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.

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)**

**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 in^{2}

l ⋅ w = 150

x ⋅ 2x = 150

2x^{2} = 150

Divide each side by 2.

x^{2} = 75

Find positive square root on both sides.

√x^{2 } = √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

AB^{2} + BC^{2} = AC^{2}

Substitute.

AB^{2} + 3^{2} = (√13)^{2}

Simplify and solve for AB.

AB^{2} + 9 = 13

Subtract 9 from each side.

AB^{2} = 4

Find positive square root on both sides.

√AB^{2} = √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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