CBSE Class 10 Math Question Paper 2016 | Q8

CBSE Maths Board Paper Solution | 2 Mark Question | Coordinate Geometry

The given question is a medium difficulty 2-mark question from the chapter Coordinate Geometry. Concept tested: Computing lengths of sides of a triangle using the coordinates of the three vertices of the triangle and applying Pythagoras Theorem to establish that the triangle is a right triangle. The question appeared in the 2016 CBSE class 10 board paper.

Question 8: Prove that the point (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle

NCERT Solution to Class 10 Maths

Explanatory Answer | CBSE 2016 Maths Paper Q8

Let (3, 0), (6, 4), and (-1, 3) be the coordinates of vertices A, B, and C respectively.

Approach: Compute lengths of AB, BC, and CA. Identify type of triangle

AB = $$sqrt{$6 - 3$^2 + (4 - 0)^2 }) = $$sqrt{3^2 + 4^2}$ = $\sqrt{25}$ = 5 BC =$\sqrt{$-1 - 6$^2 + (3 - 4)^2}) = $$sqrt{7^2 + 1^2}$ = $\sqrt{50}$ CA =$\sqrt{$-1 - 3$^2 + (3 - 0)^2}) = $\sqrt{4^2 + 3^2}$ = $\sqrt{25}$ = 5
Sides of triangle ABC measure 5, 5, and √50.

Inference 1: 2 sides are equal and therefore, the triangle is an Isosceles triangle.

Observation: 52 + 52 = $\sqrt{50^2}$
Inference 2: Sum of squares of two sides equal square of the third side. Satisfies Pythagoras theorem.
Hence, the given triangle is a right triangle.

∴ the given triangle is a Right Isosceles triangle.

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