(a) Curved Surface Area = \(\pi rl\) \(115.5 = \frac{22}{7} \times r \times 10.5\) \(115.5 = 33r\) \(r = \frac{115.5}{33} = 3.5 cm\) (b) \(\therefore h^{2} ...

This question tests the three standard measurements of a right circular cone and how they connect. A cone has a base radius \(r\), a vertical (perpendicular) height \(h\) measured from the tip straight down to the centre of the base, and a slant height \(l\) measured along the sloping surface from the tip to the edge of the base. The key idea is that \(r\), \(h\) and \(l\) form a right-angled triangle, with the right angle at the centre of the base, so \(l\) is the hypotenuse.

(a) Finding the base radius \(r\)

The curved (lateral) surface area of a cone is \(\pi r l\), not the full surface, so we use only the sloping part with \(l = 10.5\text{ cm}\):

\[\pi r l = 115.5\] \[\frac{22}{7}\times r \times 10.5 = 115.5\]

Since \(\dfrac{22}{7}\times 10.5 = 33\), this becomes

\[33\,r = 115.5 \quad\Rightarrow\quad r = \frac{115.5}{33} = 3.5\text{ cm}\]

(b) Finding the vertical height \(h\)

Because \(r\), \(h\) and \(l\) form a right-angled triangle with \(l\) as the hypotenuse, Pythagoras' theorem gives \(r^{2} + h^{2} = l^{2}\). Rearranging for \(h\):

\[h^{2} = l^{2} - r^{2} = (10.5)^{2} - (3.5)^{2} = 110.25 - 12.25 = 98\] \[h = \sqrt{98} = 9.90\text{ cm (3 s.f.)}\]

(c) Finding the volume

The volume of a cone is \(\dfrac{1}{3}\pi r^{2} h\). Note this uses the vertical height \(h\), not the slant height, so we use the value found in part (b):

\[V = \frac{1}{3}\times \frac{22}{7}\times (3.5)^{2}\times 9.90\] \[= \frac{1}{3}\times \frac{22}{7}\times 12.25 \times 9.90 = \frac{1}{3}\times 38.5 \times 9.90 = \frac{381.15}{3}\] \[V \approx 127\text{ cm}^{3}\]

So \(r = 3.5\text{ cm}\), \(h \approx 9.90\text{ cm}\) and the volume is about \(127\text{ cm}^{3}\).

A common mistake is to mix up the slant height and the vertical height: using \(l = 10.5\) inside the volume formula would give the wrong answer. The slant height belongs only to the surface-area formula \(\pi r l\); the volume formula \(\tfrac{1}{3}\pi r^{2} h\) must use the perpendicular height \(h\) that you obtain from Pythagoras. Whenever a cone problem gives you one height and asks for a quantity that needs the other, expect to use \(l^{2}=r^{2}+h^{2}\) as the bridge between them.