Question

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy( in $c{m}^2$) if the height of the conical part is 12 cm.

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Answer 1

The toy is in the shape shown below:

Radius of the hemispherical part = 5 cm.

Curved surface area of the Hemi-spherical part

= 27$\pi r^2=[27\pi \times (5{)}^2]c{m}^2=(50\pi )c{m}^2$.

Cylindrical part has radius = 5 cm and height = 13 cm.

Curved surface area of the cylindrical part = 2$\pi rh=(2\pi \times 5\times 13)c{m}^2=\left(130\pi \right)c{m}^2$.

Conical part has radius = 5 cm and height = 12 cm.

Its slant height = $\sqrt{5^2+(12{)}^2}=\sqrt{25+144}=\sqrt{169}=13cm$

Curved surface area of the conical part = $\pi$rl

= ($\pi$ x 5 x 13) $c{m}^2=\left(65\pi \right)c{m}^2$

Hence, the surface area of the toy = (50$\pi$ + 130$\pi$ + 65$\pi$) $c{m}^2$ = (245$\pi$) $c{m}^2$ .

= $(245\times \frac{22}{7})c{m}^2=770c{m}^2$
Also, volume of the toy = $(\frac{2}{3}\pi r^3+\pi r^{2}h+\frac{1}{3}\pi r^{2}H)$ $c{m}^3$ = $(\frac{250\pi }{3}+325\pi +100\pi )$ $c{m}^3$ = $\left(\frac{1525\pi }{3}\right)c{m}^3$

= $(\frac{1525}{3}\times \frac{22}{7})c{m}^3=1597.6c{m}^3$