Question

A frame around a rectangular family portrait has a perimeter of $60$ $\mathrm{in}$. The length is $15$$\mathrm{in}$ less than $2$ times the width. Find the length and width of the frame.

Answer 1

Step 1: To find the width.

Formula for the perimeter of a rectangle is $2\left(\mathrm{l}+\mathrm{w}\right)...\left(1\right)$, where $l$ is length and $w$ is width.

Given perimeter of the frame is $60$ $\mathrm{in}$.

The relationship between length and width is $l=2w-15...\left(2\right)$.

Substitute $\left(2\right)$ into $\left(1\right)$and equate with $60$.

$$\begin{gathered} 2(2w-15+w)=60 \\ \Rightarrow 2(3w-15)=60 \\ \Rightarrow (3w-15)=30[\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2] \\ \Rightarrow 3w=45[\mathrm{adding}\mathrm{both}\mathrm{sides}\mathrm{with}15] \\ \therefore w=15\mathrm{in}[\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3] \end{gathered}$$

Step 2: To find the length.

Substitute the value of w into $\left(2\right)$ to obtain the length.

$$\begin{gathered} l=2\left(15\right)-15 \\ \Rightarrow l=30-15 \\ \therefore l=15\mathrm{in} \end{gathered}$$

Hence, the length is $15$ $\mathrm{in}$ and width is $15$ $\mathrm{in}$.