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

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==Problem==
 
==Problem==
 
Let <math>PQRS</math> be a square piece of paper.  <math>P</math> is folded onto <math>R</math> and then <math>Q</math> is folded onto <math>S</math>. The area of the resulting figure is 9 square inches. Find the perimeter of square <math>PQRS</math>.
 
Let <math>PQRS</math> be a square piece of paper.  <math>P</math> is folded onto <math>R</math> and then <math>Q</math> is folded onto <math>S</math>. The area of the resulting figure is 9 square inches. Find the perimeter of square <math>PQRS</math>.
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[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("<math>P</math>",(0,2),SE); label("<math>Q</math>",(2,2),SW); label("<math>R</math>",(2,0),NW); label("<math>S</math>",(0,0),NE); [/asy]
 
  
 
==Solution==
 
==Solution==

Latest revision as of 06:57, 18 December 2024

Problem

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 $PQRS$.

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

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

Each side is $6$ inches, 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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