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Area of a Polygon

We have learnt that the area of a quadrilateral can be found by splitting it into triangles. Similarly,
we can find the area of a polygon by splitting it into triangles or quadrilaterals.

When a polygon has been split into triangles or quadrilaterals, we add the areas of these parts and
thus get the area of the given polygon.

For example, a pentagon can be split into three triangles or a trapezium and two triangles as
shown below.
A

B D
or or
G F

E C
Example 11: Find the area of the pentagon with the given dimensions: 12 cm

4.8 cm
Solution: Draw lines and complete the measurements as shown in the 4 cm
figure given alongside. The given pentagon is divided into two trapeziums 10 cm
ABCF and FCDE.
1
Area of trapezium ABCF = × BG × (AB + CF) 12 cm
2 A B
1 12 cm
= × 4.8 cm × (12 + 16) cm = 67.2 cm 2
2 4.8 cm
1 10 cm F 12 cm 4 cm C
Area of trapezium EDCF = × DG × (ED + CF) G
2 5.2 cm
1
= × 5.2 cm × (12 + 16) cm = 72.8 cm 2
2 E 12 cm D
Area of pentagon ABCDE = Area of trapezium ABCF + Area of trapezium EDCF

2
2
= 67.2 cm + 72.8 cm = 140 cm 2
Example 12: There is a hexagon MNOPQR of each side 5 cm. Raghav and Riddhi divided the figure
in two different ways, as shown below, and found the area. Try to find the area of this hexagon
using both ways and check whether their areas are same?
N
N N
M O M O M O
11 cm 5 cm
R P
R P R P
Q
8 cm Q Q
Raghav’s method Riddhi’s method
Solution: Raghav’s method: N
DMNO and DPQR are congruent triangles with altitude 3 cm as shown in 3 cm
the adjoining figure. 8 cm O
1 11 cm M 5 cm
2
So, area of DMNO = × 8 cm × 3 cm = 12 cm = Area of DPQR 8 cm
2 R P 3 cm
and area of rectangle MOPR = 8 cm × 5 cm = 40 cm 2
Q
2
2
2
Now, area of the hexagon MNOPQR = 40 cm + 12 cm + 12 cm = 64 cm 2 8 cm
239 Mensuration