question_answer

ABC is an equilateral triangle inscribed in a circle with\[AB=5\text{ }cm\]. Let the bisector of the angle A meet BC in X and the circle in Y. What is the value of AX.AY?                       

A)  \[16\,c{{m}^{2}}\]                 

B)  \[20\,c{{m}^{2}}\]

C)  \[25\,c{{m}^{2}}\]                 

D)  \[30\,c{{m}^{2}}\]

Correct Answer: C

Solution :

As ABC is equilateral triangle then '0' will be it circumcentre as well as incentre. AX = height of equilateral triangle \[\therefore \,\,\,\,AX=\frac{\sqrt{3}}{2}\times 5=\frac{5\sqrt{3}}{2}\] and \[OX=\frac{side\,of\,triangle}{2\sqrt{3}}\] and  \[AO=\frac{side\,of\,triangle}{\sqrt{3}}=OY\] AY = Diameter of circle \[\Rightarrow \,\,\,AY=2AO\] \[=2\times \frac{5}{\sqrt{3}}=\frac{10}{\sqrt{3}}\] Now, \[AX.AY=\frac{5\sqrt{3}}{2}\times \frac{10}{\sqrt{3}}=\,\mathbf{25c}{{\mathbf{m}}^{\mathbf{2}}}\]