Difference between revisions of "1998 AJHSME Problems/Problem 20"
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After both folds are completed, the square would become a triangle that has an area of <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math> of the original square. | After both folds are completed, the square would become a triangle that has an area of <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math> of the original square. | ||
| − | Since the area is <math>9</math> inches for <math>\frac{1}{4}</math> of the square, <math>9\times4=36</math> square inches is the area of square <math>PQRS</math> | + | Since the area is <math>9</math> square inches for <math>\frac{1}{4}</math> of the square, <math>9\times4=36</math> square inches is the area of square <math>PQRS</math> |
The length of the side of a square that has an area of <math>36</math> square inches is <math>\sqrt{36}=6</math> inches. | The length of the side of a square that has an area of <math>36</math> square inches is <math>\sqrt{36}=6</math> inches. | ||
Revision as of 12:06, 31 July 2011
Problem 20
Let 
be a square piece of paper. 
is folded onto 
and then 
is folded onto 
. The area of the resulting figure is 9 square inches. Find the perimeter of square .
![[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$P$",(0,2),SE); label("$Q$",(2,2),SW); label("$R$",(2,0),NW); label("$S$",(0,0),NE); [/asy]](http://latex.artofproblemsolving.com/6/e/8/6e8eea15c0d64bd2c75d70fc6db8f1c1871a98c4.png)
Solution
After both folds are completed, the square would become a triangle that has an area of 
of the original square.
Since the area is 
square inches for 
of the square, 
square inches is the area of square
The length of the side of a square that has an area of 
square inches is 
inches.
Each side is 
inches, so the total perimeter is 
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
