In `DeltaABC and DeltaDEF`, it is given that AB = DE and BC = EF. In order that `DeltaABC~=DeltaDEF`, we must have

In DeltaABC and DeltaDEF , it is given that angleB=angleE and angleC=angleF , In order that DeltaABC~=DeltaDEF , we must have

In triangleABC and triangleDEf , it is given that AB=DE and BC=EF in order that triangleABCequivtrianleDEF , we must have to

In the given figure, BA_|_AC and DE_|_DF such that AB = DE and BF = EC. Show that DeltaABC~=DeltaDEF.

If DeltaABC-DeltaDEF such that AB = 1.2 cm and DE= 1.4 cm, the ratio of the areas of DeltaABC and DeltaDEF is:

If DeltaABC ~ DeltaDEF such that A(DeltaABC) = 4A(DeltaDEF) and AC = 6 cm, then DF =

If DeltaABC ~ DeltaDEF such that A(DeltaABC) = 4A(DeltaDEF) and AC = 6 cm, then DF =

DeltaABC~DeltaDEF and 2AB=DE and BC=8cm then EF=……..cm.

DeltaABC ~ DeltaDEF such that AB = 9.1 cm and DE =6.5 cm. If the perimeter of DeltaDEF is 25 cm, what is the perimeter of DeltaABC ?

If DeltaABC ~ DeltaDEF and AB : PQ = 5 : 7 then write the ratio of A(DeltaABC) : A(DeltaPQR) .