The diagram shows the cross section of a plastic pipe. The arc AB is part of the circle with centre O and radius 45 mm. The pipe has an uniform thickness of x mm. The shaded area represents the cross-sectional area that is filled with water. AB = 65.32 mm.

1. Relate the chord length to the central angle.

Let $\angle AOB=\theta$ radians.

For chord $AB=65.32$ mm in a circle of radius $45$ mm,
$65.32=2(45)\sin\frac{\theta}{2}.$

2. Solve for the central angle and round to the required value.

$\sin\frac{\theta}{2}=\frac{65.32}{90}$
$\frac{\theta}{2}=\sin^{-1}\left(\frac{65.32}{90}\right)$
$\theta=2\sin^{-1}\left(\frac{65.32}{90}\right)=1.624291\ldots$
$\theta=1.6243 \text{ radians}.$

1. Form an equation for the cross-sectional area of the pipe.

The cross-sectional area of the pipe material is
$\pi(45+x)^2-\pi(45)^2=1398.29.$

2. Solve the equation for the pipe thickness $x$.

$\pi\left((45+x)^2-45^2\right)=1398.29$
$(45+x)^2-45^2=\frac{1398.29}{\pi}$
$(45+x)^2=45^2+\frac{1398.29}{\pi}$
$45+x=\sqrt{45^2+\frac{1398.29}{\pi}}$
$x=\sqrt{45^2+\frac{1398.29}{\pi}}-45=4.71\text{ mm} \;\text{(to 3 s.f.)}.$