P is the point in interior of a parallelogram ABCD.area of parallelogr

1.	P is the point in interior of a parallelogram ABCD.area of parallelogram is 60 cm^2 find area of triangle ADP+ triangle BPC
Answer» ONG>ANSWER: EF∥AB⟶(1) [By construction] ∵AD∥BC ∵ opposite sides of a ∥gm are ∥al. ∴ AE∥BF⟶(2) From equation (1) and (2), quadrilateral ABEF is a ∥gm. A quadrilateral is a ∥gm if its opposite sites are parallel. Similarly, quadrilateral CDEF is a ∥gm. ∵△APB and ∥gm ABFE are on the same base AB and between the same ∥als AB and EF. ∴ar(△APB)= $\large\tt { \frac {1}{2}}$ ar(∥gmABFE)⟶(3) ∵△PCD and ∥gm CDEF are on the same base DC nd between the same ∥als DC and EF. Therefore $\large ar(∆PCD = \frac {1}{2} ar ( parallelogram\:CDEF) (4)$ Adding equations (3) and (4), we get ar(△APB)+ar(△PCD)= $\large \frac {1}{2} ar$ = $\large \frac {1}{2} 60 = 30 cm^{2}$

P is the point in interior of a parallelogram ABCD.area of parallelogram is 60 cm^2 find area of triangle ADP+ triangle BPC

Answer» ONG>ANSWER:

EF∥AB⟶(1) [By construction]

∵AD∥BC

∵ opposite sides of a ∥gm are ∥al.

∴ AE∥BF⟶(2)

From equation (1) and (2), quadrilateral ABEF is a ∥gm. A quadrilateral is a ∥gm if its opposite sites are parallel. Similarly, quadrilateral CDEF is a ∥gm.

∵△APB and ∥gm ABFE are on the same base AB and between the same ∥als AB and EF.

∴ar(△APB)= $\large\tt { \frac {1}{2}}$ ar(∥gmABFE)⟶(3)