Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2009 Grade 8 CEMC Gauss Problems/Problem 15

Problem

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_Grade_8_CEMC_Gauss_Problems/Problem_15&oldid=251578"