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

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(Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
  
Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. This will result in <math>40</math> - <math>33</math> <math>\frac{1}{2}</math> = <math>\boxed{6 </math>\frac{1}{2}}$
+
Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. This will result in <math>40</math> - <math>33</math> <math>\frac{1}{2}</math> = <math>\boxed{6\frac{1}{2}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 15:39, 4 February 2016

Problem

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]

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

Solution 1

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 $\boxed{6\frac{1}{2}}$.

Solution 2

Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. This will result in $40$ - $33$ $\frac{1}{2}$ = $\boxed{6\frac{1}{2}}$.

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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
2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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