ICSE Class 10: Basic Proportionality Theorem Videos | More on BP theorem

In ∆ABC, DE is parallel to BC.

So by BPT,

AD/DB = AE/EC

(x + 2)(x − 2) = x(x − 1)

x² − 4 = x² − x

x = 4 

KP = 1.4 cm, KM = 5.6 cm PM = KM − KP = 5.6 − 1.4 = 4.2 cm

By BPT,

KP/PM = KQ/QN KQ/QN = 1/3

QN = KN − KQ = 7.2 - KQ

KQ = 1.8 cm 

In ∆ABC, DE||BC AD/DB = AE/EC

3/5 = AE/(AC − AE) 3/5 = AE/(5.6 − AE)

AE = 2.1 

In ∆KMN, PQ || MN ∴ KP/PM = KQ/QN (By BPT)⇒ KP/PM = KQ/(KN - KQ) ⇒ 4/13 = KQ/(20.4 - KQ) ⇒ 4(20.4 − KQ) = 13 KQ (cross multiply)

⇒ 81.6 - 4KQ = 13 KQ ⇒ 17KQ = 81.6⇒ KQ = 81.6/17 ∴ KQ = 4.8 cm