A geometry problem by Muhammad Fahad Khan

area of circle=9 pi by area of circle=9pi radius=3 In triangle ACD, Sin60=6/AC (as radius=3, diameter=6) AC=12/root3, area of equilateral triangle= root3/4 length^2 12 root 3 is the answer,

Sorry for not being able to type root 3....this ws my first time typing any answer like this...

Since the circle has area $9\pi$ , we know that the radius is 3, which implies that the height of the equilateral triangle is 6. This implies that $CD = \frac{6}{\sqrt{3}}$ . Thus the area is just $6(\frac{6}{\sqrt{3}}) \approx 20.784$

the height of the triangle is 6 and since it is an equilateral triangle, the rest comes from elementary geometry. The area = 12*sqrt[3].

Area = 9(pi)
AD =dia = 6
CD = AC/2 = (AC^2 - 6^2)^(1/2)
AC = (48)^(1/2)
Area of triangle = 1/2 * b * h = 1/2 * 6.9282 * 6 = 20.7846(Ans.)

As circles area is $9\pi=r^2\pi\Rightarrow$ the circle's radii is $3\Rightarrow AD=6$ .

Then, $\frac{AD}{\sin 60°}=\frac{AB}{\sin 90}\Rightarrow \frac{6}{\frac{\sqrt{3}}{2}}=\frac{AB}{1}\therefore AB=\frac{12}{\sqrt{3}}=4\sqrt{3}$

Finally, the area of $ABC$ is $\frac{4\sqrt{3}\times6}{2}=\boxed{12\sqrt{3}}$