Question

A rectangle has a length that is three feet more than twice it's width. The perimeter of the rectangle is 78 feet. State the value for the width.

Answer 1

Let the width of the rectangle be 'w'

According to the question length of the rectangle (l) is three feet more than twice it's width;

$\longrightarrow$ l = (2w + 3) ft

Perimeter of rectangle = 78 ft

Formula of perimeter of rectangle = 2(Length + Width)

So,

$\rm \implies 2(Length + Width) = 78 \\ \\ \rm \implies 2(2w + 3 + w) = 78 \\ \\ \rm \implies 2(3w + 3) = 78 \\ \\ \rm \implies 2 \times 3(w + 1) = 78 \\ \\ \rm \implies 6(w + 1) = 78 \\ \\ \rm \implies w + 1 = \dfrac{78}{6} \\ \\ \rm \implies w + 1 = 13 \\ \\ \rm \implies w = 13 - 1 \\ \\ \rm \implies w = 12 \: ft$

$\therefore$ Width of the rectangle (w) = 12 ft