Question

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1cm, PB = 3cm, AQ = 1.5cm, QC = 4.5cm, prove that area of ΔAPQ is 116 of the area of ΔABC.  

Answer 1

We have : 

`(AP)/(AB)=1/1+3=1/4` amd `(AQ)/(AC)=1.5/
(1.5+4.5)=1.5/6=1/4` 

⇒ `(AP)/(AB)=(AQ)/(AC)` 

Also, ∠𝐴= ∠𝐴
By SAS similarity, we can conclude that ΔAPQ- ΔABC.   

`(ar(Δ APQ))/(ar(Δ ABC))=(AP^2)/(AB^2)=1^2/4^2=1/16` 

⇒ `(ar(ΔAPQ))/(ar(ΔABC))=1/16` 

⇒` ar (ΔAPQ)=1/16xxar(ΔABC) ` 

Hence proved.