AMC10 2010 B

AMC10 2010 B · Q7

AMC10 2010 B · Q7. It mainly tests Triangles (properties), Pythagorean theorem.

A triangle has side lengths 10, 10, and 12. A rectangle has width 4 and area equal to the area of the triangle. What is the perimeter of this rectangle?

一个三角形边长为 10、10 和 12。一个矩形宽为 4，面积等于该三角形的面积。这个矩形的周长是多少？

(A) 16 16

(B) 24 24

(D) 32 32

(E) 36 36

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the triangle be $ABC$ with $AB=12$, and let $D$ be the foot of the altitude from $C$. Then $\triangle ACD$ is a right triangle with hypotenuse $AC=10$ and one leg $AD=\frac{1}{2}AB=6$. By the Pythagorean Theorem $CD=\sqrt{10^2-6^2}=8$, and the area of $\triangle ABC$ is $\frac{1}{2}(12)(8)=48$. The rectangle has length $\frac{48}{4}=12$ and perimeter $2(12+4)=32$.

设三角形为 $ABC$，$AB=12$，$D$ 是从 $C$ 垂到 $AB$ 的垂足。那么 $\triangle ACD$ 是直角三角形，斜边 $AC=10$，一条直角边 $AD=\frac{1}{2}AB=6$。由勾股定理，$CD=\sqrt{10^2-6^2}=8$，$\triangle ABC$ 的面积为 $\frac{1}{2}(12)(8)=48$。矩形的长为 $\frac{48}{4}=12$，周长为 $2(12+4)=32$。

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