(b) 5√3 – 5.
Given, ABC is an equilateral triangle such that AB = BC = CA = 10 m
If O is any point in the ΔABC, then
Area of ΔABC
= Area (ΔOAB) + Area (ΔOAC) + Area (ΔOBC)
= \(\frac{1}{2}\)x AB x OR + \(\frac{1}{2}\) x AC x OP + \(\frac{1}{2}\) x AC x OQ
= \(\frac{1}{2}\) X AB x (OR + OP + OQ) (∵ AB = BC = CA)
= \(\frac{1}{2}\) x 10 x 2 (2 + 3 + OQ)
∵ Area of an equilateral Δ = \(\frac{\sqrt3}{4}\) (side)2
∴ \(\frac{\sqrt3}{4}\) x (10)2 = \(\frac{1}{2}\) x 10 x (5 + OQ)
⇒ 5√3 = 5 + OQ ⇒ OQ = 5√3 – 5.