[" 10.ABCD is a parallelogram and "x" is the mid-point of AB.If ar' AxCD ) : 246"1" ".The "],[" ar (ABC)."]

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

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. (AXCD)= 24 cm^(2) , then ar (ABC) = 24 cm^(2) .

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(DPC) = 1/2 ar(EFGD)

ABCD is a parallelogram and P is a point of bar(CD) . Show that ar(/_\ABP) = ar(/_\ADP+ /_\BCP) .

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 parallelogram. X and Y are the mid-points of BC and CD respectively. Prove that ar (DeltaAXY) =3/8 ar (||)^(gm) ABCD .

ABCD is parallelogram. X and Y are the mid-points of BC and CD respectively. Prove that ar (DeltaAXY) =3/8 ar (||gm ABCD) .

P and Q are respectively the midpoints of sides AB and BC or a triangle ABC and R is the mid-point of AP, show ar(RQC) = 3/8 ar(ABC).