The product of the dimensions of a rectangle is the area. Usually questions asking about dimensions give you some relation between the dimensions and the area. You use the given information together with your knowledge of the area formula to create and solve the appropriate equations.
Example
Given: The length of a rectangle is 3 units more than the width. The area is 40 square units.
Find: What are the length and width?
Solution: Let "L" be the length, "W" be the width.
L = W + 3 (the length is 3 units more than the width)
L*W = 40 (the area is 40 square units)
The first equation provides an expression for L, so we can substitute that into the second equation to get an equation in W alone.
(W + 3)*W = 40
Now, we solve this.
W2 + 3W - 40 = 0 (multiply it out, subtract 40 from both sides)
(W+8)(W-5) = 0 (factor the expression)
W = -8 or W = 5 (identify values that make the factors =0)
We know the dimensions must be positive, so we choose the W = 5 solution. We also know that L=W+3, so that means L = 5+3 = 8
The length and width of the rectangle are 8 units and 5 units.
Example
Given: The length of a rectangle is 3 units more than the width. The area is 40 square units.
Find: What are the length and width?
Solution: Let "L" be the length, "W" be the width.
L = W + 3 (the length is 3 units more than the width)
L*W = 40 (the area is 40 square units)
The first equation provides an expression for L, so we can substitute that into the second equation to get an equation in W alone.
(W + 3)*W = 40
Now, we solve this.
W2 + 3W - 40 = 0 (multiply it out, subtract 40 from both sides)
(W+8)(W-5) = 0 (factor the expression)
W = -8 or W = 5 (identify values that make the factors =0)
We know the dimensions must be positive, so we choose the W = 5 solution. We also know that L=W+3, so that means L = 5+3 = 8
The length and width of the rectangle are 8 units and 5 units.