Question

in fig. 14.12, the measurements are: AF = 50 M , FG = 30M, AH = 120M, CH=30M, BG=50M, EF=30M, HD=20M. FIND THE AREA OF POLYGON ABCDE .​

Answer 1

Answer:

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Answer 2

Answer:

ide the polygon into different parts as AEF , DCH , ABC are the triangles we will find out the area by using formula 12× base × height and DH is parallel to EF hence the DHFE is trapezium we will find out its area by 12×(sum of parallel side) × height.Adding all the calculated areas we will get required answer.

Complete step-by-step answer:

First we have to divide the polygon in different parts as AEF , DCH , ABC are the triangles .

So for the triangle ABC

Area of the triangle of ABC is 12× base × height .

or 12×AC×GB

As GB is the height of triangle ABC that is equal to 50 m .

and the AC is the base of the triangle that is equal to AC = AH + HC that is equal to 150 m .

Hence area of triangle ABC = 12×150×50 = 3750 m2

Now for the triangle AEF

area of triangle is 12× base × height

or 12×AF×EF

As EF is the height of triangle AEF that is equal to 30 m it is given in the question .

and the AF is the base of triangle that is equal to 50 m ( Given )

Hence the area of triangle AEF = 12×50×30 = 750 m2

Now in the triangle DHC

Area of triangle is 12×DH×HC

Where DH and CH are given in the question that is equal to 20m and 30m respectively .

Area of triangle DHC is equal to 12×30×20=300 m2

Now see the figure DHFE and DH is parallel to EF because both are perpendicular to AC .

Hence DHFC is a trapezium .

Area of Trapezium is = 12×(sum of parallel side) ×height

or In DHFC it is equal to

= 12×(20+30)×70

12×(DH+EF)×FH

It is given in the question that DH = 20m and EF = 30 m

And FH = AC − (CH + FA ) = 150−(50+30)

FH = 70 m.

Area of DHFC = 12×(20+30)×70

= 1750 m2

Hence the area of polygon ABCDE = Area of DHFC + Area of triangle DHC + Area of triangle AEF + Area of triangle ABC

= 1750+300+750+3750 m2

The area of polygon ABCDE = 6550