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  • In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord joining A with the point of intersection D of the hypotenuse and the semicircle, then the

In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord joining 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}-\mathrm{AD}^{2}}}


Answers (1)

best_answer

\mathrm{\quad l \cdot \mathrm{x}=\mathrm{y} \sqrt{l^{2}+\mathrm{x}^{2}}}
l^{2} \mathrm{x}^{2}=\mathrm{y}^{2}\left(l^{2}+\mathrm{x}^{2}\right)
\mathrm{1^{2}\left(x^{2}-y^{2}\right)=x^{2} y^{2}}
l=\frac{\mathrm{xy}}{\sqrt{\mathrm{x}^{2}-\mathrm{y}^{2}}}=\frac{\mathrm{AB} \cdot \mathrm{AD}}{\sqrt{\mathrm{AB}^{2}-\mathrm{AD}^{2}}} Ans.

Posted by

Kuldeep Maurya

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