Question

A semicircle sheet of paper of diameter 14cm is bent to form a conical cup. Find the capacity of the cup.​

Answer 1

$\red{Required \: Solution - }$

$Let \: radius \: and \: height \: of \: the \: conical \: cup \: be \: 'r' \: and \: 'h' \: respectively.$

$Circumference \: of \: the \: base \: of \: the \: cone = Length \: of \: arc \: of \: the \: semi-circle$

$2\pi r = ( \frac{1}{2} )2\pi R$

$2r = R$

$2r = 7$

$r = \frac{7}{2}$

$Slant \: height \: of \: the \: conical \: cup = Radius \: of \: the \: semi-circular \: sheet$

$\blue{We \: know \: that,}$

${l}^{2} = {h}^{2} + {r}^{2}$

$( {7})^{2} = {h}^{2} + (\frac{7}{2})^{2}$

$49 = {h}^{2} + \frac{49}{4}$

${h}^{2} = 49 - \frac{49}{4}$

${h}^{2} = \frac{196 - 49}{4}$

${h}^{2} = \frac{147}{4}$

${h}^{2} = \sqrt{ \frac{147}{4} }$

$h = 6.06$

$Height \: of \: the \: conical \: cup = 6.06 cm.$

$\green{The \: capacity \: of \: the \: conical \: cup \: is \: given \: by,}$

$V = \frac{1}{3} \pi {r}^{2} h$

$V = \frac{1}{3} \times \frac{22}{7} \times ( \frac{7}{2} ) ^{2} \times 6.06$

$V = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times 6.06$

$V = 77.77 {cm}^{3}$

$\pink{Therefore,}$

$The \: capacity \: of \: the \: cup \: is \: {77.77}^{3}$