   SEARCH HOME Math Central Quandaries & Queries  Question from Lillian, a student: A rectangle is 5cm longer than twice its width. The width of another rectangle is 3cm less than the width of the first rectangle and its length is 6cm more than 3 times its width. If the perimeters are equal, find the dimensions of both rectangles Lillian,

I need a way to distinguish the lengths and widths of the two rectangles so I am going to let l and w be the length and width of the first rectangle and L and W the length and width of the second, all in centimeters. You are given four facts, each of which gives an equation involving some of the variables l, w, L and W.

The first fact is "A rectangle is 5cm longer than twice its width." This is the first rectangle so twice its width is 2w cm and 5 cm longer than this is 2w + 5 cm. You are told that this is the length of the first rectangle so the first equation is

l = 2w + 5

The second fact is "The width of another rectangle is 3cm less than the width of the first rectangle." What is the width the first rectangle? What is 3 cm less than this? Thus

W = ...

What are the other two equations?

Can you solve the problem from here? If you have difficulty write back and tell us what you did.

Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.