2019 AMC 8 Problems/Problem 4

Solution

$[asy] draw((-12,0)--(0,5)); draw((0,5)--(12,0)); draw((12,0)--(0,-5)); draw((0,-5)--(-12,0)); draw((0,0)--(12,0)); draw((0,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,-5)); dot((-12,0)); dot((0,5)); dot((12,0)); dot((0,-5)); label("A",(-12,0),W); label("B",(0,5),N); label("C",(12,0),E); label("D",(0,-5),S); label("E",(0,0),SW); [/asy]$

Because it is a rhombus all sides are equal. Implies all sides are 13. In a rhombus diagonals are perpendicular and bisect each other. Which means $\overline{AE}$ = $12$ = $\overline{EC}$ .

Consider one of the right triangles.

$[asy] draw((-12,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,5)); dot((-12,0)); dot((0,5)); label("A",(-12,0),W); label("B",(0,5),N); label("E",(0,0),SE); [/asy]$

$\overline{AB}$ = $13$ . $\overline{AE}$ = $12$ . Which means $\overline{BE}$ = $5$ .

Thus the values of the two diagonals are $\overline{AC}$ = $24$ and $\overline{BD}$ = $10$ . Which means area = $\frac{d_1*d_2}{2}$ = $\frac{24*10}{2}$ = $120$

$\boxed{\textbf{(D)}\ 120}$

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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2019 AMC 8 Problems/Problem 4

Solution

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