Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2016 AMC 10A Problems/Problem 11

Revision as of 19:03, 13 April 2016 by Manananananan (talk | contribs) (→‎Solution 1)

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

[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 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 $\boxed{\textbf{(D)}}$.

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
2016 AMC 12A (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_11&oldid=78029"