Perimeter :
Perimeter is a path that surrounds a rectangle. The term may be used either for the path or its length it can be thought of as the length of the outline of the rectangle.
Area :
Area of a rectangle is defined as the space occupied by the rectangle shaped object on a flat surface. The area of a shape can be measured by comparing the shape to squares of a fixed size.
length = l and width = w
Perimeter = 2l + 2w
Area = lw
The measurements of perimeter use units such as centimeters, meters, kilometers, inches, feet, yards, and miles. The measurements of area use units such as square centimeters (cm^{2}), square meters(m^{2}), and so on.
Example 1 :
Find the perimeter and area of a rectangle of length 12 inches and width 5 inches.
Solution :
Draw a rectangle and label the length and width.
Perimeter = 2l + 2w = 2(12) + 2(5) = 24 + 10 = 34 |
Area = lw = 12 ⋅ 5 = 60 |
So, the perimeter is 34 inches and the area is 60 square inches.
Example 2 :
The area of a rectangle is 56 square inches. If the length is 8 inches, find the width of the rectangle.
Solution :
Given : Area of the rectangle = 56 sq. in.
lw = 56
Substitute l = 8.
8w = 56
Divide both sides by 8.
w = 7 in
Example 3 :
The perimeter of a rectangle is 30 cm. The length is 3 more than twice the width. Find the length and width of the rectangle.
Solution :
Step 1 :
Let w = x.
Then, l = 2x + 3.
Step 2 :
Given : Perimeter of the rectangle = 30 cm.
2l + 2w = 30
Step 3 :
Substitute w = x and l = 2x + 3.
2(2x + 3) + 2x = 30
4x + 6 + 2x = 30
6x + 6 = 30
Subtract 6 from both sides.
6x = 24
Divide both sides by 6.
x = 4
Width = x = 4.
Length = 2x + 3 = 2(4) + 3 = 11.
So, the length is 11 cm and width is 4 cm.
Example 4 :
The diagonal of a rectangle is 5 cm and one of its sides is 4 cm. Find its area.
Solution :
Step 1 :
Let us assume that one of the sides given is length.
Then, l = 4 cm.
Step 2 :
Draw a rectangle and label the diagonal and length.
Step 3 :
To find area of a rectangle, we need the measures of length and width. We know the length and it is 4 cm. We have to find the width.
Step 4 :
In the rectangle above, let us consider the right triangle ABC and apply Pythagorean theorem.
AB^{2} + BC^{2} = AC^{2}
Substitute BC = 4 and AC = 6.
w^{2} + 4^{2} = 5^{2}
w^{2} + 16 = 25
Subtract 16 from both the sides.
w^{2} = 9
w^{2} = 3^{2}
Remove the exponent 2 on both sides.
w = 3 cm
Step 5 :
Area of the rectangle = lw
Substitute l = 4 and w = 3.
= 4 ⋅ 3
= 12 cm^{2 }
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