ABCD is a trapezium such that AB and CD are parallel and `BC bot CD`. If `angleADB = theta, BC = p and CD = q`, then AB is equal to

ABCD is a trapezium such that AB and CD are parallel and BC _|_ CD . If /_ADB= theta , BC =p and CD=q , then AB is equal to

ABCD is a trapezium such that AB and DC are parallel and BC is perpendicular to them. If BC=1 cm, CD=2 cm and angleADB=45^(@) , then the lengths of AB is (in cm)

In the figure, if AB = 8 , BC = 6, AC = 10 and CD = 9, then AD =

ABCD is a trapezium such that AB||CD. Its diagonals AC and BC intersect each other at O. Prove that (AO)/(OC) = (BO)/(OD)

If a,b,c,d are in H.P., then ab+bc+cd is equal to

If P and Q are the middle points of the sides BC and CD of the parallelogram ABCD, then AP+AQ is equal to

In triangle ABC, D is mid-point of AB and CD is perpendiicular to AB. Bisector of angleABC meets CD at E and AC at F. Prove that: F is equidistant from AB and BC.

Let ABCD be a trapezium ,in which AB is parallel to CD, AB =11 ,BC=4,CD=6 and DA=3. the distance between AB and CD is

If, in the figure above, CD=1, AB=-2, and AD=6, then BC=

AB is a diameter of a circle. CD is a chord parallel to AB and 2CD = AB . The tangent at B meets the line AC produced at E then AE is equal to -