Question

The lengths of the sides a, b, c of a ΔABC are connected by the relation a² + b² = 5c². The angle between medians drawn to the sides 'a' and 'b' is

(a) 60º 

(b) 45º 

(c) 90º 

(d) None of these

Answer 1

(c) 90º.

AD being the median to BC, 

AB² + AC² = 2(BD² + AD²) (Apollonius Th.)

⇒ c² + b² = 2\(\bigg(\frac{a^2}{4}+AD^2\bigg)\)

⇒ 2c² + 2b² = 4\(\bigg(\frac{a^2}{4}+AD^2\bigg)\)

⇒ 4AD² = 2c² + 2b² – a² 

⇒ AD = \(\frac{1}{2}\) \(\sqrt{2c^2+2b^2-a^2}\)

Now G divides AD in ratio 2 : 1 ∴ AG = \(\frac{2}{3}\) AD

⇒ AG² = \(\frac{4}{9}\) AD² = \(\frac{4}{9}\) x  \(\frac{1}{4}\) (2c² + 2b² - a²)

= \(\frac{1}{9}\)(2c² + 2b² - a²) 

Similarly, GB² = \(\frac{1}{9}\)(2c² + 2a² - b²)

AG² + GB² = \(\frac{1}{9}\) [2c² + 2b² – a² + 2c² + 2a² – b²] = 

= \(\frac{1}{9}\) [a² + b² + 4c²] 

= \(\frac{1}{9}\) [5c² + 4c²] = \(\frac{9c^2}{9}\) = c² = BC² 

⇒ AG ⊥ GB ⇒ Angle between the medians drawn to sides a and b is 90º.