Question

A solid toy is in the form of a hemisphere covered by a right circular cone. If the height of the cone is 4 cm and the radius of the base is 3 cm, find the surface area of the toy. Use 3.14

Answer 1

Given :-

• A solid toy is in the form of a hemisphere covered by a right circular cone.

• The height of the cone is 4 cm and the radius of the base is 3 cm

To Find :-

• Surface area of toy

Solution :-

Dimensions of cone and hemisphere

• Radius of cone and hemisphere, r = 3 cm

• Height of cone, h = 4 cm

We know,

Slant height of cone is given by

$\rm :\longmapsto\: {l}^{2} = {r}^{2} + {h}^{2}$

$\rm :\longmapsto\: {l}^{2} = {3}^{2} + {4}^{2}$

$\rm :\longmapsto\: {l}^{2} = 9 + 16$

$\rm :\longmapsto\: {l}^{2} = 25$

$\rm :\longmapsto\: {l}^{2} = {5}^{2}$

$\bf\implies \:l = 5 \: cm$

We know,

$\boxed{ \bf{ \: CSA_{(cone)} = \pi \: rl}}$

$\boxed{ \bf{ \: CSA_{(hemisphere)} = {2\pi \: r}^{2} }}$

Thus,

$\rm :\longmapsto\:SA_{(toy)} = CSA_{(cone)} + CSA_{(hemisphere)}$

$\rm \: = \: \: \pi \: rl \: + \: 2 \: \pi \: {r}^{2}$

$\rm \: = \: \: \pi \: r \: (l \: + \: 2r)$

$\rm \: = \: \: 3.14 \times 3 \times (5 + 2 \times 3)$

$\rm \: = \: \: 9.42(5 + 6)$

$\rm \: = \: \: 9.42 \times 11$

$\rm \: = \: \: 103.62 \: {cm}^{2}$

$\bf\implies \:SA_{(toy)} = 103.62 \: {cm}^{2}$

More information:

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²