ABC,`(DE)/(AB)=2:3` and `ar(Delta DEF)=44`sq.units then `ar ( DeltaABC) is `

DeltaABC~DeltaDEF if DE:AB=2:3 and ar DeltaDEF =44sq. Units then ar DeltaABC = ……………sq. units.

If triangle ABC ~ triangle DEF if DE : AB = 2 : 3 and area of triangle DEF = 44 sq.units then find the area of triangle ABC .

If DeltaABC and DeltaDEF are two tnangles such that (AB)/(EF)=(BC)/(FD)=(CA)/(DE)=3/4 then find ar(DeltaDEF): ar(DeltaABC) .

E is any point on median AD of /_ABC .If ar(Delta ABE)=10cm^(2) then ar(Delta ACE) is

In the given figure, DE||BC . If DE=3 cm, BC=6 and ar (Delta ADE)=15 cm^(2) , find the area of Delta ABC .

In the given figure ar (Delta ABC)=" ar "(Delta ADE) . AC is the median of Delta ADE . If ar (ACE) =14 cm^(2)" then ar "(Delta ABC) = ?

Delta ABC~ DeltaDEF such that ar (Delta ABC)= 36 cm^(2) and ar (Delta DEF) 49 cm^(2) . Then, the ratio of their corresponding sides is

A triangle ABC laying in the first quadrant has two vertices as A(1 , 2) B(3 , 1). If , angle BAC=90^@ and ar (DeltaABC)=5sqrt5 sq. units , then the abscissa of the vertex C is :

In Delta ABC and Delta DEF , we have (AB)/(DE)=(BC)/(EF)=(AC)/(DF)=5/7 , then ar (Delta ABC) : ar (Delta DFR) =?