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
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
Given :-
To Find :-
Solution :-
Dimensions of cone and hemisphere
We know,
Slant height of cone is given by
[tex]\rm :\longmapsto\: {l}^{2} = {r}^{2} + {h}^{2} [/tex]
[tex]\rm :\longmapsto\: {l}^{2} = {3}^{2} + {4}^{2} [/tex]
[tex]\rm :\longmapsto\: {l}^{2} = 9 + 16[/tex]
[tex]\rm :\longmapsto\: {l}^{2} = 25[/tex]
[tex]\rm :\longmapsto\: {l}^{2} = {5}^{2} [/tex]
[tex]\bf\implies \:l = 5 \: cm[/tex]
We know,
[tex] \boxed{ \bf{ \: CSA_{(cone)} = \pi \: rl}}[/tex]
[tex] \boxed{ \bf{ \: CSA_{(hemisphere)} = {2\pi \: r}^{2} }}[/tex]
Thus,
[tex]\rm :\longmapsto\:SA_{(toy)} = CSA_{(cone)} + CSA_{(hemisphere)}[/tex]
[tex] \rm \: = \: \: \pi \: rl \: + \: 2 \: \pi \: {r}^{2} [/tex]
[tex]\rm \: = \: \: \pi \: r \: (l \: + \: 2r)[/tex]
[tex]\rm \: = \: \: 3.14 \times 3 \times (5 + 2 \times 3)[/tex]
[tex]\rm \: = \: \: 9.42(5 + 6)[/tex]
[tex]\rm \: = \: \: 9.42 \times 11[/tex]
[tex]\rm \: = \: \: 103.62 \: {cm}^{2} [/tex]
[tex]\bf\implies \:SA_{(toy)} = 103.62 \: {cm}^{2} [/tex]
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
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