PERIMETER OF RECTANGLE

A rectangle is a four-sided closed figure where the lengths of opposite sides will be equal and each vertex angle will be right angle or 90o as shown below. 

Formula for Perimeter of Rectangle

Let l be the length and w be the width of a rectangle. 

Then, the formula for perimeter of the rectangle : 

Perimeter  =  2(l + w)

Examples

Example 1:

The length and width of a rectangle are 16 cm and 12 cm respectively. Find its perimeter. 

Solution:

Formula for perimeter of a rectangle :

=  2(l + w) 

Substitute 16 for l and 12 for w.

=  2(16 + 12)

=  2(28)

=  56

So, the perimeter of the rectangle is 56 cm.

Example 2:

If the perimeter of a rectangle is 50 cm and its length is 15 cm, then find its width. 

Solution:

Perimeter of the rectangle  =  50 cm

2(l + w)  =  50

Divide each side by 2.

 l + w  =  25

Substitute 15 for l.

15 + w  =  25

Subtract 15 from each side.

w  =  10

So, the width of the rectangle is 10 cm. 

Example 3 :

The area of the rectangle is 150 square inches. If the length is twice the width, then find its perimeter. 

Solution:

Let x be the width of the rectangle. 

Then, the length of the rectangle is 2x.

Area of the rectangle  =  150 in2

⋅ w  =  150

⋅ 2x  =  150

2x2  =  150

Divide each side by 2.

x2  =  75

Find positive square root on both sides.

√x =  √75

x  =  √(5 ⋅ 5 ⋅ 3)

x  =  5√3

Therefor, the width of the rectangle is 5√3 in.

Then, the length of the rectangle is 

=  2 ⋅ width

=  2 ⋅ 5√3

=  10√3 in

Formula for perimeter of a rectangle :

=  2(l + w)

Substitute 10√3 for l and 5√3 for w. 

=  2(10√3 + 5√3)

=  2(15√3)

=  30√3

So, the perimeter of the rectangle is 30√3 in. 

Example 4:

The length of a rectangle is 3 ft and one of the diagonal measures √13 ft. Find its perimeter. 

Solution:

To find the perimeter of a rectangle, we have to know its length and width. Length is given in the question, that is 3 ft. So, find its width. 

Draw a sketch. 

In the figure shown above, consider the right triangle ABC. 

By Pythagorean Theorem, we have

AB2 + BC2  =  AC2

Substitute.

AB2 + 32  =  (√13)2

Simplify and solve for AB. 

AB2 + 9  =  13

Subtract 9 from each side. 

AB2  =  4

Find positive square root on both sides.

 √AB2  =  √4

AB  =  2

Therefore, the width of the rectangle is 2 ft. 

Formula for perimeter of a rectangle.

=  2(l + w)

Substitute 3 for l and 2 for w.

=  2(3 + 2)

=  2(5)

=  10

So, the perimeter of the rectangle is 10 ft. 

Example 5:

The length of a rectangle is 3 yards more than its width and its perimeter is 18 yards. Find its length and width.

Solution:

Let x be the width of the rectangle.

Then, the length of the rectangle is (x + 3) yards. 

Perimeter of the rectangle  =  18 yards

2(l + w)  =  18

Divide each side by 2. 

l + w  =  9

Substitute (x + 3) for l and x for w. 

(x + 3) + x  =  9

x + 3 + x  =  9

2x + 3  =  9

Subtract 3 from each side. 

2x  =  6 

Divide each side by 2. 

x  =  3

x + 3  =  6

So, the length of width of the rectangle are 6 yards and 3 yards respectively.

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