Question

prove that area of trapezium ABCD =1/2(AB+CD)×CM where CM is the distancebetween parallel sides AB and CD

Answer 1

Let CD = b1, AB = b2 and CM = h

Area of 1st triangle = ½ xh

Area of 2nd triangle = ½ yh

Area of rectangle = b1h
Total area = ½ xh + ½ yh + b1h= ½ (x + y + 2b1)hFrom the diagram b2 = b1 + x+ y

Meaning that x + y = b2 – b1, substitute this in the above equation above

Total area = ½ (x + y + 2b1)h

                 = ½ (b2 – b1 + 2b1)h

                 = ½ (b2 + b1)h

                 = ½ (AB + CD)CM         Proved

Answer 2

Hope the answer is helpful !!

Good Day !!