Volumes of common solids 249
Total surface area of pyramid
= 2 ( 26. 80 )+ 2 ( 39. 83 )+( 3. 60 )( 5. 40 )
= 53. 60 + 79. 66 + 19. 44
=152.7cm^2
Problem 18. Calculate the volume and total
surface area of a hemisphere of diameter 5.0cm
Volume of hemisphere=
1
2
(volume of sphere)
=
2
3
πr^3 =
2
3
π
(
5. 0
2
) 3
=32.7cm^3
Total surface area
=curved surface area+area of circle
=
1
2
(surface area of sphere)+πr^2
=
1
2
( 4 πr^2 )+πr^2
= 2 πr^2 +πr^2 = 3 πr^2 = 3 π
(
5. 0
2
) 2
=58.9cm^2
Problem 19. A rectangular piece of metal having
dimensions 4cm by 3cm by 12cm is melted down
and recast into a pyramid having a rectangular base
measuring 2.5cm by 5cm. Calculate the
perpendicular height of the pyramid
Volume of rectangular prism of metal=^4 ×^3 ×^12
=144cm^3
Volume of pyramid
=
1
3
(area of base)(perpendicular height)
Assuming no waste of metal,
144 =
1
3
( 2. 5 × 5 )(height)
i.e.perpendicular height of pyramid=
144 × 3
2. 5 × 5
=34.56cm
Problem 20. A rivet consists of a cylindrical
head, of diameter 1cm and depth 2mm, and a shaft
of diameter 2mm and length 1.5cm.Determinethe
volume of metal in 2000 such rivets
Radius of cylindrical head=
1
2
cm= 0 .5cmand
height of cylindrical head=2mm= 0 .2cm.
Hence, volume of cylindrical head
=πr^2 h=π( 0. 5 )^2 ( 0. 2 )= 0 .1571cm^3
Volume of cylindrical shaft
=πr^2 h=π
(
0. 2
2
) 2
( 1. 5 )= 0 .0471cm^3
Total volume of 1 rivet= 0. 1571 + 0. 0471
= 0 .2042cm^3
Volume of metal in 2000 such rivets
= 2000 × 0. 2042 =408.4cm^3
Problem 21. A solid metal cylinder of radius
6cm and height 15cm is melted down and recast
into a shape comprising a hemisphere surmounted
by a cone. Assuming that 8% of the metal is wasted
in the process, determine the height of the conical
portion if its diameter is to be 12cm
Volume of cylinder=πr^2 h=π×^62 ×^15
= 540 πcm^3
If 8% of metal is lost then 92% of 540π gives the
volume of the new shape, shown in Figure 27.16.
h
r
12cm
Figure 27.16