Question

Area of a rectangle is 99 ft^2 , and the length of the rectangle is 7 ft more than twice the width, how do you find the dimensions of the rectangle?

Answer 1

Width `=11/2` ft and Length `=18` ft

Explanation:

Let width of the rectangle be `x` ft
Given that length is `7` ft more than the twice the width.
`:.` length `=2xx "width"+7` ft
or Length `=2x+7` ft

Area of the rectangle = Length `xx` Width
Inserting given and above values

`99=(2x+7)xxx`
`=>2x^2+7x-99=0`
Solve by finding roots `x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-7+-sqrt(7^2-4*2*(-99)))/(2*2)`
or `x=(-7+-sqrt(49+792))/4`
or `x=(-7+-sqrt(841))/4`
or `x=(-7+-29)/4`
or `x=-9, 11/2`
Ignoring `-ve` root as width can not be negative.
Width `=11/2 and` Length `2xx11/2+7=18`