The diagram shows the cross section of a cylinder, centre O, radius r, lying on its side. The cylinder contains water to a depth of 18 cm. The width, AB, of the surface of the water is 24 cm.

1. Express the half-chord and the perpendicular distance from the centre to the water surface in terms of $r$.

Let $M$ be the midpoint of $AB$.

Since the perpendicular from the centre of a circle to a chord bisects the chord,
$AM=\frac{24}{2}=12\text{ cm}.$

The distance from the water surface $AB$ to the bottom point $P$ is $18\text{ cm}$, and $OP=r$.

Hence
$OM=18-r.$

2. Apply Pythagoras' theorem and solve the equation for $r$.

In right-angled triangle $OMA$,
$OA^2=OM^2+AM^2.$

Since $OA=r$,
$r^2=(18-r)^2+12^2.$

$r^2=324-36r+r^2+144$
$36r=468$
$r=13.$

1. Form a right-angled triangle using half the chord and identify the half-angle.

Let $M$ be the midpoint of $AB$.

Since $AB=24\text{ cm}$,
$AM=\frac{24}{2}=12\text{ cm}.$

From part (a), $OA=13\text{ cm}$.

In right-angled triangle $OAM$,
$\sin \angle AOM=\frac{AM}{OA}=\frac{12}{13}.$

2. Calculate the full angle $AOB$ and round to 2 decimal places.

$\angle AOM=\sin^{-1}\left(\frac{12}{13}\right)=67.3801\ldots^\circ$

Since $OM$ bisects $\angle AOB$,
$\angle AOB=2(67.3801\ldots^\circ)=134.7602\ldots^\circ.$

Therefore,
$\angle AOB=134.76^\circ \quad \text{correct to 2 decimal places}.$

1. Find the area of sector $AOB$ using the angle from part (b).

From earlier parts, $r=13$ cm and $\angle AOB=134.76^\circ$.

$\text{Area of sector }AOB=\frac{134.76}{360}\times \pi \times 13^2$

$=198.74\text{ cm}^2\text{ approximately}$

2. Find the area of triangle $AOB$.

The perpendicular distance from $O$ to chord $AB$ is $5$ cm, and $AB=24$ cm.

$\text{Area of }\triangle AOB=\frac12\times 24\times 5$

$=60\text{ cm}^2$

3. Find the area of the small segment above the water surface.

$\text{Area of small segment}=\text{area of sector }AOB-\text{area of }\triangle AOB$

$=198.74-60$

$=138.74\text{ cm}^2$

4. Find the shaded water segment by subtracting the small segment from the whole circle.

$\text{Area of circle}=\pi\times 13^2=530.93\text{ cm}^2$

$\text{Area of shaded segment}=530.93-138.74$

$=392.18\text{ cm}^2$

$\approx 392\text{ cm}^2$

1. Write an expression for the water volume when the cylinder is upright.

When upright, the water forms a cylinder of radius $13$ cm and height $h$ cm.

$\text{Volume of water}=\pi(13)^2h=169\pi h$

2. Equate the unchanged water volume to the volume from the horizontal position and solve for h.

From part (c), the cross-sectional area of the water is $392\text{ cm}^2$.

The original length of the cylinder is $40$ cm, so

$\text{Volume of water}=392\times 40$

Equating volumes,

$169\pi h=392\times 40$

$h=\frac{392\times 40}{169\pi}=29.5\text{ cm}\quad\text{(3 s.f.)}$