SSC GD 2024 PYQ — Question 42

The correct answer is Option 4 i.e. 160 cm.

Let the side of the rhombus be a

AC = 48 cm and DB = 64 cm

Now,

⇒ AO/2 = 48/2 = 24 cm and,

⇒ DO/2 = 64/2 = 32 cm

So, In ΔAOD

By Pythagoras theorem

⇒ AD² = AO² + DO²

⇒ a = $\sqrt {{24}^2+ {32}^2}$

⇒ a = $\sqrt {{576}+ {1024}}$

⇒ a = $\sqrt {1600}$

⇒ a = 40

Now,

The perimeter of the rhombus = 4 × side

⇒ 4 × 40 = 160 cm