Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (meadsy69))
m (→‎Solution 1)
Line 60: Line 60:
  
 
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>.
−
The area of a rhombus is = <math>\frac{d_1*d_2}{2}</math> = <math>\frac{24*10}{2}</math> = <math>120</math>
+
The area of a rhombus is = <math>\frac{d_1\cdot{d_2}}{2}</math> = <math>\frac{24\cdot{10}}{2}</math> = <math>120</math>
  
 
<math>\boxed{\textbf{(D)}\ 120}</math>              ~phoenixfire
 
<math>\boxed{\textbf{(D)}\ 120}</math>              ~phoenixfire

Revision as of 17:39, 5 July 2020

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

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

Solution 2 (meadsy69)

[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); draw((13,0)--(-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] Since a rhombus has sides of equal length, $AB=BC=CD=DA=\frac{52}{4}=13$. In triangle ABC, $AB=BC=13$ and $AC=24$. Using Heron's formula, we have $[ABCD]=[ABC]+[ACD]=2[ABC]=2\sqrt{25*12*12*1}$. Simplifying, we have $[ABC]=\sqrt{3600}=60$ so $[ABCD]=2*60=\boxed{\textbf{(D)} 120}$. ~~RWhite

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_4&oldid=127532"