# Mensuration-44

### From Homeworkwiki

** The height of a solid cylinder is 10 cm and diameter 7 cm. Two equal conical holes of radius 3 cm and height 4 cm**

**are cut off from both ends of the cylinder. Find the volume and surface area of the remaining solid.**

**Solution:** Radius of the cylinder (R) = 7/2 cm

Height (H) = 10 cm

Therefore, Volume of the cylinder = Π R^{2} H = 22/7 x (7/2)^{2} x 10 cm^{3}
= 385 cm^{3}

Total surface of cylinder = 2 Π R (R + H)

= 2 x 22/7 x 7/2 (7/2 + 10) cm^{2} = 2 x 22/7 x 7/2 x 27/2 cm^{2}

= 297 cm^{2}

Radius of each conical hole (r) = 3 cm, height (h) = 4 cm

Therefore, slant height (l) = √(r^{2} + h^{2}) = √(3^{2} + 4^{2})

= 5 cm

Volume of the two conical holes = 2 x 1/3 Π r^{2} h = 2 x 1/3 x 22/7 x 3^{2} x 4 cm^{3}

= 75.43 cm^{2}

Curved surface of the two conical cavities = 2 x Π r l = 2 x 22/7 x 3 x 5 cm^{2} = 94.29 cm^{2}

Therefore, Volume of the remaining solid = Volume of cylinder - Volume of two cones

= (385 - 75.43) cm^{3} = 309.57 cm^{3}

Surface area of the remaining solid

= Total surface of the cylinder + curved surface of the two cones - 2(area of the circular base of the cones)

= [297 + 94.29 - 2 x 22/7 x 3^{2}] cm^{2} = (297 + 94.29 - 56.57) cm^{2}

= 334.72 cm^{2}