Math CentralQuandaries & Queries


Question from Thomas, a parent:

I'm trying to help my 9th grade child with their homework. The problem is the
Length of a rectangle is twice as long as it is wide. The perimeter is 68m. Find
It's length and width

Hi Thomas,

Suppose the length of the rectangle is $L$ m and the width is $W$ m.

You know two facts that you want to use to produce two equations in $L$ and $W.$

First the length is twice the width so $L = 2 W.$

The second fact is that the perimeter is 68 m. The perimeter is $L + W + L + W$ so your second equation is $2L + 2W = 68.$

From the first equation you can substitute $L = 2 W$ into the second equation and solve for $W.$

I hope this helps,

About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS