 You are already acquainted with the term area. It is defined as the region occupied inside the boundary of a flat object or figure. The measurement is done in square units with the standard unit being square metres (m2). For the computation of area, there are pre-defined formulas for squares, rectangles, circle, triangles, general quadrilaterals etc. In this article, we will learn about the area of a quadrilateral.

Before going into the calculation of area, let us define what is a quadrilateral. A quadrilateral is a four-sided polygon, having the sum of interior angles equal to 360o.

• Every quadrilateral has 4 vertices and 4 sides enclosing 4 angles.
• The sum of its interior angles is 360 degrees.
• A quadrilateral, in general, has sides of different lengths and angles of different measures. However, squares, rectangles, parallelograms, etc. are special types of quadrilaterals with some of their sides and angles being equal.

Consider a quadrilateral PQRS, of different(unequal) lengths, let us derive a formula for the area of a quadrilateral. • We can view the quadrilateral as a combination of 2 triangles, with the diagonal PR being the common base.
• h1 and h2 are the heights of triangles PSR and PQR respectively. • Area of quadrilateral PQRS is equal to the sum of the area of triangle PSR and the area of triangle PQR.
• Area of triangle PSR = (base * height)/2 = (PR * h1)/2
• Area of triangle PQR = (base * height)/2 = (PR* h2)/2
• Thus, area of quadrilateral PQRS is,
• Area of triangle PSR + Area of triangle PQR =$\frac{PR \times h_{1}}{2} + \frac{PR \times h_{2}}{2} = PR \left ( \frac{h_{1}+ h_{2}}{2} \right )$
• $= \frac{1}{2} PR \times (h_{1}+ h_{2})$

Hence, the area of a quadrilateral formula is,

Area of a general Quadrilateral $= \frac{1}{2} \times \; diagonal \times (Sum \; of \; height \; of \; two \; triangle)$

### Area of a Quadrilateral Example

Question:

In the given quadrilateral ABCD, the side BD = 15 cm and the heights of the triangles ABD and BCD are 5 cm and 7 cm respectively. Find the area of the quadrilateral ABCD. Solution:

Diagonal = BD = 15 cm

Heights, $h_{1} = 5$ cm & $h_{2} = 7$ cm

Sum of the heights of the triangles = h1 + h2 = 5 + 7 = 12 cm

Thus, area of quadrilateral ABCD =

$= \frac{1}{2} \times \; diagonal \times (Sum \; of \; height \; of \; two \; triangle)$

= (15 * 12)/2 = 90 cm2