Question

65. In the given figure, AP : PB = 2:3, PQ II BC and is extended to Q so that CQ II BA. Then ar(AAOP) ar(ACOQ) (a) 2:5 (c) 2:3 (b) 4:3 (d) 4:9​

Answer 1

Answer:

D IS THE CORECT ANSWER

Step-by-step explanation:

(i) we have to find the area △APO: area △ABC,

Then,

∠A=∠A … [common angles for both triangles]

∠APO=∠ABC … [because corresponding angles are equal]

Then, △APO∼△ABC … [AA axiom]

We know that, area of △APO/area of △ABC=AP

2 /AB

2 =AP

2 /(AP+PB) 2

=22/(2+3) 2

=4/25

Therefore, area △APO: area △ABC is 4:25

(ii) we have to find the area △APO : area △CQO

Then, ∠AOP=∠COQ … [because vertically opposite angles are equal]

∠APQ=∠OQC … [because alternate angles are equal]

Therefore, area of △APO/area of △CQO=AP 2 /CQ

2area of △APO/area of △CQO=AP

2 /PB 2area of △APO/area of △CQO=2

2 /3

2 area of △APO/area of △CQO=4/9

Therefore, area △APO : area △CQO is 4:9.