Difference between revisions of "2010 AMC 8 Problems/Problem 17"
Hithere22702 (talk | contribs) (→Solution 2(More rigorous)) |
Hithere22702 (talk | contribs) m (→Solution 2(More rigorous)) |
||
| Line 45: | Line 45: | ||
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math> | Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
| − | ==Solution 2 | + | ==Solution 2== |
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
Revision as of 17:54, 29 April 2020
Contents
Problem
The diagram shows an octagon consisting of 
unit squares. The portion below 
is a unit square and a triangle with base 
. If bisects the area of the octagon, what is the ratio 
?
![[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((0,0)--(6,0),linewidth(1.2pt)); draw((0,0)--(0,1),linewidth(1.2pt)); draw((0,1)--(1,1),linewidth(1.2pt)); draw((1,1)--(1,2),linewidth(1.2pt)); draw((1,2)--(5,2),linewidth(1.2pt)); draw((5,2)--(5,1),linewidth(1.2pt)); draw((5,1)--(6,1),linewidth(1.2pt)); draw((6,1)--(6,0),linewidth(1.2pt)); draw((1,1)--(5,1),linewidth(1.2pt)); draw((1,1)--(1,0),linewidth(1.2pt)); draw((2,2)--(2,0),linewidth(1.2pt)); draw((3,2)--(3,0),linewidth(1.2pt)); draw((4,2)--(4,0),linewidth(1.2pt)); draw((5,1)--(5,0),linewidth(1.2pt)); draw((0,0)--(5,1.5),linewidth(1.2pt)); dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); dot((0,1),ds); dot((1,1),ds); dot((1,2),ds); dot((5,2),ds); label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); dot((4,2),ds); dot((5,1.5),ds); label("$Q$", (5.14,1.51),NE*lsf); clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle); [/asy]](http://latex.artofproblemsolving.com/3/b/b/3bba290e642b0a44220ce1e8ce69c823fe134b28.png)
Solution 1
We see that half the area of the octagon is . We see that the triangle area is 
. That means that 
.
![\[\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}\]](http://latex.artofproblemsolving.com/c/5/1/c51e67a5af29dec7181d654c8f9db251979e609e.png)
Meaning, 
Solution 2
Like stated in solution 1, we know that half the area of the octagon is . That means that the area of the trapezoid is 
. 
. Solving for 
, we get 
. Subtracting 
from 
, we get 
. Therefore, the answer comes out to 
~Hithere22702
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
