A field is in a shape of a quadrilateral ABCD, in which AB= 18m, AD= 24m, BC= 40m, DC= 50m and angle A= 90 degree. Find the area of the field.
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Answered by
9
The diagonal BD^2 = 18^2 + 24^2 = 6^2 (3^2+4^2) = 30^2
So BD = 30m
Area ABD = 1/2 AB AD = 1/2 18 * 24 = 621 m^2
Now area BCD has sides BC, BD, CD. s = 1/2 (30+40+50) = 60m
area = √ s (s-a) (s-b)(s-c) = √60 *30* 20 * 10 = 600 m^2
Total area = area ABCD = 1221 m^2
So BD = 30m
Area ABD = 1/2 AB AD = 1/2 18 * 24 = 621 m^2
Now area BCD has sides BC, BD, CD. s = 1/2 (30+40+50) = 60m
area = √ s (s-a) (s-b)(s-c) = √60 *30* 20 * 10 = 600 m^2
Total area = area ABCD = 1221 m^2
Answered by
3
The answer is.......1221 m ^ 2........
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