In Fig. 15.78, ABCD is a trapezium in which AB=7 cm, AD=BC=5 cm, DC=x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Draw AL perpendicular to DC
And,
BM perpendicular DC
Then,
AL = BM = 4 cm
And,
LM = 7 cm
In Δ ADL, we have
AD2 = AL2+ DL2
25 = 16 + DL2
DL = 3 cm
Similarly,
MC =
=
= 3 cm
Therefore,
x = CD = CM + ML + LD
= 3 + 7 + 3
= 13 cm
Area of trapezium ABCD = (AB + CD) * AL
= (7 + 13) * 4
= 40 cm2