Question

In the given figure, ABC is triangle and PQ is a stright line meeting AB in P and AC in Q. if `AP=1 cm, PB=3 cm, AQ=1.5 cm, QC=4.5 cm` prove that area of `Delta APQ ` is ` (1)/(16)` of the area of `Delta ABC`.

Answer 1

`(AP)/(AB)=(1)/((1+3))=(1)/(4),(AQ)/(AC)=(1.5)/((1.5+4.5))=(1.5)/(6)=(1)/(4)`
`:. Delta APQ~Delta ABC` [ by SAS-similarity]
`rArr (ar (Delta APQ))/(ar (Delta ABC))=(AP^(2))/(AB^(2))=((AP)/(AB))^(2)=((1)/(4))^(2)=(1)/(16)`
`rArr (Delta APQ)=(1)/(16) * ar (Delta ABC)`