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Difference between revisions of "2016 AMC 10A Problems/Problem 11"

(Created page with "What is the area of the shaded region of the given <math>8 \times 5</math> rectangle? <asy> size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); fi...")
 
(Added solution)
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<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math>
 
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math>
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 +
First, split the rectangle into <math>4</math> triangles:
 +
<asy>
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 +
size(6cm);
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defaultpen(fontsize(9pt));
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draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
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filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));
 +
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label("$1$",(1/2,5),dir(90));
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label("$7$",(9/2,5),dir(90));
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 +
label("$1$",(8,1/2),dir(0));
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label("$4$",(8,3),dir(0));
 +
 +
label("$1$",(15/2,0),dir(270));
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label("$7$",(7/2,0),dir(270));
 +
 +
label("$1$",(0,9/2),dir(180));
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label("$4$",(0,2),dir(180));
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 +
draw((0,5)--(8,0));
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 +
</asy>
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The bases of these triangles are all <math>1</math>, and their heights are <math>4</math>, <math>\frac{5}{2}</math>, <math>4</math>, and <math>\frac{5}{2}</math>. Thus, their areas are <math>2</math>, <math>\frac{5}{4}</math>, <math>2</math>, and <math>\frac{5}{4}</math>, which add to the area of the shaded region, which is <math>\boxed{6\frac{1}{2}}</math>.
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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 23:03, 3 February 2016

What is the area of the shaded region of the given $8 \times 5$ rectangle?

[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$4$",(8,3),dir(0)); label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180)); [/asy]

$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$

First, split the rectangle into $4$ triangles: [asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$4$",(8,3),dir(0)); label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180)); draw((0,5)--(8,0)); [/asy]

The bases of these triangles are all $1$, and their heights are $4$, $\frac{5}{2}$, $4$, and $\frac{5}{2}$. Thus, their areas are $2$, $\frac{5}{4}$, $2$, and $\frac{5}{4}$, which add to the area of the shaded region, which is $\boxed{6\frac{1}{2}}$.

See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 AMC 10 Problems and Solutions

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