Question

Practice 2-7 Try, Test, ReviseUse the try, test, revise strategy to solve each problem.1. The length of a rectangle is 9 in. greater than the width. The area is36 in. Find the dimensions.

Answer 1

Given:

The length of a rectangle is 9 in. greater than the width.

So, l=9+w

The area is 36 square inches.

To find the dimensions:

Area formula is,

`A=l* w`

On substitution we get,

`\begin{gathered} 36=(9+w)(w) \\ 36=9w+w^2 \\ w^2+9w-36=0 \\ (w_{}+12)(w-3)=0 \\ w=-12,3 \end{gathered}`

Negative value is negligible.

Hence, the width, w=3 and the length l=9+3=12.

Hence, the dimensions are,

`l=12,w=3`