CAT Exam 2021 Slot 2 Solved Paper

All the sides of the rhombus are equal. The area of a rhombus is $\frac{1}{2}(d_1)\cdot(d_2)=12$ $d_1 \times d_2 = 24$ The length of the side of a rhombus is given by $\frac{\sqrt{d_1 + d_2}}{2}$ . This is because the two diagonals and a side from a right-angled triangle with sides $\frac{d_1}{2}, \frac{d_2}{2}$ and the side length. $\frac{\sqrt{d_1 + d_2}}{2} = 5$ Hence, $\sqrt{d_1^2 + d_2^2} = 10$ $d_1^2 + d_2^2 = 100$ Using $d_1 \times d_2 = 24, 2 \times d_1 \times d_2 = 48$ $d_1^2 + d_2^2 + 2 \cdot d_1 \cdot d_2 = 100 + 48 = 148$ $d_1^2 + d_2^2 - 2 \cdot d_1 \cdot d_2 = 100 - 48 = 52$ $d_1 + d_2 = \sqrt{148}$ $d_1 - d_2 = \sqrt{52}$ adding equation (1) and (2), we get, $2 \times d_1 = 2 \times (\sqrt{37} + \sqrt{13})$ $d_1 = \sqrt{37} + \sqrt{13}$ or In a rhombus the area of a Rhombus is given by : The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b. The area of a Rhombus is : $\frac{1}{2} \cdot (2a) \cdot (2b) = 12$ $ab = 6$ . The length of each side is : $\sqrt{a^2 + b^2} = 5, a^2 + b^2 = 25$ $(a + b)^2 = 37, (a + b) = \sqrt{37}$ $(a - b)^2 = 13, (a - b) = \sqrt{13}$ $2a = (\sqrt{37} + \sqrt{13})$ , $2b = (\sqrt{37} - \sqrt{13})$ 2a is longer diagonal which is equal to $\sqrt{37} + \sqrt{13}$