Answer : 25 cm2
∵ In an equilateral ∆; angle bisector AX bisects the base BC at X
∴ BX = CX = \(\frac{5}{2}\) cm
AX = \(\sqrt{5^2 -(5/2)^2}
\)
= \(\sqrt{25-\frac{25}{4}}
\)
= \(\sqrt{\frac{75}{4}} = \frac{5\sqrt{3}}{2}\)
AY and BC being the chords of the circle,
AX . XY = BX . XC
⇒ \(\frac{5\sqrt{3}}{2}\) ⋅ XY = \(\frac{5}{2}\). \(\frac{5}{2}\)
⇒ XY = \(\frac{5}{2\sqrt{3}}\)
∴ AX . AY = \(\frac{5\sqrt{3}}{2}\) . ( \(\frac{5\sqrt{3}}{2}\) + \(\frac{5}{2\sqrt{3}}\))
= \(\frac{75}{4}\) + \(\frac{25}{4}\) = \(\frac{100}{4}\)
= 25 cm2