Question

#13The area of a rectangle is 78 square inches. If the length of the rectangleis one more than twice the width, find the dimensions of the rectangle.

Answer 1

Given

The area of rectangle is given 78 sq.inches.

The length is one more than the twice the width.

Explanation

To determine the dimensions of the rectangle.

Let the length be l and width be w.

Then the area of rectangle is

$A=l\times w$

From the statement given, the equation formed is

$l=2w+1$

Substitute the length relation in the area of rectangle

$\begin{gathered} A=(2w+1)\cdot w \\ A=2w^2+w \\ 78=2w^2+w \\ 2w^2+w-78=0 \end{gathered}$

Solve the quadratic equation and find width w.

$\begin{gathered} (w-6)(w+\frac{13}{2})=0 \\ w=6,\frac{-13}{2} \end{gathered}$

As width cannot be negative then width is 6 in.

And length is determined by

$\begin{gathered} l=2w+1 \\ l=2\times6+1 \\ l=12+1 \\ l=13 \end{gathered}$Answer

Hence the length is 13 in and width is 6 in.