SSC CHSL 2023 PYQ — Question 71

The correct answer is option 2 i.e. 3PQ² = 4PM².

Triangle PQR is an equilateral triangle.

Each side of the triangle = a

PM is perpendicular to the QR

So, triangle PQM is and right-angle triangle 

Height PM = $\frac{\sqrt3}{2}$a

From option 2,

4PM² = 4 × ($\frac{\sqrt3}{2}$a)²

= 3a²

and 3PQ² 

= 3 (a)²

= 3a²

So, 4PM² = 3PQ²