Difference between revisions of "1953 AHSME Problems/Problem 33"
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+ | ==Problem== | ||
The perimeter of an isosceles right triangle is <math>2p</math>. Its area is: | The perimeter of an isosceles right triangle is <math>2p</math>. Its area is: | ||
<math>\textbf{(A)}\ (2+\sqrt{2})p \qquad \textbf{(B)}\ (2-\sqrt{2})p \qquad \textbf{(C)}\ (3-2\sqrt{2})p^2\ \textbf{(D)}\ (1-2\sqrt{2})p^2\qquad \textbf{(E)}\ (3+2\sqrt{2})p^2</math> | <math>\textbf{(A)}\ (2+\sqrt{2})p \qquad \textbf{(B)}\ (2-\sqrt{2})p \qquad \textbf{(C)}\ (3-2\sqrt{2})p^2\ \textbf{(D)}\ (1-2\sqrt{2})p^2\qquad \textbf{(E)}\ (3+2\sqrt{2})p^2</math> | ||
+ | ==Solution== | ||
Given leg length <math>x</math>, we can write the perimeter of this triangle to be <math>2x+x\sqrt{2}=2p</math>. Thus, <math>x(2+\sqrt{2})=2p</math>. Divide to get <math>x=\frac{2p}{2+\sqrt{2}}</math>. Multiply by the conjugate and simplify to get <math>x=p(2-\sqrt{2})</math>. Square and divide by two to get the area of the triangle, or <math>\frac{p^2*(2-\sqrt{2}^2)}{2}</math>, or <math>p^2(3-2\sqrt{2})</math>. <math>\boxed{C}</math> | Given leg length <math>x</math>, we can write the perimeter of this triangle to be <math>2x+x\sqrt{2}=2p</math>. Thus, <math>x(2+\sqrt{2})=2p</math>. Divide to get <math>x=\frac{2p}{2+\sqrt{2}}</math>. Multiply by the conjugate and simplify to get <math>x=p(2-\sqrt{2})</math>. Square and divide by two to get the area of the triangle, or <math>\frac{p^2*(2-\sqrt{2}^2)}{2}</math>, or <math>p^2(3-2\sqrt{2})</math>. <math>\boxed{C}</math> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 50p box|year=1953|num-b=32|num-a=34}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 00:40, 4 February 2020
Problem
The perimeter of an isosceles right triangle is . Its area is:
Solution
Given leg length , we can write the perimeter of this triangle to be . Thus, . Divide to get . Multiply by the conjugate and simplify to get . Square and divide by two to get the area of the triangle, or , or .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
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