UPSC CDS I 2024 Math Paper

Concept:Use the properties of tangents drawn from vertices to the incircle and the Pythagoras theorem.Explanation:Let the incircle of triangle ABC have centre O and radius $2$ cm.Since $\angle B = 90^{\circ}$, the tangents from vertex B to the incircle are equal to the inradius. Thus, $B$ to the point of contact on $AB$ equals $2$ cm, and $B$ to the point of contact on $BC$ also equals $2$ cm.Given $MA : MC = 2 : 3$, where $M$ is the point of tangency on $AC$. Let $MA = 2x$ and $MC = 3x$. Then $AC = MA + MC = 5x$.The tangents from $A$ to the incircle: $A$ to $M$ on $AC$ and $A$ to the point on $AB$ are equal. So $AB$ length = $MA + 2 = 2x + 2$.Similarly, $BC$ = $MC + 2 = 3x + 2$.In right triangle $ABC$, apply Pythagoras: $AB^2 + BC^2 = AC^2$.$(2x + 2)^2 + (3x + 2)^2 = (5x)^2$Expanding: $4x^2 + 8x + 4 + 9x^2 + 12x + 4 = 25x^2$Simplify: $13x^2 + 20x + 8 = 25x^2$Bring terms: $12x^2 - 20x - 8 = 0$Divide by 4: $3x^2 - 5x - 2 = 0$Factor: $(3x + 1)(x - 2) = 0$Thus $x = 2$ (since $x = -\frac{1}{3}$ is not possible).Hence $AB + BC = (2x + 2) + (3x + 2) = 5x + 4 = 5 \times 2 + 4 = 14$.Answer:$AB + BC = 14$ cm.