Difference between revisions of "2008 AMC 8 Problems/Problem 6"

m (minor edit)
 
Line 1: Line 1:
==Problem==
+
== Problem ==
 
In the figure, what is the ratio of the area of the gray squares to the area of the white squares?
 
In the figure, what is the ratio of the area of the gray squares to the area of the white squares?
 +
 
<asy>
 
<asy>
 
size((70));
 
size((70));
Line 14: Line 15:
 
fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray);
 
fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray);
 
</asy>
 
</asy>
 +
 
<math> \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 </math>
 
<math> \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 </math>
  
==Solution==
+
== Solution ==
 
Dividing the gray square into four smaller squares, there are <math>6</math> gray tiles and <math>10</math> white tiles, giving a ratio of <math>\boxed{\textbf{(D)}\ 3:5}</math>.
 
Dividing the gray square into four smaller squares, there are <math>6</math> gray tiles and <math>10</math> white tiles, giving a ratio of <math>\boxed{\textbf{(D)}\ 3:5}</math>.
==See Also==
+
 
 +
== See Also ==
 
{{AMC8 box|year=2008|num-b=5|num-a=7}}
 
{{AMC8 box|year=2008|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:36, 19 October 2020

Problem

In the figure, what is the ratio of the area of the gray squares to the area of the white squares?

[asy] size((70)); draw((10,0)--(0,10)--(-10,0)--(0,-10)--(10,0)); draw((-2.5,-7.5)--(7.5,2.5)); draw((-5,-5)--(5,5)); draw((-7.5,-2.5)--(2.5,7.5)); draw((-7.5,2.5)--(2.5,-7.5)); draw((-5,5)--(5,-5)); draw((-2.5,7.5)--(7.5,-2.5)); fill((-10,0)--(-7.5,2.5)--(-5,0)--(-7.5,-2.5)--cycle, gray); fill((-5,0)--(0,5)--(5,0)--(0,-5)--cycle, gray); fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray); [/asy]

$\textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1$

Solution

Dividing the gray square into four smaller squares, there are $6$ gray tiles and $10$ white tiles, giving a ratio of $\boxed{\textbf{(D)}\ 3:5}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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. AMC logo.png

Invalid username
Login to AoPS