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Difference between revisions of "1998 AJHSME Problems/Problem 20"

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==Solution==
 
==Solution==
  
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After all folds are completed, the square would be <math>\frac{1}{4}</math> of the original square.
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After both folds are completed, the square would become a triangle that has an area of <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math> of the original square.
  
 
Since the area is <math>9</math> for <math>\frac{1}{4}</math> of the square, <math>9\times4=36</math> is the area of square <math>PQRS</math>
 
Since the area is <math>9</math> for <math>\frac{1}{4}</math> of the square, <math>9\times4=36</math> is the area of square <math>PQRS</math>
  
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<math>\sqrt{36}=6</math>
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The length of the side of a square that has an area of <math>36</math> square units is <math>\sqrt{36}=6</math> units.
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<math>6\times4=24=\boxed{D}</math>
 
  
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Each side is <math>6</math> units, so the total perimeter is <math>6\times4=24=\boxed{D}</math>
  
 
== See also ==
 
== See also ==

Revision as of 11:06, 31 July 2011

Problem 20

Let $PQRS$ be a square piece of paper. $P$ is folded onto $R$ and then $Q$ is folded onto $S$. The area of the resulting figure is 9 square inches. Find the perimeter of square $PQRS$.

[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$P$",(0,2),SE); label("$Q$",(2,2),SW); label("$R$",(2,0),NW); label("$S$",(0,0),NE); [/asy]

$\text{(A)}\ 9 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 36$

Solution

After both folds are completed, the square would become a triangle that has an area of $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ of the original square.

Since the area is $9$ for $\frac{1}{4}$ of the square, $9\times4=36$ is the area of square $PQRS$

The length of the side of a square that has an area of $36$ square units is $\sqrt{36}=6$ units.

Each side is $6$ units, so the total perimeter is $6\times4=24=\boxed{D}$

See also

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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