Question

A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end . The radius of the common base is 8cm and the height of the cylinderical and conical portions are 10 cm and 6 cm respectively . Find the total surface area of the solid.(take value of pie as 3.14).

Answer 1

$\bold{Given,}$

A solid cylinder and a cone are such that they have a common base we need to find the total surface area of whole solid so,

$\large \bold{FOR \: CYLINDER}$

$\mathsf{ Radius = 8 cm}$

$\mathsf{ Height = 10 cm}$

$\huge \boxed{ Surface \: area \: of \: cylinder = 2 \times \pi \times r \times h}$

$\mathsf{ Surface \: area \: of \: cylinder = 2 \times 3.14 \times 8 \times 10}$

$\huge \boxed{ Surface \: area \: of \: cylinder = 502.4 cm}$

$\large \bold{FOR \: CONE}$

$\mathsf{ Radius(Base) = 8 cm}$

$\mathsf{ Height = 6cm}$

$\mathsf{ lenght = ?}$

By Pythagoras theorem we know

$\mathsf{lenght^{2} = base^{2} + height^{2}}$

$\mathsf{lenght^{2} = 8^{2} + 6^{2}}$

$\mathsf{lenght^{2} = 64 + 36}$

$\mathsf{lenght^{2} = 100}$

$\mathsf{lenght = \sqrt{100}}$

$\mathsf{lenght =10 cm}$

$\huge \boxed{ Surface \: area \: of \: cone = \pi \times r \times l}$

$\mathsf{ Surface \: area \: of \: cylinder = 3.14 \times 8 \times 10}$

$\huge \boxed{ Surface \: area \: of \: cone = 251.2 cm }$

$\boxed{Total \: surface \: area \: of \: solid = Surface \: area \: of \: cylinder + Surface \: area \: of \: cone}$

$\mathsf{Total \: surface \: area \: of \: solid = 502.4 + 251.2}$

$\huge \boxed{Total \: surface \: area \: of \: solid = 753.6 cm}$

Answer 2

Answer:

question is wrong please correct it