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Difference between revisions of "1960 AHSME Problems/Problem 21"

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==Problem==
 
==Problem==
  
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The diagonal of square <math>I</math> is <math>a+b</math>. The area of square <math>II</math> with twice the area of <math>I</math> is:
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The diagonal of square <math>I</math> is <math>a+b</math>. The perimeter of square <math>II</math> with twice the area of <math>I</math> is:
  
 
<math>\textbf{(A)}\ (a+b)^2\qquad
 
<math>\textbf{(A)}\ (a+b)^2\qquad

Latest revision as of 17:11, 24 September 2024

Problem

The diagonal of square $I$ is $a+b$. The perimeter of square $II$ with twice the area of $I$ is:

$\textbf{(A)}\ (a+b)^2\qquad \textbf{(B)}\ \sqrt{2}(a+b)^2\qquad \textbf{(C)}\ 2(a+b)\qquad \textbf{(D)}\ \sqrt{8}(a+b) \qquad \textbf{(E)}\ 4(a+b)$

Solution

Since the diagonal of square $I$ is $a+b$ units long, the side length of square $I$ is $\frac{a+b}{\sqrt{2}}$, so the area of square $I$ is $\frac{(a+b)^2}{2}$.

The area of square $II$ is twice as much as the area of square $I$, so the area of square $II$ is $(a+b)^2$. That means the side length of square $II$ is $a+b$, so the perimeter of square $II$ is $4(a+b)$, or answer choice $\boxed{\textbf{(E)}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions
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