2009 CEMC Gauss (Grade 8) Problems/Problem 15

Problem

In rectangle $PQRS$ , $PQ = 12$ and $PR = 13$ . The area of rectangle $PQRS$ is

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$\text{ (A) }\ 30 \qquad\text{ (B) }\ 60 \qquad\text{ (C) }\ 36 \qquad\text{ (D) }\ 78 \qquad\text{ (E) }\ 72$

Solution 1

Because $PQRS$ is a rectangle, we can notice that $\angle PQR$ is a right angle. Thus, $\Delta PQR$ is a right triangle.

This means that we can use the pythagorean theorem to find the length of $QR$ :

$QR^{2} + 12^{2} = 13^{2}$

$QR^{2} = 13^{2} - 12^{2} = 169 - 144 = 25$

$QR = \sqrt{25}$

$QR = 5$

We can now use the formula for the area of a rectangle, which is the rectangle's width multiplied by its height:

$A = 12 \times 5 = \boxed {\textbf {(B) } 60}$

~anabel.disher

Solution 1.1

We can follow the same steps as solution 1. However, when finding $QR^{2}$ , we can also use difference of squares.

$QR^{2} = 13^{2} - 12^{2} = (13 - 12) \times (13 + 12) = 1 \times 25 = 25$

This eventually leads us to the same answer, or $\boxed {\textbf {(B) } 60}$ .

~anabel.disher

Solution 1.2

Instead of having to use the pythagorean theorem in solution 1, we can remember that $5$ , $12$ , and $13$ form a pythagorean triple. This gives us $QR = 5$ , and eventually leads us to the same answer, or $\boxed {\textbf {(B) } 60}$ .

~anabel.disher

2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2009 CEMC Gauss (Grade 8) Problems/Problem 15

Contents

Problem

Solution 1

Solution 1.1

Solution 1.2