Question

please answer this question correctly otherwise I will report ​

Answer 1

[Ar(∆DFC)=5000m² + Ar(Trap.FCBH)=8250m² + Ar(∆AHB)=1250m² + Ar(∆DEG)=10800m² + Ar(∆AGE)=4800m²] = 30050m²

Answer 2

Answer:

first finding

area of triangle DCF

area of triangle = 1/2× base × height

= 1/2 × 100 × 100 (dividing 100 by 2)

= 1 × 50 × 100

= 5000 sq.m

now considering ADE as one triangle

area of triangle = 1/2 × base × height

here base = AH + HG + GF + FD

= 50 + 30 + 80 + 100

= 260 .

= 1/2 × base × height

= 1/2 × 260 × 120 ( dividing 260 by 2 )

= 1 × 130 × 120

= 15600 sq.m

now finding area of triangle HAB

area of triangle = 1/2 × base × height

= 1/2 × 50 × 50 ( dividing 50 by 2 )

= 1/2 × 25 × 50

= 1 × 25 × 50

= 25 × 50

= 1250 sq.m

FHBC is the trapezium

in which FC and HB are parallel side

and height is FH

FH = FG + GH

FH = 80 + 30

FH = 110

area of trapezium = 1/2 × (sum of lenght of parallel side ) × height

1/2 = (100+50) × 110

1/2 = 150 × 110 (dividing 150 by 2)

=1 × 75 × 110

=8250 sq.m

Area of field = DCF + ADE + HAB + FHBC

Area of field = 5000 + 15600 + 1250 + 8250

Area of field = 30100 sq.m