The diagram shows a cardboard in the shape of a major sector, centre O and radius 15 cm. The total area of the major sector OPRQ is 450 cm².

1. Substitute the given area and radius into the sector area formula.

For a sector, $A=\frac{1}{2}r^2\theta.$ Given $A=450$ and $r=15$, $450=\frac{1}{2}(15)^2\theta.$

2. Solve for the reflex angle $\theta$.

$450=112.5\theta$ $\theta=\frac{450}{112.5}=4$

1. Find the length of the major arc using $s=r\theta$.

Since the reflex angle is $4$ radians and the radius is $15$ cm,

$\text{arc length}=r\theta=15\times 4=60\text{ cm}.$

2. Add the two radii to the major arc length to get the perimeter.

$\text{perimeter}=15+15+60=90\text{ cm}.$

1. Find the radius of the cone's base from the arc length of the sector.

From earlier parts, the length of the major arc is $60\text{ cm}$. This becomes the circumference of the base of the cone.

$2\pi r=60$
$r=\frac{60}{2\pi}=\frac{30}{\pi}\text{ cm}$

2. Use the slant height and base radius to form a right-angled triangle equation.

The radius of the sector, $15\text{ cm}$, becomes the slant height of the cone.

Using Pythagoras' theorem,

$r^2+h^2=15^2$
$\left(\frac{30}{\pi}\right)^2+h^2=225$

3. Solve for the vertical height of the cone.

$h^2=225-\left(\frac{30}{\pi}\right)^2$
$h=\sqrt{225-\left(\frac{30}{\pi}\right)^2}$
$h=11.6\text{ cm}\;\text{(to 3 s.f.)}$

1. Substitute the cone radius and height into the cone volume formula.

From earlier parts, the base radius is $r=\dfrac{30}{\pi}$ and the height is
$h=\sqrt{225-\left(\dfrac{30}{\pi}\right)^2}$.

$V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi\left(\frac{30}{\pi}\right)^2\sqrt{225-\left(\frac{30}{\pi}\right)^2}$

2. Evaluate the volume and round suitably.

$V\approx \frac{1}{3}\pi(9.549\ldots)^2(11.579\ldots)$

$V\approx 1105.7\ldots$

To $3$ significant figures,
$V=1110\text{ cm}^3$