Question

Please solve q 90?

Answer 1

option (1)

Explanation:

`DeltaAPF and DeltaBPF` have same height
So
`(AF)/(FB)=(DeltaAPF)/(DeltaBPF)=43/30=4/3`

For similar reason

`(DeltaACF)/(DeltaBCF)=(AF)/(FB)=4/3`

`=>(40+x+84)/(30+35+y)=4/3`

`=>372+3x=260+4y`

`=>4y-3x=112.......[1]`

Again

`(DeltaAPC)/(DeltaDPC)=(AP)/(PD)=(DeltaAPB)/(DeltaDPB)`

`=>(84+x)/y=(40+30)/35=2`

So `2y=84+x....[2]`

Comparing [1] and [2] we get

`2(84+x)-3x=112`

`=>168+2x-3x=112`

`=>x=56`

Inserting in [2] we get
`2y=84+56`
`=>y=70`

So area of `DeltaABC=84+56+40+70+35+30=315`