SSC CGL 2021 PYQ — Question 53

The correct answer is option 4 i.e.  4$\sqrt{14}$ + 2$\sqrt{39}$.

Here AB act as bisector of OO'.

$\angle$ACO = $\angle$ACO' = 90° 

AC = 20/2 = 10 cm.

In Triangle AOC,

OC² = OA² - AC²

OC² = 324 - 100 = 224.

OC² = 4$\sqrt{14}$ cm.

Now in Triangle AO'C,

O'C² = O'A² - AC²

O'C² = 256 - 100 = 156

O'C = 2$\sqrt{39}$ cm.

OO' = OC + O'C =  4$\sqrt{14}$ +  2$\sqrt{39}$.