The total surface area of a cone of slant height 1cm and base radius rcm...

MATHEMATICS
WAEC 2020

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 \(\pi\) = \(\frac{22}{7}\)]

Explanation

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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