#### Question

In a ΔABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find AB and PQ.

#### Solution

We have || BC

Therefore, by BPT

We have,

`"AP"/"PB"="AQ"/"QC"`

`2.4/"PB"=2/3`

`rArr"PB"=(3xx2.4)/2=(3xx24)/2=(3xx6)/5=18/5`

⇒ PB = 3.6 cm

Now, AB = AP + PB

= 2.4 + 3.6 = 6cm

Now, In ΔAPQ and ΔABC

∠A = ∠A [common]

∠APQ = ∠ABC [∵ PQ || BC ⇒ Corresponding angles are equal]

⇒ ΔAPQ ~ ΔABC [By AA criteria]

`rArr"AB"/"AP"="BC"/"PQ"` [corresponding sides of similar triangles are proportional]

`rArr"PQ"=(6xx2.4)/6`

⇒ PQ = 2.4 cm

Hence, AB = 6 cm and PO = 2.4 cm

Is there an error in this question or solution?

#### APPEARS IN

Solution In a δAbc, P and Q Are Points on Sides Ab and Ac Respectively, Such that Pq || Bc. If Ap = 2.4 Cm, Aq = 2 Cm, Qc = 3 Cm and Bc = 6 Cm, Find Ab and Pq. Concept: Basic Proportionality Theorem Or Thales Theorem.