Question 1 Report
The total surface area of a cone of slant height 1cm and base radius rcm is 224\(\pi\) cm\(^2\). If r : 1 = 2.5, find:
(a) correct to one decimal place, the value of r
(b) correct to the nearest whole number, the volume of the cone [Take \(\pi\) = \(\frac{22}{7}\)]
Setting up. The total surface area of a cone is \(\pi r(r + l) = 224\pi\), so
\[r(r + l) = 224.\]
The given ratio of slant height to base radius is \(l : r = 2.5\), i.e. \(l = 2.5r\) (the slant height must exceed the radius). Substituting,
\[r(r + 2.5r) = 224 \Rightarrow 3.5r^2 = 224 \Rightarrow r^2 = 64.\]
(a) \(r = 8.0\text{ cm}\), and \(l = 2.5\times8 = 20\text{ cm}\).
(b) Height \(h = \sqrt{l^2 - r^2} = \sqrt{20^2 - 8^2} = \sqrt{336} = 18.33\text{ cm}.\)
\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times64\times18.33 = 1229\text{ cm}^3\text{ (to the nearest whole number)}.\]
Everything you need to excel in your exams