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 90^o 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

l ⋅ w  =  150

x ⋅ 2x  =  150

2x²  =  150

Divide each side by 2.

x²  =  75

Find positive square root on both sides.

√x²  =  √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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