UPSC CDS II 2024 Math Paper

Concept:A median of a triangle divides it into two triangles of equal area.Explanation:Given: $AB = 10\ \mathrm{cm}$, $BP = CP$, $\angle BAP = 45^\circ$.Since $P$ is the midpoint of $BC$, $AP$ is a median.A median splits the triangle into two equal areas.Thus $\triangle APC$ has half the area of $\triangle ABC$. Statement I is correct.From the figure (or given data), $BC = BP + CP = 2 \times BP = 20\ \mathrm{cm}$.Using Pythagoras theorem in right $\triangle ABC$ (assuming $\angle B = 90^\circ$):$AC^2 = AB^2 + BC^2 = 10^2 + 20^2 = 500$$AC = \sqrt{500} = 22.36\ \mathrm{cm}$ (approx).Perimeter of $\triangle ABC = AB + BC + AC = 10 + 20 + 22.36 = 52.36\ \mathrm{cm}$, which is more than $46\ \mathrm{cm}$.Hence the perimeter of $\triangle APC$ is also greater than $46\ \mathrm{cm}$. Statement II is correct.Area of $\triangle ABC = \frac{1}{2} \times BC \times AB = \frac{1}{2} \times 20 \times 10 = 100\ \mathrm{cm}^2$.Area of $\triangle APC = \frac{100}{2} = 50\ \mathrm{cm}^2$. Statement III is correct.Answer:All three statements I, II, and III are correct. Option D.