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Difference between revisions of "1994 AJHSME Problems/Problem 16"

(Created page with "==Problem== The perimeter of one square is <math>3</math> times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?...")
 
 
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<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9</math>
 
<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9</math>
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==Solution==
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Let <math>a</math> be the sidelength of one square, and <math>b</math> be the sidelength of the other, where <math>a>b</math>. If the perimeter of one is <math>3</math> times the other's, then <math>a=3b</math>. The area of the larger square over the area of the smaller square is
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<cmath>\frac{a^2}{b^2} = \frac{(3b)^2}{b^2} = \frac{9b^2}{b^2} = \boxed{\text{(E)}\ 9}</cmath>
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==See Also==
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{{AJHSME box|year=1994|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 00:13, 5 July 2013

Problem

The perimeter of one square is $3$ times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$

Solution

Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$. If the perimeter of one is $3$ times the other's, then $a=3b$. The area of the larger square over the area of the smaller square is

\[\frac{a^2}{b^2} = \frac{(3b)^2}{b^2} = \frac{9b^2}{b^2} = \boxed{\text{(E)}\ 9}\]

See Also

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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

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