In ΔABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find: (i) area ΔAPO : area Δ ABC. (ii) area ΔAPO : area Δ CQO.

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

(i) area ΔAPO : area Δ ABC.

(ii) area ΔAPO : area Δ CQO.

ML Aggarwal Avichal Publication class 10, similarity, chapter 13.3, question no 10

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=AP2/AB2 =AP2/(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=AP2/CQ2
area of △APO/area of △CQO=AP2/PB2
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.

## 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=AP2/AB2

=AP2/(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=AP2/CQ2

area of △APO/area of △CQO=AP2/PB2

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.