UPSC CDS I 2024 Math Paper

Concept:Use geometry of circles inside a rectangle to find the unknown radius, then sum their areas.Explanation:Let the bigger circle have radius $R = 4$ cm (it touches the top and bottom of the rectangle of height $8$ cm).Let the smaller circle have radius $r$ cm.Both circles touch each other externally, so the distance between centers $O_1O_2 = 4 + r$.In the right triangle $O_1EO_2$:• Vertical leg $O_1E = 4 - r$ (difference in distances from the top).• Horizontal leg $EO_2 = 5 - r$ (since the big circle’s center is $4$ cm from the left wall, the small circle’s center is $r$ cm from the right wall, and the rectangle width is $9$ cm; gap = $9 - 4 - r$).Apply Pythagoras: $(4+r)^2 = (4-r)^2 + (5-r)^2$.Expand: $16 + 8r + r^2 = (16 - 8r + r^2) + (25 - 10r + r^2)$.Simplify: $16 + 8r + r^2 = 41 - 18r + 2r^2$.Rearrange: $0 = 41 - 18r + 2r^2 - 16 - 8r - r^2$ → $0 = 25 - 26r + r^2$ → $r^2 - 26r + 25 = 0$.Factor: $(r-1)(r-25)=0$ → $r=1$ or $r=25$ (reject $25$ as too large).Thus $r=1$ cm.Area of bigger circle: $\pi R^2 = \pi \times 16 = 16\pi$ sq cm.Area of smaller circle: $\pi r^2 = \pi \times 1 = \pi$ sq cm.Sum of areas: $16\pi + \pi = 17\pi$ sq cm.Answer:$17\pi$ square unit (Option A).