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?
\(559\)
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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Select the answer using the code given below:
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