Question

The point A $(2a,4a)$, B$(2a,6a)$ and C$(2a+\sqrt{3}a,5a)$ (when $a>0$) are vertices of

• A. an obtuse angled triangle
• B. an equilateral triangle
• C. an isosceles obtuse angled triangle
• D. a right angled triangle

Answer 1

Answer: (B)

an equilateral triangle

Distance AB = $\sqrt{{(2a-2a)}^2+{\left(6a-4a\right)}^2}=\sqrt{4a^2}=2a$

Distance AC = $\sqrt{{(2a+\sqrt{3a}-2a)}^2+{\left(5a-4a\right)}^2}$

$\sqrt{3a^2+a^2}=2a$

Distance BC= $\sqrt{{\left(2a+\sqrt{3a}-2a\right)}^2+{\left(5a-6a\right)}^2}$

$\sqrt{3a^2+a^2}=2a$

AB = BC = CA

Equilateral Triangle