Question

O is any point on the diagonal AC of a parallelogram ABCD. Prove that area of triangle ADO=area of triangle ABO​

Answer 1

Answer :

We have a parallelogram ABCD , So AB  =  CD and BC  =  DA

And we know diagonal of parallelogram bisect each other , We assume these diagonal bisect at " P "

So

AP  =  PC   And  BP  =  PD                               ----------- ( 1 )

We know diagonal of parallelogram divide it in four same area part . SO

Area of Δ∆APB  = Area of Δ∆APD                                                          ---------------- ( A )

For Δ∆ POB and Δ∆ POD

And We know from equation 1

BP  =  PD

And Height of both triangle ( Δ∆ POB and Δ∆ POD ) will be same As they have same vertex " P "

So let height of both triangle  =  h

We know Area of triangle  = 12× Base × Height12× Base × Height

So,

Area of Δ∆ POB  = 12× BP × h12× BP × h                                            --------- ( 2 )

And

Area of Δ∆ POD  = 12× PD × h12× PD × h      

We know BP  =  PD , So

Area of Δ∆ POD  = 12× BP × h12× BP × h      , So from equation 1 , we get

Area of Δ∆ POB  = Area of Δ∆ POD                                                       --------(B)

And

Area of Δ∆ AOB   = Area of Δ∆ APB  + Area of Δ∆ POB  

And

Area of Δ∆ AOD   = Area of Δ∆ APD  + Area of Δ∆ POD  

So from equation A and B , we get

Area of Δ∆ AOB   = Area of Δ∆ AOD                                                                                     ( Hence proved )

PLS  MARK AS BRAINLIEST!!!!!

Answer 2

Is this the question of Class 8th ?