AMC8 2019

AMC8 2019 · Q4

AMC8 2019 · Q4. It mainly tests Triangles (properties), Pythagorean theorem.

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 $ABCD$?

四边形 $ABCD$ 是边长相等的四边形，周长 52 米。对角线 $\overline{AC}$ 长 24 米。菱形 $ABCD$ 的面积（平方米）是多少？

(A) 60 60

(B) 90 90

(D) 120 120

(E) 144 144

Answer

Correct choice: (D)

正确答案：（D）

Solution

A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$, each side is $\frac{52}{4}=13$. In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ = $12$ = $\overline{EC}$. Consider one of the $\overline{AB}$ = $13$, and $\overline{AE}$ = $12$. Using the Pythagorean theorem, we find that $\overline{BE}$ = $5$. You know the Pythagorean triple, (5, 12, 13). Thus the values of the two diagonals are $\overline{AC}$ = $24$ and $\overline{BD}$ = $10$. The area of a rhombus is = $\frac{d_1\cdot{d_2}}{2}$ = $\frac{24\cdot{10}}{2}$ = $120$ $\boxed{\textbf{(D)}\ 120}$

菱形边长相等。周长 52，每边 $\frac{52}{4}=13$。对角线垂直平分，$\overline{AE} = \overline{EC} = 12$。考虑右三角形 $\triangle ABE$：$\overline{AB} = 13$，$\overline{AE} = 12$，由勾股定理，$\overline{BE} = 5$。（5-12-13 三元组）对角线 $\overline{AC} = 24$，$\overline{BD} = 10$。面积 $= \frac{d_1 \cdot d_2}{2} = \frac{24 \cdot 10}{2} = 120$。 $\boxed{\textbf{(D)}\ 120}$

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