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Quantitative Aptitude

Triangles

1.The co-ordi...

Question

1.The co-ordinates of vertices of triangle ABC are A (0, 0) B (0, 2) and C (2, 0).Prove that triangle ABC is an isosceles triangle. Also find its area

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Solution

$By distance formula, find length of AB, BC and CA . AB = \sqrt{{(0 - 0)}^{2} + {(0 - 2)}^{2}} = 2 units AC = \sqrt{{(0 - 2)}^{2} + {(0 - 0)}^{2}} = 2 units BC = \sqrt{{(0 - 2)}^{2} + {(2 - 0)}^{2}} = 2 \sqrt{2} units Hence, △ ABC, is isoceles triangle Draw AD perpendicular to BC, therefore, D is midpoint of BC . By mid point formula, coordinates of D are (1, 1) Length AD = \sqrt{{(1 - 0)}^{2} + {(1 - 0)}^{2}} = \sqrt{2} units Hence, area of triangle = \frac{1}{2} \times base \times height = \frac{1}{2} \times 2 \sqrt{2} \times \sqrt{2} = 2 sq units$

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