Question

In ΔABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

Answer 1

Step-by-step explanation:

From the question it is given that,

PB=2:3

PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

(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

=4/5

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

2

area of △APO/area of △CQO=AP

2

/PB

2

area 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