`DeltaABC` is an acute angled triangle. CD be the altitude through C. If `AB=8, CD=6` , then the distance between the mid-points of AD and BC is

Let ABC be an acute angled triangle and CD be the altitude through C. If AB = 8 cm and CD = 6 cm, then the distance between the midpoints AD and BC is :

In Delta ABC,AD perp BC,AD=6,BC=8 then find distance between mid-points of BD and AC

DeltaABC is an acute angle triangle. AD is perpendiculra on BC and DB = 3 CD. Find BC^(2) -

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

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

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