AMC10 2020 A

AMC10 2020 A · Q12

AMC10 2020 A · Q12. It mainly tests Triangles (properties), Area & perimeter.

Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV = CU = 12$. What is the area of $\triangle AMC$?

三角形 $AMC$ 是等腰三角形，$AM = AC$。中线 $\overline{MV}$ 和 $\overline{CU}$ 相互垂直，且 $MV = CU = 12$。三角形 $AMC$ 的面积是多少？

(A) 48 48

(B) 72 72

(D) 144 144

(E) 192 192

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $G$ be the point of intersection of $\overline{MV}$ and $\overline{CU}$. Right triangle $CVG$ has legs with lengths $GC = \frac{3}{2} \cdot 12 = 8$ and $GV = \frac{1}{2} \cdot 12 = 4$, so it has area $\frac{1}{2} \cdot 8 \cdot 4 = 16$. Because the medians of a triangle divide the triangle into six triangular regions of equal area, the area of $\triangle AMC$ is $6 \cdot 16 = 96$.

令 $G$ 为 $\overline{MV}$ 和 $\overline{CU}$ 的交点。直角三角形 $CVG$ 的两条直角边长度为 $GC = \frac{3}{2} \cdot 12 = 8$ 和 $GV = \frac{1}{2} \cdot 12 = 4$，因此其面积为 $\frac{1}{2} \cdot 8 \cdot 4 = 16$。因为三角形的中线将三角形分成六个面积相等的三角区域，三角形 $AMC$ 的面积是 $6 \cdot 16 = 96$。

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