### Solved Examples

**Question 1: **A rectangle is 5m long and 4 m wide. What is its area?

** Solution: **

Length of the rectangle, l = 5m.

Width of the rectangle, w = 4m.

Area of the rectangle = length $\times$ width = 5 $\times$ 4 = 20$m^2$.

**Question 2: **A rectangle is 5m long and 4 m wide. What is its perimeter?

** Solution: **

Length of the rectangle, l = 5m.

Width of the rectangle, w = 4m.

Perimeter of the rectangle = 2(l +w) = 2(5+4) = 2$\times$9 = 18 m.

**Question 3: **A rectangle is 4m long and 3 m wide. What is the length of its diagonal?

** Solution: **

Length of the rectangle, l = 4m.

Width of the rectangle, w = 3m.

The length of its diagonal, d = $\sqrt{l^2 + w^2}$, where l is its length and w is its width.

Diagonal length, d = $\sqrt{4^2 + 3^2}$ = $\sqrt{16 + 9}$ = $\sqrt{25}$ = 5m.

**Question 4: **A rectangle has a length which is 4 less than 5 times the width. If the
area of this rectangle is 9 units, find the dimensions of this rectangle.

** Solution: **

Let the width of the rectangle = w.

Given, the length of the rectangle is 4 less than 5 times its width. Then, l = 5w – 4.

The area of the rectangle = 9 units. …………………(1)

Also by the definition of area of rectangle, Area of rectangle = l $\times$ w = (5w - 4)$\times$ w = 5w$^2$ – 4w ……(2)

Equating (1) and (2),

5w$^2$ - 4w = 9

5w$^2$ - 4w - 9 = 0

5w$^2$ + 5w - 9w - 9 = 0

5w(w + 1) -9 (w + 1) = 0

(5w - 9)(w + 1) = 0

Using the zero product rule, we have, 5w – 9 = 0 or w + 1 = 0

5w = 9 or w = -1

Discarding the negative values as width is a measurement, we have,

Width of the rectangle, w = $\frac{9}{5}$ = 1.8 units

Length of the rectangle, l = 5w – 4 = 5(1.8) - 4 = 9 - 4 = 5 units

**Question 5: **In the diagram as shown below, a rectangle's length is 3x + 2 and its
width is 3x – 2. If its area is 32 cm$^2$, what are the rectangle's
dimensions and what is its perimeter?

** Solution: **

Length of the rectangle, l = 3x + 2.

Width of the rectangle, w = 3x - 2.

The area of the rectangle = 32 $cm^2$. ………………….(1)

Also by the definition of area of rectangle, Area of rectangle = l $\times$ w = (3x + 2)$\times$(3x - 2) …………(2)

Equating (1) and (2),

(3x + 2)$\times$(3x - 2) = 32

9x$^2$ – 4 = 32

9x$^2$ = 36

x$^2$ = $\frac{36}{9}$ = 4

x = $\sqrt{4}$ = ±2

Considering x = 2 and discarding x = -2 as it gives negative dimensions, we have

Length of the rectangle, l = 3x + 2 = 3(2) + 2 = 6 + 2 = 8 cms.

Width of the rectangle, w = 3x - 2 = 3(2) – 2 = 6 - 2 = 4 cms.

Perimeter of the rectangle = 2(l + w) = 2(8 + 4) = 2(12) = 24 cms.

**Question 6: **A rectangle has a width of 12 units and a diagonal of 13 units. Find the area and perimeter of this rectangle.

** Solution: **

Let the length of the rectangle, l = l units.

The width of the rectangle = 12 units.

The length of its diagonal, d = $\sqrt{l^2 + w^2}$, where l is its length and w is its width.

Diagonal length, d = $\sqrt{l^2 + 12^2}$ = $\sqrt{l^2 + 144}$.

But given diagonal, d = 13 units, so we have,

13 = $\sqrt{l^2 + 144}$

13$^2$ = l$^2$ + 144

l$^2$ = 169 – 144 = 25

Thus the length of the rectangle, l = $\sqrt{25}$ = 5 units.

Area of the rectangle = length $\times$ width = 5 $\times$ 12 = 60 units$^2$.

Perimeter of the rectangle = 2(l + w) = 2(5 + 12) = 2 $\times$ 17 = 34 units.