Triangle ABC and parallelogram ABEF are on the same base, AB as in between the same parallels AB and EF. Prove that ar `(DeltaABC) = 1/2` ar(|| gm ABEF)

In the figure, ΔABC and ΔABD are two triangles on the same base AB. If line segment CD is bisected by bar(AB) at O, show that ar (DeltaABC) = ar (DeltaABD).

Draw two triangles ABC and DBC on the same base and between the same parallels as shown in the figure with P as the point of intersection of AC and BD. Draw CE || BA and BF || CD such that E and F lie on line AD. Can you show ar(DeltaPAB) = ar(DeltaPDC) Hint : These triangles are not congruent but have equal areas.

A parallelogram, a rectangle and a triangle are on the same base and between same parallels and their areas are P, R and T respectively, then

D is the mid-point of the side BC of the DeltaABC . The parallelogram CDEF lie between the same parallels CD and PQ, a parallel st. line of PQ through A. Prove that DeltaABC =parallelogram CDEF.

ABCD is parallelogram and ABEF is a rectangle and DG is perpendicular on AB. Prove that (i) ar (ABCD) = ar(ABEF) (ii) ar (ABCD) = AB xx DG

ABC is an isosceles triangle right angled at C. Prove that AB^(2) = 2AC^(2) .

D is the mid-point of the side BC of the DeltaABC and P is any point on BC. Let us join P, A. The line segment drawn through D, parallel to PA intersects AB at Q. Prove that DeltaBPQ=1/2DeltaABC .

The area of parallelogram ABCD is 36 cm^(2) . Calculate the height of parallelogram ABEF if AB = 4.2 cm