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2019 AMC 8 Problems/Problem 4

Revision as of 02:21, 20 November 2019 by Phoenixfire (talk | contribs) (Created page with "== Solution == <asy> draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); draw((0,0)--(13,0)); draw((0,0)--(0,5)); draw((0,0)--(-13,0)); dra...")
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Solution

[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); draw((0,0)--(13,0)); draw((0,0)--(0,5)); draw((0,0)--(-13,0)); draw((0,0)--(0,-5)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),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((-13,0)--(0,5)); draw((0,0)--(-13,0)); draw((0,0)--(0,5)); dot((-13,0)); dot((0,5)); label("A",(-13,0),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 = $\frac{d_1*d_2}{2}$ = $\frac{24*10}{2}$ = $120$

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