Area of Rhombus

Consider the trapezium with constructed lines as shown in the figure below:

In order to find the area of a trapezium, first, let us find the area of the triangles $ABD$ and $BDF$ with the same height $h$.

The base of triangle $ABD$ is $b_1$ and the base of triangle $BDF$ is $b_2$.

The area of triangle $ABD = \frac{1}{2} b_1 h$ and the area of triangle $BDF = \frac{1}{2} b_2 h$.

The total area of the trapezium is:

$\text{Area of trapezium} = \frac{1}{2} (b_1 + b_2) h$

Generally, the area of the trapezium is given by the product of half the sum of the parallel sides (bases) and the perpendicular distance between them (height).

Example 1

Find the height of the trapezium with area 90 square units and bases of 6 units and 14 units.

Solution:

Consider the trapezium below:

The area of the trapezium is given by:

$\text{Area} = \frac{1}{2} (b_1 + b_2) h$

Substitute the given values:

$90 = \frac{1}{2} (6 + 14) h$

$90 = 10h$

$h = 9 \, \text{units}$

Therefore, the height of the trapezium is 9 units.

Consider the parallelogram below with constructed lines as shown in the figure:

The area of the parallelogram can be derived from the formula for the area of the trapezium. The key observation is that the bases for a parallelogram are equal.

The area of parallelogram $ABCD$ is:

$\text{Area of parallelogram} = \frac{1}{2} (AB + DC) h$

Since $AB = DC$, we can simplify this to:

$\text{Area of parallelogram} = AB \times h$

If $AB = b$, then:

$\text{Area of parallelogram} = b \times h$

Therefore, the area of a parallelogram is equal to the product of the base and the perpendicular height.

Consider the rectangle below:

The rectangle $ABCD$ is divided into two congruent triangles, triangle $ABD$ and triangle $ACD$, by the diagonal $AD$.

The area of $ABCD =$ area of triangle $ABD +$ area of triangle $ACD$.

Since the triangles are equal, the area of $ABCD$ is double the area of one of the triangles. The area of $ABCD$ is:

$\text{Area of rectangle} = l \times w$

Where $l$ is the length and $w$ is the width of the rectangle.

Therefore, the area of a rectangle is the product of its length and width.

A square is a special rectangle with equal sides. Therefore, the area of the square is the product of its sides:

$\text{Area of square} = l \times l = l^2$

We can also find the area of a square by using the length of its diagonals. Consider the square below with diagonals $AC$ and $BD$:

Each of the diagonals of a square bisects at a right angle. The area of triangle $ABC$ is equal to the area of triangle $ADC$.

Since the length of the diagonals are equal, then $AC = BD$. So, the area of square $ABCD$ is:

$\text{Area of square} = \frac{1}{2} (AC)^2$

Therefore, the area of a square is equal to half of the product of the lengths of the diagonals.

Example 2

Find the area of a parallelogram $ABCD$ if $AC = 7 \, \text{cm}$, $AB = 9 \, \text{cm}$, and the angle $\theta = 60^\circ$.

Solution:

Using the formula for the area of a parallelogram:

$\text{Area of parallelogram} = AB \times h$

We need to find the height $h$. To do this, consider triangle $ACE$ where $h = AC \times \sin \theta$. Thus,

$h = 7 \times \sin 60^\circ = 7 \times 0.866 = 6.062 \, \text{cm}$

Now, substitute into the area formula:

$\text{Area of parallelogram} = 9 \times 6.062 = 54.558 \, \text{cm}^2$

Therefore, the area of the parallelogram $ABCD$ is 53.427 cm².