Name: Area of Polygon by Triangulation
Uploaded: 2019-07-19
Duration: 11 min 5 s
Description: How can you calculate the area of polygons when you forget its formula? Simple! Break it down into shapes that you know the formula for. Watch to learn how!

Question

Find the area of the pentagon shown in the fig, if $AD = 8 cm$ , $AH=6 cm$ , $AG = 4 cm$ , $AF = 3 cm$ , $BF = 2 cm$ , $CH = 3 cm$ and $EG = 2.5 cm$ .

Accepted Answer

Given that in pentagon $A B C D E$ , $EG ⊥ AD, EF ⊥ AD & CH ⊥ AD$ .

$CH = 3 cm$ and $EG = 2.5 cm$ .

We are asked to find the area of the pentagon.

We can see that the pentagon is divided into different triangles and trapezium. So the area of the pentagon will be equal to the sum of the areas of these figures.

That is,

Area of pentagon $A B C D E =$ Area of $△ A B F +$ Area of trapezium $BCFH +$ Area of $△ CHD +$ Area of $△ ADE$

Consider $△ ABF$

Area of a triangle $= \frac{1}{2} \times base \times height$

So, the area of $△ ABF = \frac{1}{2} \times AF \times BF$

$= \frac{1}{2} \times 3 cm \times 2 cm$

$= \frac{1}{2} \times 3 \times 2 {cm}^{2}$

$= 3 {cm}^{2}$

Similarly, area of $△ C H D = \frac{1}{2} \times C H \times D H$

$AD - AH = DH \Rightarrow 8 cm - 6 cm \Rightarrow DH = 2 cm$

$∴$ Area of $△ CHD = \frac{1}{2} \times 3 cm \times 2 cm$

$= \frac{1}{2} \times 3 \times 2 {cm}^{2}$

$= 3 {cm}^{2}$

Now, area of $△ ADE = \frac{1}{2} \times AD \times GE$

$= \frac{1}{2} \times 8 cm \times 2.5 cm$

$= 4 \times 2.5 {cm}^{2}$

$= 10 {cm}^{2}$

Now, the area of the trapezium $= \frac{1}{2} \times (Sum of parellel sides) \times (height)$

So, the area of trapezium $BCFH = \frac{1}{2} \times (BF + CH) \times FH$

$FH = AH - AF = 6 cm - 3 cm = 3 cm$

So the area of trapezium $= \frac{1}{2} \times (2 cm + 3 cm) \times 3 cm$

$= \frac{1}{2} \times 5 cm \times 3 cm$

$= \frac{15}{2} {cm}^{2}$

$= 7.5 {cm}^{2}$

Now, area of the pentagon $A B C D E = area (△ A B F) + area (trapezium B C F H) + area (△ C H D) + area (△ A D E)$

$= 3 {cm}^{2} + 7.5 {cm}^{2} + 3 {cm}^{2} + 10 {cm}^{2}$

$= (3 + 7.5 + 3 + 10) {cm}^{2}$

$= 23.5 {cm}^{2}$

Hence, the area of the pentagon $A B C D E$ is $23.5 {cm}^{2}$ .

Find the area of the pentagon shown in the fig, if $AD = 8 cm$ , $AH=6 cm$ , $AG = 4 cm$ , $AF = 3 cm$ , $BF = 2 cm$ , $CH = 3 cm$ and $EG = 2.5 cm$ .

Important Questions on Area of Polygons

Learn and Prepare for any exam you want !

Find the area of the pentagon shown in the fig, if AD=8cm, AH=6cm, AG=4cm, AF=3cm, BF=2cm,CH=3cm and EG=2.5cm.

Important Questions on Area of Polygons

Learn and Prepare for any exam you want !

Find the area of the pentagon shown in the fig, if $AD = 8 cm$ , $AH=6 cm$ , $AG = 4 cm$ , $AF = 3 cm$ , $BF = 2 cm$ , $CH = 3 cm$ and $EG = 2.5 cm$ .