Hi, there--
THE PROBLEM:
A square and a rectangle each has a perimeter of 48 m. The difference between the areas of
two figures is 4 square meters. Find the dimensions of the rectangle.
A SOLUTION:
Consider the square with perimeter 48m.
Let x be the length of each side.
The perimeter is the distance around, or 4x.
4x = 48
x = 12
The length of each side is 12m.
The area of the square is the side length times the side length, or x*x.
12*12 = 144
The area of the square is 144 square meters.
We are told that the difference between the area of the square and the rectangle is 4 square
meters. Then the area of the rectangle must be 4 more or 4 less.
For now, assume that the rectangle has the lesser area, or 140 square meters.
The formula for the area of a rectangle is length times width, or L*W. We have
L*W = 140.
We also know that the perimeter of the rectangle is 48. We have
2L + 2W = 48
We have two equations with two variables. Solve for L and W.
Solve the first equation for L.
L*W = 140
L = 140/W
Substitute 140/W for L in the second equation.
2L + 2W = 48
2(140/W) + 2W = 48
Solve for W.
280/W + 2W = 48
Subtract 2W from both sides.
280/W = -2W + 48
Multiply both sides by W.
280 = -2W^2 + 48W
This is a quadratic equation. Solve by factoring.
2W^2 - 48W + 280 = 0
Divide each term by 2.
W^2 - 24W + 140 = 0
(W - 14)(W - 10) = 0
W = 14 or W = 10
The width of the rectangle is 10m or 14m.
If the width of the rectangle is 10m, then its length is 14m because
2L + 2W = 2(10) + 2(14) = 48
The dimensions of the rectangle are 10m by 14m.
You can work out the dimensions of the rectangle if it has the larger area. Then your
equations will be
L*W = 148
2L + 2W = 48
I'll leave that to you.
Hope this helps! Feel free to email if you have any questions about the solution.
Good luck with your math,
Mrs. F
math.in.the.vortex@gmail.com
