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

Latest revision as of 23:16, 30 March 2015

Problem

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

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

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

Solutions

Solution 1

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.

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

Solution 2

Answer: C

Divide the square into 16 smaller squares as shown. The shaded square is formed from 4 half-squares, so its area is 2. The ratio 2 to 16 is 1/8.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 <math>\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}</math>
-==Solution==
+==Solutions==
+=== Solution 1 ===
 We can divide the large square into quarters by diagonals.
@@ Line 21: / Line 22: @@
 <math>\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}=\boxed{C}</math>
+=== Solution 2 ===
+Answer: '''C'''
+[[File:1998ajhsme-13-2.png]]
+Divide the square into 16 smaller squares as shown. The shaded square is formed from 4 half-squares, so its area is 2. The ratio 2 to 16 is 1/8.
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