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> for <math>\frac{1}{4}</math> of the square, <math>9\times4=36</math> is the area of square <math>PQRS</math> | + | 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> |
− | The length of the side of a square that has an area of <math>36</math> square | + | 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. |
− | Each side is <math>6</math> | + | Each side is <math>6</math> inches, so the total perimeter is <math>6\times4=24=\boxed{D}</math> |
== See also == | == See also == |
Revision as of 11:06, 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 .
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 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 |