Question

In the given figure, DF || AB, BC ⊥ AB, BC = 5, BG = 4, BA = 12, DA = 3. Then CE is:

(a) 5.44 

(b) 0.54 

(c) 4.9 

(d) 0.42

Answer 1

(c) 4.9 

By Pythagoras' Theorem, 

GC² = (BC)² + (GB)² 

= 25 + 16 = 41 

⇒ GC =\(\sqrt{41}\) 

AC² = (BC)² + (AB)² 

= 24 + 144 = 169 

⇒ AC = 13

∴ DC = AC – AD = 13 – 3 = 10 

ΔDEC ~ ΔAGC               (AA similarly ∵ DF || AB) 

∴ \(\frac{DC}{AC}\) = \(\frac{CE}{GC}\)

∴ \(\frac{10}{13}\) = \(\frac{CE}{\sqrt{41}}\) ⇒ CE = \(\frac{10\sqrt{41}}{13}\) = 4.9 (approx)