Question

find the area of the pentagon as shown in the figure,if AD = 15cm, AH = 10cm, AG = 8cm, AF =5cm, BF = 6cm, GE = 3cm and CH = 9cm.​

Answer 1

The Area of Pentagon ABCDE is 97.5 sq. cm.

Step-by-step explanation:

Given,

AD = 15 cm, AH = 10 cm, AG = 8 cm, AF =5 cm, BF = 6 cm, GE = 3 cm and CH = 9 cm.​

We have to find out the area of Pentagon ABCDE.

Solution,

The pentagon ABCDE is composed of three triangles ΔABF, ΔCHD and ΔADE and a trapezium BFHC.

So the area of petagon is the sum of areas of three triangles and area of trapezium.

In  ΔABF,

AF =5 cm     BF = 6 cm

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

Here, AF = base  and BF = height

On putting the values, we get;

Area of ΔABF =$\frac{1}{2}\times5\times6=5\times3=15\ cm^2$

In trapezium BFHC,

BF = 6 cm   CH = 9 cm

$FH=AH-AF=10-5=5\ cm$

Area of trapezium = $\frac{1}{2}\times sum\ of\ parallel\ sides\times distance\ between\ them$

here BF and CH are parallel sides and FH is the distance between them.

Area of trapezium BFHC=$\frac{1}{2}\times(6+9)\times5=\frac{1}{2}\times15\times5=\frac{75}{2}=37.5\ cm^2$

In  ΔCHD,

CH = 9 cm  

$HD=AD-AH=15-10=5\ cm$

Area of ΔCHD =$\frac{1}{2}\times5\times9=\frac{45}{2}=22.5\ cm^2$

In  ΔADE,

AD = 15 cm     GE = 3 cm

Area of ΔADE =$\frac{1}{2}\times15\times3=\frac{45}{2}=22.5\ cm^2$

Area of Pentagon ABCDE=Area of ΔABF+Area of trapezium BFHC+Area of ΔCHD+Area of ΔADE

Area of Pentagon ABCDE= $15+37.5+22.5+22.5=97.5\ cm^2$

Hence The Area of Pentagon ABCDE is 97.5 sq. cm.

Answer 2

Step-by-step explanation:

this is the answer ☺️☺️☺️☺️☺️☺️☺️