Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD = 3$, and $DC = 4$. What is the area of $\triangle ABC$?

三角形 $ABC$ 在 $B$ 处为直角。$D$ 是 $B$ 垂足，$AD = 3$，$DC = 4$。$ riangle ABC$ 的面积是多少？

(A) $4\sqrt{3}$ $4\sqrt{3}$

(B) $7\sqrt{3}$ $7\sqrt{3}$

(D) $14\sqrt{3}$ $14\sqrt{3}$

(E) 42 42

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): By the Pythagorean Theorem, $AB^2 = BD^2 + 9$, $BC^2 = BD^2 + 16$, and $AB^2 + BC^2 = 49$. Adding the first two equations and substituting gives $2 \cdot BD^2 + 25 = 49$. Then $BD = 2\sqrt{3}$, and the area of $\triangle ABC$ is $\frac{1}{2}\cdot 7 \cdot 2\sqrt{3} = 7\sqrt{3}$.

答案（B）：由勾股定理，$AB^2 = BD^2 + 9$，$BC^2 = BD^2 + 16$，且 $AB^2 + BC^2 = 49$。将前两个方程相加并代入得 $2 \cdot BD^2 + 25 = 49$。因此 $BD = 2\sqrt{3}$，并且 $\triangle ABC$ 的面积为 $\frac{1}{2}\cdot 7 \cdot 2\sqrt{3} = 7\sqrt{3}$。

Topics

GE.005 Pythagorean theorem