Question

A solid toy is in the form of a right circular cylinder with hemispherical shape at one end and a cone
at the other end. Their common diameter is 4.2 cm and the height of the cylindrical and conical
portions are 12cm and 7cm respectively. Find the volume of the solid toy.​

Answer 1

firstly,

write the given data and draw diagram

write the formula which has to be applied

then slove

Answer 2

$\mathfrak \red{Given:-}$

• Let height of the conical portion h(1) =7 cm.

• The height of cylinderical portion h(2) = 12 cm

$\mathfrak\red{Find:-}$

• Find the volume of the solid toy.

$\mathfrak\red{Solution:-}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \: \: \: \: \: \: \begin{gathered}\boxed {\begin{array}{cc} \\ \\ \dashrightarrow\rm \: volume \: of \: the \: solid \: toy = volum \: of \: the \: cone \: + \: volume \: of \: the \: cylinder \: + \: volume \: of \: the \: hemisphere. \\ \\ \mapsto \rm \: \frac{1}{3} \pi {r}^{2} h_{1} + \pi {r}^{2} h_{2} + \frac{2}{3} \pi {r}^{2} \\ \\ \mapsto\rm \pi {r}^{2} ( \frac{1}{3} h_{1} + h_{2} + \frac{2}{3}r) \\ \\ \mapsto \rm \frac{22}{7} \times ( \frac{21}{10} ) ^{2} \times ( \frac{1}{3} \times 7 + 12 + \frac{2}{3} \times \frac{21}{10} ) \\ \\ \mapsto \rm \: \frac{22}{7} \times \frac{441}{100} \times ( \frac{7}{3} + \frac{12}{1} + \frac{7}{5} ) \\ \\ \mapsto \rm \: \frac{22}{7} \times \frac{441}{100} \times ( \frac{35 + 180 + 21}{15} ) \\ \\ \mapsto \rm \: \frac{22}{7} \times \frac{441}{100} \times \frac{236}{15} \\ \\ \mapsto \rm \: \frac{27258}{125} \\ \\ \rm⇒ \fbox\red{218.064 \: cm} \: ^{(3)} \end{array}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

• ✓ Hence proved

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@Shivam