Difference between revisions of "1985 AJHSME Problems/Problem 12"
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==Problem== | ==Problem== | ||
| − | A [[square]] and a [[triangle]] have equal [[Perimeter|perimeters]]. The [[length|lengths]] of the three [[ | + | A [[square]] and a [[triangle]] have equal [[Perimeter|perimeters]]. The [[length|lengths]] of the three [[edge|sides]] of the triangle are <math>6.2 \text{ cm}</math>, <math>8.3 \text{ cm}</math> and <math>9.5 \text{ cm}</math>. The [[area]] of the square is |
<math>\text{(A)}\ 24\text{ cm}^2 \qquad \text{(B)}\ 36\text{ cm}^2 \qquad \text{(C)}\ 48\text{ cm}^2 \qquad \text{(D)}\ 64\text{ cm}^2 \qquad \text{(E)}\ 144\text{ cm}^2</math> | <math>\text{(A)}\ 24\text{ cm}^2 \qquad \text{(B)}\ 36\text{ cm}^2 \qquad \text{(C)}\ 48\text{ cm}^2 \qquad \text{(D)}\ 64\text{ cm}^2 \qquad \text{(E)}\ 144\text{ cm}^2</math> | ||
Revision as of 12:03, 21 July 2009
Problem
A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 
, 
and 
. The area of the square is
Solution
We are given the three side lengths of the triangle, so we can compute the perimeter of the triangle to be 
. The square has the same perimeter as the triangle, so its side length is 
. Finally, the area of the square is 
, which is choice 
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
