The length of a rectangular poster is 9 inches longer than its width. If the diagonal is 15 inches, what are the dimensions of the poster?

To solve the problem, we need to find the dimensions of the rectangular poster. Let's denote the width of the poster as \( w \) inches. Since the length is 9 inches longer than the width, the length can be expressed as \( w + 9 \) inches.

Given that the diagonal of the rectangle is 15 inches, we can use the Pythagorean theorem, which states that:

\[
\text{Diagonal}^2 = \text{Length}^2 + \text{Width}^2
\]

Substituting the given values into the equation:

\[
15^2 = (w + 9)^2 + w^2
\]

Now, let's solve for \( w \).

\[
225 = (w + 9)^2 + w^2
\]

Expand the square:

\[
225 = w^2 + 18w + 81 + w^2
\]

Combine like terms:

\[
225 = 2w^2 + 18w + 81
\]

Subtract 225 from both sides to set the equation to zero:

\[
0 = 2w^2 + 18w + 81 - 225
\]

\[
0 = 2w^2 + 18w - 144
\]

Divide the entire equation by 2 to simplify:

\[
0 = w^2 + 9w - 72
\]

Now, we solve this quadratic equation using the quadratic formula:

\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation, \( a = 1 \), \( b = 9 \), and \( c = -72 \):

\[
w = \frac{-9 \pm \sqrt{9^2 + 4 \times 72}}{2}
\]

\[
w = \frac{-9 \pm \sqrt{81 + 288}}{2}
\]

\[
w

Discover how to find the dimensions of a rectangular poster when given its diagonal length of 15 inches. This problem involves applying the Pythagorean theorem in geometry to solve for the width and length of the poster. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation