Schel.
43.
The base radius and height of a cylindrical block of wood are 8 centimetres and 15
centimetres. A cone of maximum size is carved out of this.
(a) What are the radius and height of the cone?
(b)
Find its slant height.
(c)
Find the curved surface area of this cone.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
and
According to statement,
Therefore,
[tex]\rm :\longmapsto\:Radius_{(cone)} = Radius_{(cylinder)} = 8 \: cm[/tex]
[tex]\rm :\longmapsto\:Height_{(cone)} = Height_{(cylinder)} = 15 \: cm[/tex]
So,
Dimensions of Cone are
Now,
We know that
[tex] \rm :\longmapsto\:l \: = \sqrt{ {r}^{2} + {h}^{2} } [/tex]
[tex] \rm :\longmapsto\:l \: = \sqrt{ {8}^{2} + {15}^{2} } [/tex]
[tex]\rm :\longmapsto\:l = \sqrt{64 + 225} [/tex]
[tex]\rm :\longmapsto\:l = \sqrt{289} [/tex]
[tex]\bf\implies \: \boxed{ \bf{l \: = \: 17 \: cm}}[/tex]
[tex]\bf\implies \: \boxed{ \bf{Slant \: height \: of \: cone, \: l \: = \: 17 \: cm}}[/tex]
Now,
We know that,
[tex] \rm :\longmapsto\:\bf \: Curved \: Surface \: Area_{(cone)} = \pi \: rl[/tex]
[tex]\rm :\longmapsto\:Curved \: Surface \: Area_{(cone)} = \dfrac{22}{7} \times 8 \times 17[/tex]
[tex]\rm :\longmapsto\:Curved \: Surface \: Area_{(cone)} = \dfrac{2992}{7} \: {cm}^{2} [/tex]
Additional Information :-
Perimeter of rectangle = 2(length× breadth)
Diagonal of rectangle = √(length²+breadth²)
Area of square = side²
Perimeter of square = 4× side
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