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Question

In a triangle ABC, AD is the bisector of angle A that meets BC at D. If AB = 21 cm, DC = 20 cm and AB : AC = 3 : 4, then what is \(AC^2-BD^2\) equal to?

This question was previously asked in
CDS 2 2026 Maths Question Paper (13-Sep-2026)
The correct answer is

\(559\)

Given \(AB=21\) cm and \(AB:AC=3:4\), so \(AC=\dfrac{21\times4}{3}=28\) cm. By the angle bisector theorem, \(\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{3}{4}\). Since \(DC=20\) cm, \(BD=\dfrac{3}{4}\times20=15\) cm. Hence \(AC^2-BD^2=28^2-15^2=784-225=559\).

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