Difference between revisions of "1985 AJHSME Problem 12"
Coolmath34 (talk | contribs) (Created page with "== Problem 12 == A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are <math>6.2 \text{ cm}</math>, <math>8.3 \text{ cm}</math> a...") |
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<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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| + | ==In-depth Solution by BoundlessBrain!== | ||
| + | https://youtu.be/rIEgWFY_afM | ||
== Solution == | == Solution == | ||
The perimeter of the triangle is <math>6.2+8.3+9.5 = 24</math> cm, so the side length of the square is <math>\frac{24}{4} = 6</math> cm. The area of the square is <math>6^2 = \boxed{\text{(B)} 36}</math> square centimeters. | The perimeter of the triangle is <math>6.2+8.3+9.5 = 24</math> cm, so the side length of the square is <math>\frac{24}{4} = 6</math> cm. The area of the square is <math>6^2 = \boxed{\text{(B)} 36}</math> square centimeters. | ||
Latest revision as of 14:58, 4 July 2023
Problem 12
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
In-depth Solution by BoundlessBrain!
Solution
The perimeter of the triangle is 
cm, so the side length of the square is 
cm. The area of the square is 
square centimeters.
