A container, in the form of a cone resting on its vertex, is full when 4.158 litres of water is poured into it.
(a) If the radius of its base is 21 cm,
(i) represent the information in a diagram;
(ii) calculate the height of the container.
(b) A certain amount of water is drawn out of the container such that the surface diameter of the water drops to 28 cm. Calculate the volume of the water drawn out. (Take \(\pi\) = \(\frac{22}{7}\))
Explanation
(a) (i) The diagram can be represented as:
(ii) first they were expected to convert from litres to cm\(^3 \) by using the fact that 1000 cm\(^3\) = 1 litre, then 4.158 litres = 4158 cm\(^3\).
If the volume and height of the container were V\(_1\) and h\(_1\) respectively, then, V\(_1\) = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 21 x 21 x h\(_1\) = 4,158 which simplified to 462h\(_1\) = 4,158. Therefore, h\(_1\) = 9cm.
(b) let h\(_2\) be the height of the water remaining in the container.
Then by similar triangles, \(\frac{h_2}{9}\) = \(\frac{14}{21}\), h\(_2\) = 6 cm.
If we let V\(_2\) be the volume of water remaining in the container, V\(_2\) = \(\frac{1}{3} \times \frac{22}{7}\) x 14 x 14 x 6 = 1,232 cm\(^2\).
Volume of water drawn out = (4,158 — 1,232) cm\(^3\) = 2,926 cm\(^3\)