Difference between revisions of "1995 AJHSME Problems/Problem 18"
(→Problem) |
|||
| Line 17: | Line 17: | ||
The area taken up by the L's is <math>4*\frac{3}{16}=\frac{3}{4}</math> of the area of the whole square. What remains has <math>\frac{1}{4}</math> of the area of the larger square. <math>\frac{100*100}{4}=\frac{100}{2}*\frac{100}{2}=50*50</math> is the area of the smaller square, so its side length is 50. <math>\text{(C)}</math> | The area taken up by the L's is <math>4*\frac{3}{16}=\frac{3}{4}</math> of the area of the whole square. What remains has <math>\frac{1}{4}</math> of the area of the larger square. <math>\frac{100*100}{4}=\frac{100}{2}*\frac{100}{2}=50*50</math> is the area of the smaller square, so its side length is 50. <math>\text{(C)}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1995|num-b=17|num-a=19}} | ||
Revision as of 03:16, 23 December 2012
Problem
The area of each of the four congruent L-shaped regions of this 100-inch by 100-inch square is 3/16 of the total area. How many inches long is the side of the center square?
![[asy] draw((2,2)--(2,-2)--(-2,-2)--(-2,2)--cycle); draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle); draw((0,1)--(0,2)); draw((1,0)--(2,0)); draw((0,-1)--(0,-2)); draw((-1,0)--(-2,0)); [/asy]](http://latex.artofproblemsolving.com/a/1/a/a1a3a520aef5349391bdc7054d86229bbbe3bfcc.png)
Solution
The area taken up by the L's is 
of the area of the whole square. What remains has 
of the area of the larger square. 
is the area of the smaller square, so its side length is 50. 
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
