Difference between revisions of "1985 AJHSME Problems/Problem 12"
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==Problem== | ==Problem== | ||
| − | A | + | A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.2, 8.3 and 9.5. 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> | ||
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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> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/C1_dFnM-G00 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1985|num-b=11|num-a=13}} | {{AJHSME box|year=1985|num-b=11|num-a=13}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 12:03, 4 July 2023
Contents
Problem
A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.2, 8.3 and 9.5. 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. 
