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In a right traiangle \mathrm{ABC}, right angled at \mathrm{A},on the leg \mathrm{AC}  as diameter, a semicircle is described. The chord joining \mathrm{A}  with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to

Option: 1

\frac{\mathrm{AB} \cdot \mathrm{AD}}{\sqrt{\mathrm{AB}^{2}+\mathrm{AD}^{2}}}


Option: 2

\frac{\mathrm{AB} \cdot \mathrm{AD}}{\mathrm{AB}+\mathrm{AD}}


Option: 3

\sqrt{\mathrm{AB} \cdot \mathrm{AD}}


Option: 4

\frac{\mathrm{AB} \cdot \mathrm{AD}}{\sqrt{\mathrm{AB}^{2}-\mathrm{AD}^{2}}}


Answers (1)


\therefore \quad \frac{\mathrm{AC}}{\mathrm{AD}}=\frac{1}{\sin \theta} \Rightarrow \frac{\mathrm{AD}}{\mathrm{AB}}=\sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta \quad \Rightarrow \quad \frac{\mathrm{AD}^{2}}{\mathrm{AC}^{2}}+\frac{\mathrm{AD}^{2}}{\mathrm{AB}^{2}}=1

\mathrm{Aliter:} \quad \Delta=\frac{1}{2} \cdot \sqrt{\mathrm{x}^{2}+l^{2}} \cdot \mathrm{y}=\mathrm{x}^{2} l^{2} \quad \Rightarrow \quad l=\frac{\mathrm{xy}}{\sqrt{\mathrm{x}^{2}-\mathrm{y}^{2}}}=\frac{\mathrm{AB} \cdot \mathrm{AD}}{\sqrt{\mathrm{AB}^{2}-\mathrm{AD}^{2}}}.

Posted by

Ramraj Saini

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