The points A(2a, 4a),B(2a, 6a),and C(2a + sqr | vedclass

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(A) Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$AB = \sqrt{(2a - 2a)^2 + (6a - 4a)^2} =

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An acute angled triangle

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(A) Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$AB = \sqrt{(2a - 2a)^2 + (6a - 4a)^2} = \sqrt{0 + (2a)^2} = 2a$.$BC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 6a)^2} = \sqrt{(\sqrt{3}a)^2 + (-a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.$CA = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 4a)^2} = \sqrt{(\sqrt{3}a)^2 + a^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.Since $AB = BC = CA = 2a$,the triangle is an equilateral triangle.Every equilateral triangle is an acute angled triangle,as all its angles are $60^\circ$.

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

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