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 \(\pi\) = \(\frac{22}{7}\)]
TSA = 224\(\pi\) cm\(^2\)
r : 1 = 2: 5
\(\frac{r}{l} = \frac{2}{5}\)
r = \(\frac{2l}{5}\)
\(\frac{3r}{2}\) = l
TSA \(\pi\)rl + \(\pi ^2\)
TSA = \(\pi\)(rl + r\(^2\))
224\(\pi\) = \(\pi\)r(l + r)
224 = r (\(\frac{5r}{2} + \frac{r}{1}\))
224 = r(\(\frac{5r + 2r}{2}\))
224 = \(\frac{7r^2}{2}\)
r\(^2\) = \(\frac{448}{7}\)
r = \(\sqrt{448}{7}\)
r = \(\sqrt{64cm}\)
l = \(\frac{5 \times 8}{2}\)
r = 8cm
l = 20cm
(b) Volume of cone = \(\frac{1}{3}\)\(\pi ^2\)h
h\(^2\) = l\(^2\) - r\(^2\)
h\(^2\) = (20)\(^2\) - (8)\(^2\)
h\(^2\) = 400 - 64
h = \(\sqrt{336}\)
= 18.33cm
Volume = \(\frac{1}{3} \times \frac{22}{7}\) x 8\(^2\) x 18.33
= \(\frac{25,809.06}{21}\)
= 1229 cm\(^3\)
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