The points $$A(2a$$,$$4a)$$,$$B(2a$$,$$6a)$$ and $$C(2a+3a$$,$$5a)$$ (when$$ a>$$0) are vertices of

Correct option is 2an equilateral triangle

Questions

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

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an obtuse angled triangle

an equilateral triangle

an isosceles obtuse angled triangle

a right angle triangle

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detailed solution

Correct option is B

AB=(2a−2a)2+(4a−6a)2=2a BC=(2a+3a−2a)2+(5a−6a)2=(3a)2+a2=2a and CA=(2a−2a−3a)2+(5a−4a)2=(3a)2+a2=2a Since AB=BC=CA, the triangle is an equilateral.

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The points A(2a,4a),B(2a,6a) and C(2a+3a,5a) (when a>0) are vertices of

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