Question

21. In the given figure, if PQ || BC and AP/PB = 3/2, then ar(∆POQ)/ar(COB) is (a) 25/9 (b) 4/9 (c) 9/4 (d) 9/25

Answer 1

Step-by-step explanation:

In ∆ ABC and ∆ APQ

< APQ= <ABC ( corresponding <s)

<PAQ = <BAC ( Common)

∆ABC ~ ∆APQ ( By AA Similarity)

=> AP/AB = AQ/AC = PQ/BC = 3/5 ( CPST)

Now, In ∆ POQ and ∆ BOC

<POQ = <BOC (V.O.A.)

<PQB = < QBC ( Alt. Int. <s)

∆ POQ ~ BOC ( By AA)

=>PO / BO = OQ/OC = PQ/BC ( CPST)

Ar ( ∆POQ)/ Ar ( ∆ BOC) = ( PQ) ²/ (BC) ² ( By Theorem)

= 3² /5²

= 9/25

Answer 2

Given: In the given figure, PQ || BC and AP/PB = 3/2.

To find: ar(∆PAQ)/ar(CAB)

Solution:

The angle ∠A is common to both the triangles (PAQ and CAB). The angles ∠B and ∠P are equal because they are corresponding angles. Similarly,  angles ∠C and ∠Q are also equal. So, the triangles PAQ and CAB are similar by the AAA (Angle-Angle-Angle) postulate.

The ratio of AP to PB is 3:2. So, the ratio of AP to AB is 3:5. The sides AP and AB are the corresponding sides of the two similar triangles. Thus, the ratio of the areas of the two triangles can be calculated as follows.

$\frac{ ar(PAQ)}{ar(CAB)} = (\frac{AP}{AB} )^{2}$

             $= \frac{9}{25}$

Therefore, ar(∆PAQ)/ar(CAB) is 9/25

Although part of your question is missing, you might be referring to this full question:

In the given figure, if PQ || BC and AP/PB = 3/2, then ar(∆PAQ)/ar(CAB) is

(a) 25/9 (b) 4/9 (c) 9/4 (d) 9/25

Also, a figure of your question is missing, you might be referring to the one attached.