Question

A field in the shape of a quadrilateral ABCD in which side AB=18m,AD=24m,side BC=40m,DC=50m and angle A=90 degree. Find the area of the field
Please give me the diagram.

Answer 1

construction: join DtoB
In rt ∆DAB
AB^2+AD^2=DB^2
DB^2=18^2+24^2
DB^2=324+576
DB^2=900
DB=30m
Area of ∆DAB=1/2×base×altitude
=1/2×18×24
=216m^2
Now,Area of ∆ DCB=√S(S-a)(S-b)(S-c). [By heron's formula]
S= DB+DC+BC/2
=30+50+40/2
=60m
So area =√ 60(60-30)(60-50)(60-40)
=10√6×3×3
=10×2√3×3
=10×2×3
=30m^2

Therefore , Area of quad.ABCD = Areaof ∆DAB+Area of ∆DCB
=(216+30)m^2
Ans-246m^2