Question

what is the area of isoceles trapezium?.​

Answer 1

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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