ABCD is a parallelogram and X is the mid-point of AB. `(AXCD)= 24 cm^(2)`, then `ar (ABC) = 24 cm^(2)`.

ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm^2 , then ar (ABC) = 24 cm^2 .

In the figure ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar(DPC) = 1/2 ar(EFGD)

If ABCD is a parallelogram,E is the mid-point of AB and CE bisects /_BCD, then /_DEC is

In the given fig. ABCD is a parallelogram and ABE is a triangle. Also AB||CE . If ar (ABCD) = 60 cm^2 then ar (DeltaABE) is :

In the figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then, ar (DeltaDPC) = (1)/(2) ar (EFGD) .

ABCD is a parallelogram, E is the mid-point of AB and CE bisects angleBCD . Then angleDEC is

ABCD is a parallelogram. P is any point on CD. If ar(triangleDPA) = 15 cm^2 and ar(triangleAPC) = 20 cm^2 , the ar (triangleAPB) =

In the figure ABCD and ABFE are Parallelograms then find ar (Delta BCF) . If ar (ABCE) =18 cm^(2) ar (ABCD) =25 cm^(2)

ABCD is a parallelogram. If ar (ABC) = 18 cm^(2) , then are (ABCD) = …………….. cm^(2)