Question

The points $A\left(2a,4a\right),B\left(2a,6a\right)$ and $C\left(2a+\sqrt{3}a,5a\right)$ (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

The correct option is B an equilateral triangle

Consider the given point

$A(2a,4a),B(2a,6a)andC(2a+\sqrt{3}a,5a)$

Distance between AB,BC and CA respectively.

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

$=\sqrt{0+{(-2a)}^2}$

$=\sqrt{4a^2}=2a$

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

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

$=\sqrt{4a^2}=2a$

$CA=\sqrt{{(2a+\sqrt{3}a-2a)}^2+{(5a-4a)}^2}$

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

$=\sqrt{4a^2}=2a$

Hence, $AB=BC=CA$

It is an equilateral triangle.

Option (B) is correct answer.