Question

A solid consisting of a cylinder with a one cone at a hemisphere at the other end . The total length of solid 20cm and the common diameter is 7cm if the cylindercal portion has height 4.5cm find the total surface area of solid.​

Answer 1

Given :-

A solid consisting of a cylinder with a cone at one end and a hemisphere at the other end .

total length of solid = 20 cm

diameter of solid = 7 cm or radius = 7/2 cm

height of cylindrical portion = 4.5 cm

To find :-

the total surface area of solid

Solution :-

total surface area of Solid = CSA of cone + CSA of cylinder + CSA of hemisphere

or

total surface area of Solid = πrl + 2πrh + 2πr²

for finding CSA of cone we need the value of l (slant height)

$\sf \: l = \sqrt{ {(r)}^{2} + {(h)}^{2} }$

here we have the value of r (radius) i.e 7/2 cm or 3.5 cm

h (height of cone) = total height of solid - height of cylinder - radius of hemisphere

$\sf \: h \: = 20 - 4.5 - \dfrac{7}{2} \\ \\ \sf \: h \: = \dfrac{40 - 9 - 7}{2} \\ \\ \sf \: h \: = \dfrac{24}{2} = 12 \: cm$

now we have the value of height of cone .

$\therefore \sf \: l \: = \sqrt{ {(3.5)}^{2} + {(12)}^{2} } \\ \\ \sf \: l \: = \sqrt{156.25} \\ \\ \sf \: l \: =12.5 \: cm \:$

now we have the value of l (slant height) also .

then total surface area of Solid :-

$\sf \: total \: surface \: area = πrl + 2πrh + 2πr² \\ \\ \bf \: taking \:common \\ \\ \sf \: = \pi \: r(l + 2h + 2r) \\ \\ \sf\dfrac{11 \: \: \cancel{22}}{ \cancel7} \times \dfrac{ \cancel7}{ \cancel2} (12.5 + 9 + \cancel2 \times \dfrac{7}{ \cancel2} ) \\ \\ = \sf11 \times 28.5 \\ \\ \sf \: = 313.5 \: {cm}^{2}$

therefore total surface area of Solid = 313.5 cm²

additional information :-

CONE :-

• CSA of cone = πrl
• TSA of cone = πr(r + l)
• volume of cone = 1/3 πr²h

CYLINDER :-

• CSA of cylinder = 2πrh
• TSA of cylinder = 2πr (r + h)
• volume of cylinder = πr²h

HEMISPHERE :-

• CSA of hemisphere = 2πr²
• TSA of hemisphere = 3πr²
• volume of hemisphere = 2/3 πr³

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