AMC12 2012 B

AMC12 2012 B · Q10

AMC12 2012 B · Q10. It mainly tests Area & perimeter, Coordinate geometry.

What is the area of the polygon whose vertices are the points of intersection of the curves $x^2 + y^2 =25$ and $(x-4)^2 + 9y^2 = 81 ?$

由曲线 $x^2 + y^2 =25$ 与 $(x-4)^2 + 9y^2 = 81$ 的交点作为顶点所形成的多边形的面积是多少？

(A) 24 24

(B) 27 27

(D) 37.5 37.5

(E) 42 42

Answer

Correct choice: (B)

正确答案：（B）

Solution

The first curve is a circle with radius $5$ centered at the origin, and the second curve is an ellipse with center $(4,0)$ and end points of $(-5,0)$ and $(13,0)$. Finding points of intersection, we get $(-5,0)$, $(4,3)$, and $(4,-3)$, forming a triangle with height of $9$ and base of $6.$ So the area of this triangle is $9 \cdot 6 \cdot 0.5 =\boxed{27 \textbf{ (B)}}.$

第一条曲线是以原点为圆心、半径为 $5$ 的圆，第二条曲线是以 $(4,0)$ 为中心的椭圆，其端点为 $(-5,0)$ 和 $(13,0)$。求交点得 $(-5,0)$、$(4,3)$、$(4,-3)$，它们构成一个三角形，高为 $9$，底为 $6$。因此该三角形面积为 $9 \cdot 6 \cdot 0.5 =\boxed{27 \textbf{ (B)}}.$

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