Question

四边形 $ABCD$ 是一个周长为 $52$ 米的菱形。对角线 $\overline{AC}$ 的长度为 $24$ 米。菱形 $ABCD$ 的面积是多少平方米？

Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD?$

$60$

$90$

$105$

$120$

$144$

答案：D

知识点：菱形勾股定理

难度评级：1020

视频讲解：

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文字解答：

可以把菱形分成 $4$ 个如图所示的直角三角形。

菱形对角线互相垂直平分，且菱形边长为 $52 \div 4 = 13$ ，所以这些直角三角形的斜边长为十三，一条直角边长为十二。

用勾股定理，另一条直角边长为 $\sqrt{13^2 - 12^2} = \sqrt{25} = 5.$

因此菱形面积为正确答案是 D。 $4 \cdot \dfrac{5 \cdot 12}{2} = 4 \cdot 30 = 120.$

We can split the rhombus up into $4$ triangles, each of which looks like this.

We know that this is a right triangle, since the diagonals of a rhombus are perpendicular, and we know the hypotenuse is $52 \div 4 = 13.$

Using the Pythagorean theorem, we get the other leg to be $\sqrt{13^2 - 12^2} = \sqrt{25} = 5.$

This means that the area of the rhombus is $4 \cdot \dfrac{5 \cdot 12}{2} = 4 \cdot 30 = 120.$ Thus, the correct answer is D.

2019 AMC 8 第 4 题

答案：D

难度评级：1020

视频讲解：

文字解答：

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