1998 AJHSME Problems/Problem 13

Revision as of 23:16, 30 March 2015 by 5849206328x (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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.

See also

1998 AJHSME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS