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Difference between revisions of "2019 AMC 8 Problems/Problem 4"

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Thus the values of the two diagonals are <math>\overline{AC}</math> = <math>24</math> and <math>\overline{BD}</math> = <math>10</math>.
 
Thus the values of the two diagonals are <math>\overline{AC}</math> = <math>24</math> and <math>\overline{BD}</math> = <math>10</math>.
 
Which means area = <math>\frac{d_1*d_2}{2}</math> = <math>\frac{24*10}{2}</math> = <math>120</math>
 
Which means area = <math>\frac{d_1*d_2}{2}</math> = <math>\frac{24*10}{2}</math> = <math>120</math>
 +
 
<math>\boxed{\textbf{(D)}\ 120}</math>
 
<math>\boxed{\textbf{(D)}\ 120}</math>
 +
 +
==See Also==
 +
{{AMC8 box|year=2019|num-b=1|num-a=3}}
 +
 +
{{MAA Notice}}

Revision as of 02:26, 20 November 2019

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 = $\frac{d_1*d_2}{2}$ = $\frac{24*10}{2}$ = $120$

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

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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