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

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A rhombus has sides of equal length. Because the perimeter of the rhombus is <math>52</math>, each side is <math>\frac{52}{4}=13</math>. In a rhombus, diagonals are perpendicular and bisect each other, which means <math>\overline{AE}</math> = <math>12</math> = <math>\overline{EC}</math>.
 
A rhombus has sides of equal length. Because the perimeter of the rhombus is <math>52</math>, each side is <math>\frac{52}{4}=13</math>. In a rhombus, diagonals are perpendicular and bisect each other, which means <math>\overline{AE}</math> = <math>12</math> = <math>\overline{EC}</math>.
  
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Consider one of the right triangles.
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Consider one of the right triangles:
  
 
<asy>
 
<asy>
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</asy>
 
</asy>
  
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<math>\overline{AB}</math> = <math>13</math>.
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<math>\overline{AB}</math> = <math>13</math>, and <math>\overline{AE}</math> = <math>12</math>. Using Pythagorean theorem, we find that <math>\overline{BE}</math> = <math>5</math>.
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<math>\overline{AE}</math> = <math>12</math>.
 
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Which means <math>\overline{BE}</math> = <math>5</math>.
 
  
 
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>.
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Which means area = <math>\frac{d_1*d_2}{2}</math> = <math>\frac{24*10}{2}</math> = <math>120</math>
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The area of a rhombus is = <math>\frac{d_1*d_2}{2}</math> = <math>\frac{24*10}{2}</math> = <math>120</math>
  
 
<math>\boxed{\textbf{(D)}\ 120}</math>              ~phoenixfire
 
<math>\boxed{\textbf{(D)}\ 120}</math>              ~phoenixfire

Revision as of 18:53, 20 November 2019

Problem 4

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 $ABCD$?

[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); 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); [/asy]

$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$


Solution 1

[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]

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 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]

$\overline{AB}$ = $13$, and $\overline{AE}$ = $12$. Using Pythagorean theorem, we find that $\overline{BE}$ = $5$.

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

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

See Also

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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