A field is in the shape of a quadrilateral ABCD in which side AB=18cm,side AD=24cm,side BC=40cm and side DC=50 cm .what is the area of the quadrilateral
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hey mate !!!!!
your answer ⬇️⬇️⬇️
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
your answer ⬇️⬇️⬇️
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
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