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

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== Solution ==
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== Problem 4 ==
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Quadrilateral <math>ABCD</math> is a rhombus with perimeter <math>52</math> meters. The length of diagonal <math>\overline{AC}</math> is <math>24</math> meters. What is the area in square meters of rhombus <math>ABCD</math>?
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<asy>
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draw((-13,0)--(0,5));
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draw((0,5)--(13,0));
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draw((13,0)--(0,-5));
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draw((0,-5)--(-13,0));
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dot((-13,0));
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dot((0,5));
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dot((13,0));
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dot((0,-5));
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label("A",(-13,0),W);
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label("B",(0,5),N);
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label("C",(13,0),E);
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label("D",(0,-5),S);
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</asy>
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<math>\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144</math>
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== Solution 1 ==
 
<asy>
 
<asy>
 
draw((-12,0)--(0,5));
 
draw((-12,0)--(0,5));

Revision as of 13:09, 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]

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

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

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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