SSC CGL 2021 PYQ — Question 68

 The correct answer is option 3 i.e. 2

Given :

DF || AC and DE || AF

BE = 4cm, CF = 3 cm

Calculations :

Let EF be x

In △ ABC, DF || AC

⇒ $\frac{BD}{AB} = \frac{BF}{BC}$         ---- (1)

Now, in △ BAF, DE || AF

⇒ $\frac{BE}{BF} = \frac{BD}{AB}$        ---- (2)

Equating equation (1) and (2), we get

⇒ $\frac{BE}{BF} = \frac{BF}{BC}$

⇒ BF² = BE × BC

⇒ (x + 4)² = 4(7 + x)

⇒ x² + 16 + 8x = 28 + 4x

⇒ x² + 4x - 12 = 0

⇒ x² + 6x - 2x - 12 = 0

⇒ (x + 6)(x - 2) = 0

⇒ x = 2

Hence, The value of EF is 2 cm.