Question

In ∆ABC, PQ||BC, AP=2cm , PB=6cm, BC=12cm, PQ=?​

Answer 1

In triangle APQ and triangle ABC

angle A = angle A... (common angle)

angle APQ = angle ABC..... (Corresponding angles)

Triangle APQ similar to triangle ABC

.... ( By AA test of similarly.)

AP/ AB = PQ/BC

2/6 = PQ/12

12 × 2 = 6 × PQ

24 = 6 × PQ

24 / 6 = PQ

PQ = 4cm

Answer 2

Given:

In ΔABC, PQ║BC

AP = 2cm

PB = 6cm

BC = 12cm

To Find:

PQ

Solution:

In ΔAPQ & ΔABC

∠A = ∠A [ common angle ]

∠APQ = ∠ABC [ complementary angles ]

∴ ΔAPQ ~ ΔABC [ By AA criteria ]

∴ $\frac{AP}{AB}$ = $\frac{AQ}{AC}$ = $\frac{PQ}{BC}$

⇒ $\frac{AP}{AB}$ = $\frac{PQ}{BC}$

⇒ $\frac{2}{6}$ = $\frac{PQ}{12}$

⇒ PQ = $\frac{12 * 2}{6}$

⇒ PQ = 4cm.

Therefore, the value of the PQ is 4cm.