Question

ABCD is a paarallelogram whose area is 100 sq. cm. P is any point inside the parallelogram. find the area of arAPB +arCPD

Answer 1

see the picture
see the photo for answer

Answer 2

Area of ΔABP +  Area of ΔCPD  = 50 cm²

Step-by-step explanation:

ABCD is a parallelogram whose area is 100 sq. cm

Lets Draw PM ⊥AB   & PN ⊥CD

=> Height = Distance between AB & CD) = MN  = PM + PN

Area of ABCD parallelogram  = AB * MN = 100

Area of ΔABP  = (1/2)(AB) PM

Area of ΔCPD  = (1/2)(CD) PN    

AB = CD

=> Area of ΔCPD  = (1/2)(AB) PN  

Area of ΔABP +  Area of ΔCPD  =  (1/2)(AB) PM  + (1/2)(AB) PN  

= (1/2)(AB) (PM + PN)

=  (1/2)(AB) (MN)

=  (1/2)100

= 50 cm²

Area of ΔABP +  Area of ΔCPD  = 50 cm²

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