Question

Prove that area of a trapezium is half the product of the sum of the lengths of the parallel sides and distance between them

Answer 1

Let ABCD be a trapezium. Let AB and DC be the parallel sides with AB being the longer one of the two.

Drop a perpendicular from C to AB and call the point of intersection with AB as X.

Similarly drop a perpendicular from D to AB and call the point of intersection with AB as Y.

Note now that we can divide the trapezium into two triangles AYD and BXC and a rectangle YXCD.

Let AB = a, DC = b and perpendicular height = DY = CX = h.

Then, area of rectangle YXCD = bh.

Now merge the two triangles AYD and BXC at the edges DY and CX to give one triangle. The area of this triangle  = 1/2 x (a - b) x h (Draw a diagram to see this). Note that the area of this triangle is the sum of the areas of AYD and BXC.

Hence area of trapezium = Area of AYD + Area of BXC + Area of YXCD

                                          = 1/2 x (a-b) x h + bh

                                          = ah/2 - bh/2 + bh

                                          = ah/2 + bh/2

                                          = (a+b)h/2.