# A field is in the shape of a quadrilateral ABCD in which side AB=18m, side AD=24m, side BC=40m, DC=50m and angle A=90 degree. Find the area of the field. explain it please.

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by vershamahli

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by vershamahli

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The quadrilateral ABCD is divided into two triangles. ABD and BCD.

ABD is a right angle triangle.

Area = 1/2 AB * AD as angle A is 90, AB is the altitude and AD is the base.

= 1/2 * 18 * 24 = 216 cm^2

BC^2 = AB^2 + AD^2 => BC = 30 cm

BCD triangle

s = semi perimeter = 60 cm

area = sqroot(60 * 30 * 20 * 10 ) = 600 cm^2

total area = 816 cm^2

ABD is a right angle triangle.

Area = 1/2 AB * AD as angle A is 90, AB is the altitude and AD is the base.

= 1/2 * 18 * 24 = 216 cm^2

BC^2 = AB^2 + AD^2 => BC = 30 cm

BCD triangle

s = semi perimeter = 60 cm

area = sqroot(60 * 30 * 20 * 10 ) = 600 cm^2

total area = 816 cm^2