Find the volume of air to be filled in a basket ball of radius 10.5 cm.
[2 MARKS]

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Solution

Formula: 1 Mark
Application: 1 Mark

$r = 10.5$

Volume of air to be filled

$= \frac{4}{3} π r^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times {10.5}^{3}$

$= 4851 {c m}^{3}$

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Find the volume of air to be filled in a basket ball of radius 10.5 cm. [2 MARKS]

Formula: 1 Mark Application: 1 Mark r=10.5 Volume of air to be filled =43πr3 =43×227×10.53 =4851cm3

Find the volume of air to be filled in a basket ball of radius 10.5 cm.
[2 MARKS]

Formula: 1 Mark
Application: 1 Mark

$r = 10.5$

Volume of air to be filled

$= \frac{4}{3} π r^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times {10.5}^{3}$

$= 4851 {c m}^{3}$