AMC12 2017 B

AMC12 2017 B · Q18

AMC12 2017 B · Q18. It mainly tests Triangles (properties), Similarity.

The diameter AB of a circle of radius 2 is extended to a point D outside the circle so that BD = 3. Point E is chosen so that ED = 5 and line ED is perpendicular to line AD. Segment AE intersects the circle at a point C between A and E. What is the area of $\triangle ABC$?

半径为 2 的圆的直径 AB 被延长到圆外一点 D，使得 BD = 3。选择点 E 使得 ED = 5 且直线 ED 垂直于直线 AD。线段 AE 与圆相交于 A 和 E 之间的点 C。求 $\triangle ABC$ 的面积。

(A) \frac{120}{37} \frac{120}{37}

(B) \frac{140}{39} \frac{140}{39}

(D) \frac{140}{37} \frac{140}{37}

(E) \frac{120}{31} \frac{120}{31}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Because $\angle ACB$ is inscribed in a semicircle, it is a right angle. Therefore $\triangle ABC$ is similar to $\triangle AED$, so their areas are related as $AB^2$ is to $AE^2$. Because $AB^2 = 4^2 = 16$ and, by the Pythagorean Theorem, $AE^2 = (4 + 3)^2 + 5^2 = 74,$ this ratio is $\frac{16}{74} = \frac{8}{37}$. The area of $\triangle AED$ is $\frac{35}{2}$, so the area of $\triangle ABC$ is $\frac{35}{2}\cdot\frac{8}{37}=\frac{140}{37}$.

答案（D）：因为$\angle ACB$是半圆内接角，所以它是直角。因此$\triangle ABC$与$\triangle AED$相似，所以它们的面积之比等于$AB^2$与$AE^2$之比。因为$AB^2=4^2=16$，并且根据勾股定理， $AE^2=(4+3)^2+5^2=74,$ 这个比值是$\frac{16}{74}=\frac{8}{37}$。$\triangle AED$的面积是$\frac{35}{2}$，所以$\triangle ABC$的面积是$\frac{35}{2}\cdot\frac{8}{37}=\frac{140}{37}$。

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