what is the area of isoceles trapezium?.
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Answer:
= [(a+c)/2][4b^2 -(2c-a)^2]^0.5.
Step-by-step explanation:
Let ABCD be an isosceles trapezium such that: AB = a, BC = DA = b and CD = c. Let AF and BE be perpendiculars from A and B on to CD.
CE = DF = (c - a/2) = (2c-a)/2.
AF = BE = [BC^2-CE^2]^0.5 = h = [b^2-(2c-a)^2/4]^0.5
= 1/2[4b^2 -(2c-a)^2]^0.5
Area of ABCD = [(a+c)/2]*h
= [(a+c)/2][4b^2 -(2c-a)^2]^0.5.
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