Difference between revisions of "1998 AJHSME Problems/Problem 13"

Revision as of 10:22, 31 July 2011

Problem 13

What is the ratio of the area of the shaded square to the area of the large square? (The figure is drawn to scale)

$[asy] draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); draw((0,0)--(4,4)); draw((0,4)--(3,1)--(3,3)); draw((1,1)--(2,0)--(4,2)); fill((1,1)--(2,0)--(3,1)--(2,2)--cycle,black); [/asy]$

$\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{7} \qquad \text{(C)}\ \dfrac{1}{8} \qquad \text{(D)}\ \dfrac{1}{12} \qquad \text{(E)}\ \dfrac{1}{16}$

Solution

We can divide the large square into quarters by diagonals.

Then, in $\frac{1}{4}$ the area of the big square, the little square would have $\frac{1}{2}$ the area.

$\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}=\boxed{C}$

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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

@@ Line 23: / Line 23: @@
 == See also ==
-{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
+{{AJHSME box|year=1998|num-b=12|num-a=14}}
 * [[AJHSME]]
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]

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Difference between revisions of "1998 AJHSME Problems/Problem 13"

Revision as of 10:22, 31 July 2011

Problem 13

Solution

See also