question_answer

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

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

B)  \[20\,\,\text{c}{{\text{m}}^{2}}\]

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

D)  \[30\,\,\text{c}{{\text{m}}^{2}}\]

Correct Answer: C

Solution :

In \[\Delta ABC,\]

\[BX=\frac{5}{2}\text{cm,}\]\[CX=\frac{5}{2}\,\,\text{cm}\]

and \[AX=\frac{\sqrt{3}}{2}\times 5=\frac{5\sqrt{3}}{2}\,\,\text{cm}\]

AY and BC are the chord of circle

\[\therefore \]      \[AX\cdot XY=BX\cdot XC\]

\[\Rightarrow \] \[\frac{5\sqrt{3}}{2}\cdot XY=\frac{5}{2}\cdot \frac{5}{2}\]\[\Rightarrow \]\[XY=\frac{5}{2\sqrt{3}}\]

\[\therefore \] \[AX\cdot AY=\left( \frac{5\sqrt{3}}{2}+\frac{5}{2\sqrt{3}} \right)\times \frac{5}{2\sqrt{3}}=25\,\,\text{c}{{\text{m}}^{2}}\]