UPSC CDS 1 2022 Math Paper

Given: The diagonal of a square is $12\sqrt{2}\ \text{cm}$ The area of an equilateral triangle $(A T)$ is $64\sqrt{3}\ \text{cm}^{2}\ \cdots$-(i) Formula Used: Perimeter of square $=4x$ Area of square $=x^{2}$ Diagonal of square $=\sqrt{}2.x$ Where, $x=$ side length of square Perimeter of equilateral $\Delta=3a$ Area of equilateral $\Delta=\frac{\sqrt{3}a^{2}}{4}$ Calculation: Let teh length of side of square be $x$ And the length of side of equilateral $\Delta$ be a According to the question $\sqrt{2}\cdot x=12\sqrt{2}$ $\Rightarrow x=12$ So, Perimeter of square $=4\times12=48\ \text{cm}\;\;-$ (ii) Area of square $(A_{S})=12^{2}=144\ \text{cm}^{2}\ \cdots$ (iii) Again according to the question $\frac{\sqrt{3}a^{2}}{4}=64\sqrt{3}$ $\Rightarrow a^{2}=256\ \text{cm}^{2}$ $\Rightarrow a=16\ \text{cm}$ So, Perimeter of equilateral $\Delta=3\times16=48\ \text{cm}\ \cdots$-(iv) From (ii) and (iv), we get Perimeter of square $=$ Perimeter of equilateral $\Delta$ Statemetn 1 is correct And from (i) and (iii), we get $\;\frac{A_{S}}{A_{T}}=\;\frac{144}{64\sqrt{3}}$ $\Rightarrow\;\frac{A_{S}}{A_{T}}=\;\frac{3\sqrt{3}}{4}$ $\Rightarrow 4A_{S}=3\sqrt{3}A_{T}$ Statement 2 is also correct. $\therefore$ Both the statements are correct.