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

Revision as of 20:41, 27 November 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 $\overline{BE}$ = $5$ . "You may recall the famous Pythagorean triple, (5, 12, 13), that's how I did it" - Zack2008

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

Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s

2019 AMC 8 (Problems • Answer Key • Resources)
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Difference between revisions of "2019 AMC 8 Problems/Problem 4"

Revision as of 20:41, 27 November 2020

Problem 4

Solution 1

See also

@@ 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>
 draw((-13,0)--(0,5));
@@ Line 5: / Line 8: @@
 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));
-draw((0,0)--(0,-5));
 dot((-13,0));
 dot((0,5));
@@ Line 16: / Line 15: @@
 label("B",(0,5),N);
 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("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.
+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 right triangles.
+Consider one of the right triangles:
 <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));
-dot((-13,0));
+dot((-12,0));
 dot((0,5));
-label("A",(-13,0),W);
+label("A",(-12,0),W);
 label("B",(0,5),N);
 label("E",(0,0),SE);
 </asy>
-<math>\overline{AB}</math> = <math>13</math>.
+<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>.
-<math>\overline{AE}</math> = <math>12</math>.
+"You may recall the famous Pythagorean triple, (5, 12, 13), that's how I did it" - Zack2008
-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>.
-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>              ~phoenixfire
+Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s
+==See also==
+{{AMC8 box|year=2019|num-b=3|num-a=5}}
+{{MAA Notice}}