Answer: \(38.6\text{ cm}^2\) to 3 significant figures.

For a sector, use radians throughout with arc length \(s=r\theta\) and area \(A=\frac12r^2\theta\).

The perimeter of the sector is made from two radii and the arc \(AB\).

So

\(20+20+20\theta=65.\)

Hence

\(20\theta=25,\)

so

\(\theta=1.25.\)

Since \(CA\) and \(CB\) are tangents from the same point, \(AC=BC\), and \(OC\) bisects angle \(AOB\).

In right triangle \(AOC\),

\(\angle AOC=\frac{\theta}{2}=0.625.\)

Also \(OA=20\), so

\(\tan\frac{\theta}{2}=\frac{AC}{20}.\)

Therefore

\(AC=20\tan(0.625)\approx14.43.\)

The area of kite \(OACB\) is the sum of the two congruent right triangles:

\(2\left(\frac12\times20\times AC\right)=20AC.\)

So

\(\text{area of kite}\approx20(14.43)=288.6.\)

The area of sector \(OAB\) is

\(\frac12 r^2\theta=\frac12(20)^2(1.25)=250.\)

The shaded area is therefore

\(288.6-250=38.6\text{ cm}^2\)

to 3 significant figures.