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

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== Problem 4 ==
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==Problem==
  
 
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>?
 
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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<math>\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144</math>
 
<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 ==
 
== Solution 1 ==
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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
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<asy>
 
<asy>
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== Solution 2 ==
 
== Solution 2 ==
 
Right off the bat, we can see that the perimeter of the figure is 52. Dividing this by four, we can get that each side is equal to 13. By drawing a line perpendicular to the one given, we can split the figure into four right triangles. 12 (24/2) is equal to the height of one small right triangle, and 13 is the slanted side. Using the Pythagorean theorem we can find that 169 (13 squared) - 144 (12 squared) = 25 (five squared). With this, we can determine that each small right triangle equals 30. Multiplying that by four we can get <math>\boxed{\textbf{(D)}\ 120}</math>
 
Right off the bat, we can see that the perimeter of the figure is 52. Dividing this by four, we can get that each side is equal to 13. By drawing a line perpendicular to the one given, we can split the figure into four right triangles. 12 (24/2) is equal to the height of one small right triangle, and 13 is the slanted side. Using the Pythagorean theorem we can find that 169 (13 squared) - 144 (12 squared) = 25 (five squared). With this, we can determine that each small right triangle equals 30. Multiplying that by four we can get <math>\boxed{\textbf{(D)}\ 120}</math>
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  ~mathisawesome
+
  ~math4all
 +
 
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/IgpayYB48C4?si=CGr1gRwXKdjrhl24&t=678
 +
 
 +
~Math-X
  
 
== Video Solution ==
 
== Video Solution ==
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~savannahsolver
 
~savannahsolver
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 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/UCaEXbe7mN0
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 +
~Hayabusa1
  
 
==See also==
 
==See also==

Latest revision as of 09:30, 9 November 2024

Problem

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


[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 the Pythagorean theorem, we find that $\overline{BE}$ = $5$. You know the Pythagorean triple, (5, 12, 13).

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

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

Solution 2

Right off the bat, we can see that the perimeter of the figure is 52. Dividing this by four, we can get that each side is equal to 13. By drawing a line perpendicular to the one given, we can split the figure into four right triangles. 12 (24/2) is equal to the height of one small right triangle, and 13 is the slanted side. Using the Pythagorean theorem we can find that 169 (13 squared) - 144 (12 squared) = 25 (five squared). With this, we can determine that each small right triangle equals 30. Multiplying that by four we can get $\boxed{\textbf{(D)}\ 120}$ 

~math4all

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=CGr1gRwXKdjrhl24&t=678

~Math-X

Video Solution

The Learning Royal: https://youtu.be/IiFFDDITE6Q

Video Solution 2

Solution detailing how to solve the problem:https://www.youtube.com/watch?v=-yHfOUapg7I&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=5

Video Solution 3

https://youtu.be/mL6gIb5y3B0

~savannahsolver

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/UCaEXbe7mN0

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

See also

2019 AMC 8 (Problems • Answer Key • Resources)
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Problem 3 		Followed by
Problem 5
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All AJHSME/AMC 8 Problems and Solutions

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