Question

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

• A. An acute angled triangle
• B. A right angled triangle
• C. An isosceles triangle
• D. None of these

Answer 1

Answer: (A), (C), (D)

The correct options are
A

An acute angled triangle

C

An isosceles triangle

D

None of these

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

BC = $\sqrt{(\sqrt{3}a{)}^2+a^2}$ = 2a

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

Since AB = BC = CA, hence triangle is equilateral.

Therefore, it is an acute angled triangle.