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 $(1,5), (0,5), (0,4), \left(4,\frac{5}{2}\right)$.

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

[asy]  size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(0,5)--(1,5)--cycle,gray(0.7)); filldraw((8,0)--(8,1)--(0,4)--(0,5)--cycle,gray(0.9));  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));  [/asy]

Split the region into four parts by the diagonal from the top left to the bottom right. Slide the top and bottom pieces next to each other to form a parallelogram with base $1$ and height $\frac{5}{2}$, and slide the left and right pieces next to each other to form a parallelogram with base $1$ and height $\frac{8}{2}$. The total area is then $1\cdot\frac{5}{2}+1\cdot\frac{8}{2} = \frac{(5+8)(1)}{2} = \boxed{\textbf{(C)}\ 6\frac{1}{2}}$. ~emerald_block


Solution 5

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

We can subtract the 4 triangular areas from the overall rectangle. It can be noticed that two triangles have area $\frac{6 \cdot \frac{5}{2}}{2} = \frac{15}{2}$, and the other two triangles have area $\frac{3 \cdot \frac{8}{2}}{2} = 6$ $\Rightarrow$ the shaded area is $(8 \cdot 5) - 2(\frac{15}{2}) - 2(6) = 40 - 15 - 12 = 13$. Since the desired area is half the shaded region, our area is $\boxed{\textbf{(D)}\ 6\frac{1}{2}}$.

~Champion1234

Video Solution

https://youtu.be/dHY8gjoYFXU

~IceMatrix

https://www.youtube.com/watch?v=WojyKGOEk_g

Video Solution

https://youtu.be/4_x1sgcQCp4?t=652

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS