UPSC NDA 2 2021 Mathematics Solved Paper

Concept:An equilateral triangle has all sides equal.Its area is given by $\frac{\sqrt{3}}{4}a^2$, where $a$ is the side length.The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Explanation:Let $A = (0, 0)$, $B = (2, 2)$, and $C = (x, y)$ be the vertices of an equilateral triangle.Side $AB = \sqrt{(2 - 0)^2 + (2 - 0)^2} = 2\sqrt{2}$.Since $\triangle ABC$ is equilateral, $AB = BC = AC = 2\sqrt{2}$.Set $AC = BC$: $\sqrt{x^2 + y^2} = \sqrt{(x-2)^2 + (y-2)^2}$.Squaring gives $x^2 + y^2 = x^2 - 4x + 4 + y^2 - 4y + 4$ $\implies 8 - 4x - 4y = 0$ $\implies y = 2 - x$.Now use $AC = 2\sqrt{2}$: $\sqrt{x^2 + y^2} = 2\sqrt{2}$ $\implies x^2 + y^2 = 8$.Substitute $y$: $x^2 + (2 - x)^2 = 8$ $\implies x^2 + 4 - 4x + x^2 = 8$ $\implies 2x^2 - 4x - 4 = 0$ $\implies x^2 - 2x - 2 = 0$.Solve: $x = \frac{2 \pm \sqrt{4 + 8}}{2} = 1 \pm \sqrt{3}$.Then $y = 2 - (1 \pm \sqrt{3}) = 1 \mp \sqrt{3}$.So $C$ has coordinates $(1+\sqrt{3}, 1-\sqrt{3})$ or $(1-\sqrt{3}, 1+\sqrt{3})$. Both contain irrational numbers. Hence statement 1 is correct.Area of equilateral triangle: $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (2\sqrt{2})^2 = \frac{\sqrt{3}}{4} \cdot 8 = 2\sqrt{3}$ square units.Since $\sqrt{3}$ is irrational, the area is irrational. Hence statement 2 is correct.Answer:Both statements 1 and 2 are correct. Therefore, option C is the correct answer.