यदि ABCD एक समांतर चतुर्भुज है, तो सिद्ध कीजिए कि
`ar(DeltaABD)=ar(DeltaBCD)=ar(DeltaABC)=ar(DeltaACD)=(1)/(2)ar("||"^(gm)ABCD)`

If ABCD is a parallelogram,the prove that ar(ABD)=ar(BCD)=ar(ABC)=ar(ACD)=1/2ar(||^(gm)ABCD)

Evaluate :ar (DeltaBCD)

DeltaABC~DeltaPQR , AB:PQ=3:4 then ar DeltaABC : ar DeltaPQR =……

In DeltaABC,AD is the bisector of angleA . Then , (ar(DeltaABD))/(ar(DeltaACD))=

In the adjoining figure, P is the point in the interior of a parallelogram ABCD.Show that ar(DeltaAPB) ar(DeltaPCD) =1/2ar(||gm ABCD)

P is a point in the interior of a parallelogram ABCD. Show that (i) ar (DeltaAPB) +ar (DeltaPCD)=1/2ar (ABCD) (ii) ar(DeltaAPD)+ar (DeltaPBC)=ar(DeltaAPB)+ar(DeltaPCD) (Hint : Throught , P draw a line parallel to AB)

P is a point in the interior of a parallelogram ABCD. Show that (i) ar (DeltaAPB) +ar (DeltaPCD)=1/2ar (ABCD) (ii) ar(DeltaAPD)+ar (DeltaPBC)=ar(DeltaAPB)+ar(DeltaPCD) (Hint : Throught , P draw a line parallel to AB)

P is a point in the interior of a parallelogram ABCD. Show that (i) ar (DeltaAPB) +ar (DeltaPCD)=1/2ar (ABCD) (ii) ar(DeltaAPD)+ar (DeltaPBC)=ar(DeltaAPB)+ar(DeltaPCD) (Hint : Throught , P draw a line parallel to AB)

In Fig. ABCD is a parallelogram. Prove that ar (DeltaBCP) = ar (DeltaDPQ) .