Difference between revisions of "1985 AJHSME Problems/Problem 12"
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We are given the three side lengths of the triangle, so we can compute the perimeter of the triangle to be <math>6.2+8.3+9.5=24</math>. The square has the same perimeter as the triangle, so its side length is <math>\frac{24}{4}=6</math>. Finally, the area of the square is <math>6^2=36</math>, which is choice <math>\boxed{\text{B}}</math> | We are given the three side lengths of the triangle, so we can compute the perimeter of the triangle to be <math>6.2+8.3+9.5=24</math>. The square has the same perimeter as the triangle, so its side length is <math>\frac{24}{4}=6</math>. Finally, the area of the square is <math>6^2=36</math>, which is choice <math>\boxed{\text{B}}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/C1_dFnM-G00 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Revision as of 08:12, 13 January 2023
Contents
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 
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
