Question

What is the area of the sector ?​

Answer 1

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

$x^2 = 3^2 + 5^2 - 2(3)(5) \cos 81^\circ = 34 - 30\cos 81^\circ$

We have an isosceles triangle formed by two radii of 9 cm and x=PQ.  By the Law of Cosines again,

$x^2 = 9^2 + 9^2 - 2(9)(9) \cos B$

$x^2 = 162 - 162 \cos B$

$\cos B = \dfrac{162 - x^2}{162} = 1 - \dfrac{x^2}{162}$

$\cos B = 1 - \dfrac{34 - 30\cos 81^\circ}{162}$

$B = \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ }{162} \right)$

The area of the circle is the fraction given by the angle,

$A = \dfrac{B}{360^\circ} \pi r ^2$

$A \approx \dfrac{ \arccos \left(1 - \dfrac{34 - 30\cos 81^\circ}{162} \right) }{360} (3.142)9^2$

$A\approx 24.7474332960707$

Answer: 24.7 sq cm