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"

(Created page with "== Solution == <asy> draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); draw((0,0)--(13,0)); draw((0,0)--(0,5)); draw((0,0)--(-13,0)); dra...")
 
(→‎Solution 1)
 
(43 intermediate revisions by 29 users not shown)
Line 1: Line 1:
−
== Solution ==
+
== Problem 4 ==
 +
 
 +
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>?
 +
 
 
<asy>
 
<asy>
 
draw((-13,0)--(0,5));
 
draw((-13,0)--(0,5));
Line 5: Line 8:
 
draw((13,0)--(0,-5));
 
draw((13,0)--(0,-5));
 
draw((0,-5)--(-13,0));
 
draw((0,-5)--(-13,0));
−
draw((0,0)--(13,0));
 
−
draw((0,0)--(0,5));
 
−
draw((0,0)--(-13,0));
 
−
draw((0,0)--(0,-5));
 
 
dot((-13,0));
 
dot((-13,0));
 
dot((0,5));
 
dot((0,5));
Line 16: Line 15:
 
label("B",(0,5),N);
 
label("B",(0,5),N);
 
label("C",(13,0),E);
 
label("C",(13,0),E);
 +
label("D",(0,-5),S);
 +
</asy>
 +
 +
<math>\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144</math>
 +
 +
== 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("D",(0,-5),S);
 
label("E",(0,0),SW);
 
label("E",(0,0),SW);
 
</asy>
 
</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.
+
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>.
−
Which means <math>\overline{AE}</math> = <math>12</math> = <math>\overline{EC}</math>.
+
 
 +
Consider one of the
  
−
Consider one of the right triangles.
 
  
 
<asy>
 
<asy>
−
draw((-13,0)--(0,5));
+
draw((-12,0)--(0,5));
−
draw((0,0)--(-13,0));
+
draw((0,0)--(-12,0));
 
draw((0,0)--(0,5));
 
draw((0,0)--(0,5));
−
dot((-13,0));
+
dot((-12,0));
 
dot((0,5));
 
dot((0,5));
−
label("A",(-13,0),W);
+
label("A",(-12,0),W);
 
label("B",(0,5),N);
 
label("B",(0,5),N);
 
label("E",(0,0),SE);
 
label("E",(0,0),SE);
 
</asy>
 
</asy>
  
−
<math>\overline{AB}</math> = <math>13</math>.
+
<math>\overline{AB}</math> = <math>13</math>, and <math>\overline{AE}</math> = <math>12</math>. Using the Pythagorean theorem, we find that <math>\overline{BE}</math> = <math>5</math>.
−
<math>\overline{AE}</math> = <math>12</math>.
+
You know the Pythagorean triple, (5, 12, 13).
−
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>.
−
Which means area = <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>
 +
 
 +
== 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>
 +
~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(Problem 1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
 +
 
 +
==See also==
 +
{{AMC8 box|year=2019|num-b=3|num-a=5}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 12:25, 3 August 2024

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


[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(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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=224367"