# 2016 AMC 10A Problems/Problem 11

## Problem

Find the area of the shaded region. $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("1",(1/2,5),dir(90)); label("7",(9/2,5),dir(90)); label("1",(8,1/2),dir(0)); label("4",(8,3),dir(0)); label("1",(15/2,0),dir(270)); label("7",(7/2,0),dir(270)); label("1",(0,9/2),dir(180)); label("4",(0,2),dir(180)); [/asy]$ $\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$

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

The bases of these triangles are all $1$, and by symmetry, 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. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown. $[asy] size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("1",(1/2,5),dir(90)); label("4",(6,5),dir(90)); label("3",(5/2,5),dir(90)); label("1",(8,1/2),dir(0)); label("5/2",(8,15/4),dir(0)); label("3/2",(8,7/4),dir(0)); label("1",(15/2,0),dir(270)); label("4",(2,0),dir(270)); label("3",(11/2,0),dir(270)); label("1",(0,9/2),dir(180)); label("5/2",(0,5/4),dir(180)); label("3/2",(0,13/4),dir(180)); draw((0,5/2)--(8,5/2)); draw((4,0)--(4,5)); [/asy]$

Notice that the two added lines bisect each of the $4$ sides of the large rectangle.

Subtracting the unshaded area from the total area gives us $40-33\frac{1}{2}=\boxed{6\frac{1}{2}}$, so the correct answer is $\boxed{\textbf{(D)}}$.

## Solution 3

Notice that we can graph this on the coordinate plane.

The top-left shaded figure has coordinates of $(1,5), (0,5), (0,4), (4,\frac{5}{2})$.

Notice that we can apply the shoelace method to find the area of this polygon.

We find that the area of the polygon is $\frac{13}{4}$.

However, notice that the two shaded regions are two congruent polygons.

Hence, the total area is $\frac{13}{2}\implies \boxed{6\frac{1}{2}}$ or $\boxed{D}$.

## Solution 4

The sum of the heights on each axis is $5$ and $8$. The bases are $1$. Hence the shaded area is $\frac{(5+8)(1)} 2 = \boxed{6\frac{1}{2}}$

~IceMatrix

~ pi_is_3.14

## See Also

 2016 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 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 (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 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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