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refer to shape 2

question 17 , d ) In ∆ABC , PQ II BC . If AP = 2.4 cm , AQ = 2cm , QC = 3cm and BC = 6cm , AB and PQ are respectively ,

A) AB = 6cm , PQ = 2.4

B) AB = 4.8cm , PQ = 8.2

C) AB = 4cm , PQ = 5.3

D) AB = 8.4cm , PQ = 2.8cm

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**Step-by-step explanation:**

Δ ABC, AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm. Also, PQ ∥ BC.

To find: AB

By using Thales Theorem, we have [As it’s given that PQ ∥ BC]

AP/AB = AQ/QC

2.4/PB = 2/3

2 x PB = 2.4 x 3

PB = (2.4×3)2(2.4×3)2 cm

⇒ PB = 3.6 cm Now finding, AB = AP + PB AB 2.4 + 3.6

⇒ AB = 6 cm Now, considering Δ APQ and Δ ABC We have, ∠A = ∠A ∠APQ = ∠ABC (Corresponding angles are equal, PQ||BC and AB being a transversal) Thus, Δ APQ and Δ ABC are similar to each other by AA criteria. Now, we know that Corresponding parts of similar triangles are propositional.

⇒ APABAPAB = PQBCPQB

⇒ PQ = (APABAPAB) x BC = (2.462.46) x 6 = 2.4

** ∴ PQ = 2.4 ****cm**

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