Solving Problems Involving the Perimeter of Rectangles with Known Area
Geometry often involves working with shapes and their associated measurements. One common problem is finding the perimeter of a rectangle when the area and a relationship between the dimensions are given. This article will explore several examples of such problems and demonstrate how to solve them using algebraic methods.
Example 1: Area and Proportion
Consider a rectangle where the length is four times the width. If the area of the rectangle is 100 cm2, what is its perimeter?
To start, let the width of the rectangle be w cm. The length, therefore, is l 4w cm. The area A of the rectangle is given by the formula:
A l × w
Substituting the expression for length:
100 4w × w
This simplifies to:
100 4w2
25 w2
Taking the square root of both sides:
w 5 cm
Now substituting back to find the length:
l 4w 4 × 5 20 cm
Using the formula for the perimeter P of the rectangle:
P 2l 2w
Substituting the values of length and width:
P 2 × 20 2 × 5 50 cm
Therefore, the perimeter of the rectangle is 50 cm.
Example 2: Area and Calculation
Consider a different example where the area of the rectangle is 80 cm2. If the length is twice the width, how do we find the perimeter?
Let x be the width, then the length 2x. The area is:
A width × length 80 x × 2x 2x2
So, 2x2 80, which simplifies to:
x2 40
x √40 6.325 cm
Length 2x 2 × 6.325 12.65 cm
The perimeter is given by:
P 2(width length) 2(6.325 12.65) 37.95 cm
Example 3: Area and Substitution
Assume the area of the rectangle is 64 cm2. If the length is four times the width, how do we determine the perimeter?
Let b be the width, then the length is 4b. The area is:
A width × length 64 b × 4b 4b2
So, 4b2 64, which simplifies to:
b2 16
b 4 cm
Length 4b 4 × 4 16 cm
The perimeter is given by:
P 2(4 cm 16 cm) 40 cm
Therefore, the perimeter of the rectangle is 40 cm.