They forget to say "not to scale". I'm guessing this is trig because I don't see another way to do it.
Let's consider x=chord PQ first. By the Law of Cosines
[tex]x^2 = 3^2 + 5^2 - 2(3)(5) \cos 81^\circ = 34 - 30\cos 81^\circ[/tex]
We have an isosceles triangle formed by two radii of 9 cm and x=PQ. By the Law of Cosines again,
[tex]x^2 = 9^2 + 9^2 - 2(9)(9) \cos B[/tex]
[tex]x^2 = 162 - 162 \cos B[/tex]
[tex]\cos B = \dfrac{162 - x^2}{162} = 1 - \dfrac{x^2}{162}[/tex]
[tex]\cos B = 1 - \dfrac{34 - 30\cos 81^\circ}{162}[/tex]
[tex]B = \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ }{162} \right)[/tex]
The area of the circle is the fraction given by the angle,
[tex]A = \dfrac{B}{360^\circ} \pi r ^2[/tex]
[tex]A \approx \dfrac{ \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ}{162} \right) }{360} (3.142)9^2[/tex]
[tex]A\approx 24.7474332960707[/tex]
Answer: 24.7 sq cm
