From a solid cylinder of height 12 centimetre and diameter 10 cm a conical cavity of same height and same diameter is hollowed out find the total surface area of remaining solid
Answers
Answer:
Volume of remaining solid = Volume of cylinder - Volume of Cone
In the given case,
r = 5cm
h = 12cm
Volume of Cylinder = pi * r * r * h = 3.14 * 5 * 5 * 12 = 942 cm^3
Volume of Cone = (1/3) * pi * r * r * h = (1/3) * 3.14 * 5 * 5 * 12 = 314 cm^3
Remaining Volume = 942 - 314 = 628cm^3
Total Surface of Remaining Solid = Conical surface area of Cone + Area of Circle + Cylindrical area of Cylinder
Conical surface area of Cone = pi * r * ((r^2 + h^2)^(1/2)) = 282.74cm^2
Area of Circle = pi * r * r = 78.54cm^2
Cylindrical area of Cylinder = 2 * pi * r * h = 376.8cm^2
Total Surface Area = 738.08cm^2
Explanation:
Volume of remaining solid = volume of cylinder - volume of cone.
r=5 cm (given)
h=12 cm
Volume of cylinder = πr
2
h
=3.14×5
2
×12
=942cm
2
Volume of cone =
3
1
πr
2
h=
3
1
×3.14×5
2
×12
=314cm
3
Remaining volume = 942-314
=628cm
3
Total surface of remaining solid = conical sutface area of cone + Area of circle + Cylindrical area of cylinder.
Conical sutface area of cone =πr
r
2
+h
2
=204.203 cm²
Conical of cirle = πr² =78.54cm²
Cylindrical area of cylinder = 2πrh
=376.8cm²
Total surface area 659.543cm²