UPSC NDA 2 2021 Mathematics Solved Paper

Concept:The diagonal of a parallelogram divides it into two triangles of equal area.Area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$.Explanation:Let the vertices of the parallelogram be $A(1,2)$, $B(-1,5)$, $C(3,3)$, and $D$.Diagonal $AC$ divides parallelogram $ABCD$ into two equal triangles: $\triangle ABC$ and $\triangle ADC$.Thus, $\text{Area}(ABCD) = 2 \times \text{Area}(\triangle ABC)$.Compute area of $\triangle ABC$ using the formula:$\text{Area}(\triangle ABC) = \frac{1}{2} \left| 1(2 - 5) + (-1)(5 - 3) + 3(3 - 2) \right|$.Simplify each term: $1(-3) = -3$, $(-1)(2) = -2$, $3(1) = 3$.Sum: $-3 - 2 + 3 = -2$.Absolute value: $|{-2}| = 2$.Thus, $\text{Area}(\triangle ABC) = \frac{1}{2} \times 2 = 1$ square unit.Now, $\text{Area}(ABCD) = 2 \times 1 = 2$ square units.Answer:$2$ square units (Option C).