The radius of the sector equals the slant height of the cone \(l\).

The arc length of the sector equals the circumference of the base:

\(\dfrac{90^\circ}{360^\circ}\cdot 2\pi l=2\pi r\).

Hence \(\dfrac{\pi l}{2}=2\pi r\Rightarrow l=4r\).

Therefore, \(l:r=4:1\).

Answer: \(4:1\).

In the axial section, the base equals the diameter and the equal sides are the slant heights.

Hence \(2r=6\Rightarrow r=3\text{ cm}\), \(l=6\text{ cm}\).

The total surface area is

\(S=\pi r^2+\pi rl\).

For \(\pi=3\): \(S=3\cdot 3^2+3\cdot 3\cdot 6=27+54=81\text{ cm}^2\).

Answer: \(81\text{ cm}^2\).

From \(\pi r^2=9\pi\), we get \(r=3\text{ cm}\).

The area of the axial section equals \(rh\), so \(3h=12\Rightarrow h=4\text{ cm}\).

The slant height is \(l=\sqrt{r^2+h^2}=\sqrt{9+16}=5\text{ cm}\).

Answer: \(5\text{ cm}\).

We have \(2r=4\Rightarrow r=2\text{ cm}\), and the slant height is \(l=2\sqrt{10}\text{ cm}\).

The height is \(h=\sqrt{l^2-r^2}=\sqrt{40-4}=6\text{ cm}\).

The volume is \(V=\dfrac13\pi r^2h\). For \(\pi=3\): \(V=\dfrac13\cdot 3\cdot 4\cdot 6=24\text{ cm}^3\).

Answer: \(24\text{ cm}^3\).

The axial section of a cone is an isosceles triangle. If it is right-angled, then the base radius equals the height: \(r=h\).

The area of the axial section is \(rh\), so \(r^2=36\), hence \(r=h=6\text{ cm}\).

The volume is \(V=\dfrac13\pi r^2h\). For \(\pi=3\): \(V=\dfrac13\cdot 3\cdot 36\cdot 6=216\text{ cm}^3\).

Answer: \(216\text{ cm}^3\).