Quadrilateral is a closed convex polygon bounded by four sides. We have already seen that the quadrilaterals are classified according to the interior angles and the sides. Therefore, the names of the quadrilaterals vary according to their properties. Areas of rectangle, square, parallelograms can be found by the suitable formulas derived. If we are given a quadrilateral whose sides or angles are unequal, we can find the areas by splitting it into triangles which do not overlap.

Hence in this section let us see different methods of finding the areas of quadrilaterals.

Quadrilateral : A quadrilateral is a closed figure bounded by four sides. What is area? Area is the amount of space occupied by a plane figure.

(1) Since a diagonal divides a quadrilateral into two non-overlapping triangles, we can say that the area of a quadrilateral is the sum of the areas of each triangle.

In the above Quadrilateral TRUE, the diagonal TU divides it into two triangles, $\Delta TRU$ and $\Delta TEU$.
Area of the Quadrilateral TRUE = Area of $\Delta TRU$ + Area of $\Delta TEU$

= $\frac{1}{2}$ x TU x h1 + $\frac{1}{2}$ x TU x h1

= $\frac{1}{2}$ x TU x [h1 + h2]

(2) If we are given the length of one of the diagonal and the length of each side of the quadrilateral we can find the area of the quadrilateral by finding the area of each triangle using Heron's Formula which is $\sqrt{s\;(s-a)\;(s-b)\;(s-c)\;}$

(3) Parallelogram : Area of a parallelogram = Base x Height

(4) Rectangle : Area of a rectangle = length x width

(5) Square : Area of a Square = side x side

(6) Trapezoid : Area of a Trapezoid = $\frac{1}{2}$ x Height x [sum of the parallel sides]

= $\frac{1}{2}$ x h x (a + b)

(7) Rhombus : Area of a Rhombus = $\frac{1}{2}$ x Diagonal1 x Diagonal2

How do you find the Area of a Quadrilateral

Step 1: Divide the quadrilateral into two triangles which do not overlap.
Step 2. Draw perpendiculars from other vertices to the diagonal drawn.
Step 3: Find the area of each triangle using the formula $\frac{1}{2}$ x Base x Height
Step 4 : Add the areas of the triangles found in step3, to find the area of the quadrilateral.

Solved Example

Question: Find the area of a quadrilateral PQRS, whose diagonal and the perpendicular distances are as shown below.

Solution:

From the above quadrilateral PQRS, PR = 10 cm, h1 = 7 cm and h2 = 6 cm.
Area of the quadrilateral PQRS   = Area of $\Delta PSR$ + Area of $\Delta PQR$

= $\frac{1}{2}$ x 10 x 7 + $\frac{1}{2}$ x 10 x 6

= $\frac{70}{2}$ + $\frac{60}{2}$

= 35 + 30
Area of the quadrilateral PQRS      = 65 cm2

Solved Examples

Question 1: Find the area of a quadrilateral one of whose diagonal is 18 cm and the perpendiculars from the other vertices to the given diagonal are 7 cm and 5 cm respectively.
Solution:

We are given that length of the diagonal = 18 cm
h1 = 7 cm and h2 = 5 cm

Using the above formula, we have Area of a quadrilateral = $\frac{1}{2}$ x Diagonal1  x [ h1 + h2

= $\frac{1}{2}$ x 18 x [ 7 + 5 ]

= $\frac{1}{2}$ x 18 x 12

= 9 x 12
Area of the quadrilateral   = 108 cm2

Question 2: Find the area of a quadrilateral whose sides are 9 m, 40 m, 28 m and 15 m respectively and the angle between first two sides is a right angle.
Solution:

According to the given question the diagram will be as follows.

Solution: In the above quadrilateral PQRS, $\angle PQR$ = 90o
Area of $\Delta PQR$ = $\frac{1}{2}$ x Base x Height

= $\frac{1}{2}$ x 9 x 40

= $\frac{360}{2}$

Area of $\Delta PQR$     = 180 cm2 - - - - - - - - - - - - - - - - ( 1 )
Applying Pythagoras Theorem, to find PR, we get,
PR2 = PQ2 + QR2
= 92 + 402
= 81 + 1600
=  1681
= 412
=> PR = 41 cm
Area of $\Delta PSR$ can be found using Heron's Formula,
Let a = 15, b = 18 and c = 41
Therefore, s = $\frac{15+28+41}{2}$
= 42 cm
s - a = 42 - 15 = 27
s - b = 42 - 28 = 14
s - c = 42 - 41 = 1
Area of $\Delta PSR$ = $\sqrt{s\;(s-a)\;(s-b)\;(s-c)\;}$
= $\sqrt{42\;\times27\;\times14\;\times1}$
= 126 m2  - - - - - - - - - - - - - -  -( 2 )
Area of Quadrilateral PQRS = Area of $\Delta PQR$ + Area of $\Delta PSR$
= 180 + 126
= 306 m2

Question 3: The length of a rectangular plot is twice its width. If the perimeter of the plot is 270 m, find its area.
Solution:

Let the width of the plot be x m
Therefore, the length of the plot = 2x m
We are given that the perimeter = 270 m
=>         2 ( length + width ) = 270 m
=>                    2 ( 2x + x ) = 270
=>                          2 ( 3x ) = 270
=>                                6x = 270
=>                                  x = $\frac{270}{6}$

= 45 m
Therefore, length of the plot = 2x = 2 ( 45 ) = 90 m
width of the plot = 45 m
Area of the Rectangle = Length $\times$ width
= 90 $\times$ 45
= 4050

Question 4: Find the area of a parallelogram one of whose sides is 35 cm and the corresponding height is 8 cm.
Solution:

We have the base of the parallelogram = 35 cm
Height = 8 cm
Area = Base x Height
= 35 x 8
Area of the parallelogram = 280 cm2

Question 5: A rectangular plot 25 m long and 20 m wide is to be covered with grass leaving 1.5 m all around it. Find the area of the grass lawn.

Solution:

The length of the plot = 25 m
Width of the plot = 20 m
Outer area of the plot = 25 x 20
= 500 m2
Inner dimensions of the plot leaving the grass portion is given by,
length of the plot = 22 m
width of the plot = 17 m
Area of the inner rectangular plot = 22 x 17
= 374 m2
Area of the lawn = Area of the outer Rectangle - Area of the Inner Rectangle
= 500 - 374
Area of the lawn = 126 m2