UPSC CDS I 2019 Math Paper

Concept:The surface area of a cuboid is given by $2(lb + bh + hl)$, and it can be found using the sum of the three dimensions and the space diagonal.Explanation:Let the length, breadth, and height of the cuboid be $a$, $b$, and $c$ respectively.Given: $a + b + c = 22$ cm and the space diagonal $= 14$ cm.For a cuboid, diagonal $\sqrt{a^2 + b^2 + c^2} = 14$, so $a^2 + b^2 + c^2 = 196$.Square the sum: $(a + b + c)^2 = 484$.Expand: $a^2 + b^2 + c^2 + 2(ab + bc + ca) = 484$.Substitute $a^2 + b^2 + c^2 = 196$: $196 + 2(ab + bc + ca) = 484$.Thus $2(ab + bc + ca) = 288$.Surface area of the cuboid is $2(ab + bc + ca)$.Hence the surface area is $288$ cm$^2$.Answer:$\boxed{288 \text{ cm}^2}$ (Option A)