Difference between revisions of "2016 AMC 10A Problems/Problem 11"
(Added solution) |
m (Added section titles + 12A box) |
||
| Line 1: | Line 1: | ||
| + | == Problem == | ||
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle? | What is the area of the shaded region of the given <math>8 \times 5</math> rectangle? | ||
| Line 23: | Line 24: | ||
<math>\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</math> | <math>\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</math> | ||
| + | |||
| + | == Solution == | ||
First, split the rectangle into <math>4</math> triangles: | First, split the rectangle into <math>4</math> triangles: | ||
| Line 52: | Line 55: | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2016|ab=A|num-b=10|num-a=12}} | ||
| + | {{AMC12 box|year=2016|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:01, 4 February 2016
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
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 
.
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. 
