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 (m

^{2}). 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 quadrilateral.

**What is a Quadrilateral?**

Before going into the calculation of area, let us define what is quadrilateral. A quadrilateral is a four-sided polygon, having the sum of interior angles equal to \(360^{\circ}\).

Properties of a quadrilateral:

- 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.

**Area of Quadrilateral**

Consider a quadrilateral PQRS, of different(unequal) lengths, let us derive a formula for the area of quadrilateral.

- We can view the quadrilateral as a combination of 2 triangles, with the diagonal PR being the common base.
*h*and_{1 }*h*are the heights of triangles PSR and PQR respectively._{2}

- 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 *
*h*)/2_{1} - Area of triangle PQR = (base * height)/2 = (PR*
*h*)/2_{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 formula,

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

**Let us solve few examples-**

**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 cm^{2}

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