# Area of Rectangle: Worked Examples

In the previous post, we have learned some problems on how to solve area of squares. In this post, we are going to solve problems involving area of rectangles. The area of rectangle is one of the most common questions in the Geometry part of the Civil Service Examination. In this post, we are going to learn how to solve problems involving area of rectangles. The formula for the Area (A) can be calculated by multiplying the length (l) and the width (w). That is, $A = l \times w$.

Problem 1

Find the area of a rectangle with length 8cm and width 5 cm.

Solution $A = l \times w$ $A = 8 \times 5$ $A = 40$

So, the area of the rectangle is $40 cm^2$

Problem 2

The area of a rectangle is $12 cm^2$ and its width is $3$ cm. What is its length?

Solution $A = l \times w$

Substituting the given, we have $12 = l \times 3$.

To get $l$, we have to divide both sides by 3. $\dfrac{12}{3} = \dfrac{l \times 3}{3}$ $4 = l$

Therefore, the length is equal to 3.

Problem 3

The length of a rectangle is twice its width. Its area is 32 square units. What are the dimensions of the rectangle?

Solution

Let $x$ = width of the rectangle

Since the length is twice, it is $2x$.

So, $w = x$ $l = 2x$ $A = l \times w$ $A = 2w \times w$ $32 = 2w^2$

Dividing both sides by 2, we have $16 = w^2$.

Getting the square root of both sides, we have $4 = w$

So, $l = 2w = 2(4) = 8$.

So, the dimension of the rectangle is 4 by 8.

Problem 4

The length of a rectangle is two more than its width. Its area is 48 square units. What are the dimensions of the rectangle?

Solution

Let length be equals x and width be equals x + 2. That is $w = x$ $l = x + 2$ $A = lw$ $48 = x(x + 2)$ $48 = x^2 + 2x$

Subtracting 48 from both sides, we have $x^2 + 2x - 48 = 0$.

Factoring, we have $(x + 8)(x - 6) = 0$ $x = -8, x = 6$

So, the width is 6 units and the length is 6 + 2 = 8 units.

Check:
Is the length two more than $6$?
Is $6 \times 8$ equals $48$? 