Difference between revisions of "2016 AMC 10A Problems/Problem 11"
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| − | Notice that the two added lines bisect each of the 4 sides of the large rectangle. | + | Notice that the two added lines bisect each of the <math>4</math> sides of the large rectangle. |
Subtracting the unshaded area from the total area gives us <math>40</math> - <math>33</math> <math>\frac{1}{2}</math> = <math>\boxed{6\frac{1}{2}}</math>, so the correct answer is <math>\boxed{\textbf{(D)}}</math>. | Subtracting the unshaded area from the total area gives us <math>40</math> - <math>33</math> <math>\frac{1}{2}</math> = <math>\boxed{6\frac{1}{2}}</math>, so the correct answer is <math>\boxed{\textbf{(D)}}</math>. | ||
Revision as of 19:00, 8 February 2016
Contents
Problem
What is the area of the shaded region of the given 
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]](http://latex.artofproblemsolving.com/9/5/f/95f8b885091c3cb0d7a2cb9325def6a059bfb982.png)
Solution 1
First, split the rectangle into 
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]](http://latex.artofproblemsolving.com/d/2/9/d29efd298290e080f87df84c3615c9d4288eef4b.png)
The bases of these triangles are all 
, and their heights are , 
, , and . Thus, their areas are 
, 
, , and , which add to the area of the shaded region, which is 
.
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]](http://latex.artofproblemsolving.com/8/2/c/82cdaaa75ab582149da4c5e56880faabde9f14bd.png)
Notice that the two added lines bisect each of the sides of the large rectangle.
Subtracting the unshaded area from the total area gives us 
- 
= , so the correct answer is 
.
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. 
