AMC10 2017 B

AMC10 2017 B · Q22

AMC10 2017 B · Q22. It mainly tests Similarity, Pythagorean theorem.

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 △ABC ?

半径为2的圆的直径AB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。△ABC的面积是多少？

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

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

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

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

Answer

Correct choice: (D)

正确答案：（D）

Solution

Because ∠ACB is inscribed in a semicircle, it is a right angle. Therefore △ABC is similar to △AED, so their areas are related as AB² is to AE². Because AB² = 4² = 16 and, by the Pythagorean Theorem, AE² = (4 + 3)² + 5² = 74, this ratio is 16/74 = 8/37. The area of △AED is 35/2, so the area of △ABC is (35/2) · (8/37) = 140/37.

因为∠ACB是内接于半圆，因此是直角。因此△ABC~△AED，它们的面积之比为AB²:AE²。因为AB²=4²=16，且由勾股定理，AE²=(4+3)²+5²=74，此比值为16/74=8/37。△AED的面积是35/2，因此△ABC的面积是(35/2)·(8/37)=140/37。

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