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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)
Line 1: Line 1:
 
== Solution ==
 
== Solution ==
 
<asy>
 
<asy>
−
draw((-13,0)--(0,5));
+
draw((-12,0)--(0,5));
−
draw((0,5)--(13,0));
+
draw((0,5)--(12,0));
−
draw((13,0)--(0,-5));
+
draw((12,0)--(0,-5));
−
draw((0,-5)--(-13,0));
+
draw((0,-5)--(-12,0));
−
draw((0,0)--(13,0));
+
draw((0,0)--(12,0));
 
draw((0,0)--(0,5));
 
draw((0,0)--(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));
−
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("C",(13,0),E);
+
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);
Line 26: Line 26:
  
 
<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);
Line 42: Line 42:
 
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>
 
Which means area = <math>\frac{d_1*d_2}{2}</math> = <math>\frac{24*10}{2}</math> = <math>120</math>
 +
<math>\boxed{\textbf{(D)}\ 120}</math>

Revision as of 02:24, 20 November 2019

Solution

[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 = $\frac{d_1*d_2}{2}$ = $\frac{24*10}{2}$ = $120$ $\boxed{\textbf{(D)}\ 120}$

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