SSC CGL Tier-1 16-08-2015 Shift 2 Solved Paper

AB = 10 cm.
∴ AF = FB = 5 cm.
CD = 24 cm.
∴ CE = DE = 12 cm.
Let OE = x cm
∴ OF = (17 – x) cm
From D ODE,
OD = √OE2+DE2
= √x2+122 ... (i)
From Δ OAF
OA = √OF2+AF2
= √(17−x)2+52 ... (ii)
Since OA = OD
∴ √x2+122 = √(17−x)2+52
⇒ x2 + 144 = 289 - 34x + x2 + 25
⇒ 34x = 289 + 25 - 144 = 170
⇒ x =

170

34

= 5
∴ From equation (i),
OD = √x2+122 = √52+144
= √169
= 13 cm