UPSC CDS 2 2021 Math Paper

Concept:Use properties of an isosceles triangle and Pythagoras theorem to find the missing side length.Explanation:Given $BD = 4$ cm, $CD = 6$ cm.So $BC = BD + CD = 4+6 = 10$ cm.Since $AB = AC$, triangle $ABC$ is isosceles with base $BC$.Draw perpendicular $AE$ from $A$ to $BC$.In an isosceles triangle, the altitude from the vertex bisects the base.Thus $BE = CE = \frac{10}{2} = 5$ cm.Now $DE = BE - BD = 5 - 4 = 1$ cm.In right $triangle ADE$: $AE^2 = AD^2 - DE^2 = 5^2 - 1^2 = 25-1=24$.In right $triangle ACE$: $AE^2 = AC^2 - CE^2 = AC^2 - 5^2$.Equate the two expressions for $AE^2$:$AC^2 - 25 = 24$$\Rightarrow AC^2 = 49$$\Rightarrow AC = 7$ cm (positive length).Since $AB = AC$, we have $AB = 7$ cm.Answer:$AB = 7$ cm, corresponding to option A.