Question

Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the distance between them

Answer 1

draw a trapezium say abcd and join the diagonals you will obtain two triangle draw two perpendiculars on diagonals from both parallel sides 
now area of quadrilateral(trapezium)=sum of area of both the triangles
        let a and b be two parallel sides and h be the perpendiculars 
area of 1st triangle=1/2 x a  x h  ....1
 area of 2nd triangle=1/2 x b x h .........2
                      sum of 1 and 2 = area of quadrilateral(trapezium)
1/2 x a x h +1/2 x b x h= area of trapezium       
1/2 x (a+b) x h= area of trapezium             (taking out 1/2 as common)

thus proved............

Answer 2

Let ABCD be a parallelogram, and BD be the diagonal forming two triangles ABD and BCD.
Let DE be the height of trapezium ABCD.
Area of trapezium ABCD=Area of tr. ABD+Area of tr. BCD
So, area of tr. ABD= 1/2×base×height
=1/2×AB×DE..(i)

Similarly, area of tr. BCD=1/2×DC×DE ..(ii) [Both triangles will have the same height as they lie between the same parallel lines]
Now, area of trap.= sum of eq(i) and (ii)
So, (1/2×AB×DE) + (1/2×DC×DE)
1/2×DE(AB+DC)
☆Hence, proved
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