1. A rectangular block of metal has dimensions of 40 mm by 25 mm by 15 mm. Determine its volume. Find also its mass if the metal has a density of 9 g/cm3.
2. Determine the maximum capacity, in litres, of a fish tank measuring 50 cm by 40 cm by 2.5 m (1 litre=1000 cm3).
3. Determine how many cubic metres of concrete are required for a 120 m long path, 150 mm wide and 80 mm deep.
4. Calculate the volume of a metal tube whose outside diameter is 8 cm and whose inside diameter is 6 cm, if the length of the tube is 4 m.
5. The volume of a cylinder is 400 cm3. If its radius is 5.20 cm, find its height. Determine also its curved surface area.
6. If a cone has a diameter of 80 mm and a perpendicular height of 120 mm calculate its volume in cm3and its curved surface area.
7. A cylinder is cast from a rectangular piece of alloy 5 cm by 7 cm by 12 cm. If the length of the cylinder is to be 60 cm, find its diameter.
8. Find the volume and the total surface area of a regular hexagonal bar of metal of length 3 m if each side of the hexagon is 6 cm.
9. A square pyramid has a perpendicular height of 4 cm. If a side of the base is 2.4 cm long find the volume and total surface area of the pyramid.
10. A sphere has a diameter of 6 cm. Determine its volume and surface area.
11. Find the total surface area of a hemisphere of diameter 50 mm.
12. How long will it take a tap dripping at a rate of 800 mm3/s to fill a 3-litre can?
24.3 Further worked problems on volumes and surface areas of regular solids
Problem 8. A wooden section is shown in Fig. 24.5. Find (a) its volume (in m3), and (b) its total surface area.
3 m
12 cm r⫽
8 mm
r
Fig. 24.5
(a) The section of wood is a prism whose end comprises a rect- angle and a semicircle. Since the radius of the semicircle is 8 cm, the diameter is 16 cm.
Hence the rectangle has dimensions 12 cm by 16 cm.
Area of end=(12×16)+1
2π82=292.5 cm2 Volume of wooden section
=area of end×perpendicular height
=292.5×300=87750 cm3=87750 m3 106
=0.08775 m3
(b) The total surface area comprises the two ends (each of area 292.5 cm2), three rectangles and a curved surface (which is half a cylinder), hence total surface area
=(2×292.5)+2(12×300)+(16×300) +1
2(2π×8×300)
=585+7200+4800+2400π
=20125 cm2 or 2.0125 m2
Volumes of common solids 183
Problem 9. A pyramid has a rectangular base 3.60 cm by 5.40 cm. Determine the volume and total surface area of the pyramid if each of its sloping edges is 15.0 cm.
The pyramid is shown in Fig. 24.6. To calculate the volume of the pyramid the perpendicular heightEFis required. Diagonal BDis calculated using Pythagoras’ theorem,
i.e. BD=
[3.602+5.402]=6.490 cm
H
B C
A
F
G D
E 15.0
cm 15.0
cm 15.0 cm
15.0 cm
5.40 cm 3.60
cm
Fig. 24.6
HenceEB=1
2BD=6.490
2 =3.245 cm.
Using Pythagoras’ theorem on triangleBEFgives BF2=EB2+EF2
from which, EF=
(BF2−EB2)
=√
15.02−3.2452=14.64 cm.
Volume of pyramid
=13(area of base)(perpendicular height)
=13(3.60×5.40)(14.64)=94.87 cm3
Area of triangleADF(which equals triangleBCF)=12(AD)(FG), whereGis the midpoint ofAD. Using Pythagoras’ theorem on triangleFGAgives
FG=
[15.02−1.802]=14.89 cm
Hence area of triangleADF=12(3.60)(14.89)=26.80 cm2
Similarly, ifHis the mid-point ofAB, then FH=
15.02−2.702=14.75 cm,
hence area of triangleABF(which equals triangleCDF)
= 12(5.40)(14.75)=39.83 cm2 Total surface area of pyramid
=2(26.80)+2(39.83)+(3.60)(5.40)
=53.60+79.66+19.44
=152.7 cm2
Problem 10. Calculate the volume and total surface area of a hemisphere of diameter 5.0 cm.
Volume of hemisphere=1
2(volume of sphere)
=2 3πr3=2
3π 5.0
2 3
=32.7 cm2 Total surface area
=curved surface area+area of circle
=1
2(surface area of sphere)+πr2
=1
2(4πr2)+πr2
=2πr2+πr2=3πr2=3π 5.0
2 2
=58.9 cm2
Problem 11. A rectangular piece of metal having dimen- sions 4 cm by 3 cm by 12 cm is melted down and recast into a pyramid having a rectangular base measuring 2.5 cm by 5 cm. Calculate the perpendicular height of the pyramid.
Volume of rectangular prism of metal=4×3×12=144 cm3 Volume of pyramid
= 13(area of base)(perpendicular height) Assuming no waste of metal,
144= 1
3(2.5×5)(height) i.e. perpendicular height=144×3
2.5×5=34.56 cm
Problem 12. A rivet consists of a cylindrical head, of diam- eter 1 cm and depth 2 mm, and a shaft of diameter 2 mm and length 1.5 cm. Determine the volume of metal in 2000 such rivets.
184 Basic Engineering Mathematics
Radius of cylindrical head=1
2cm=0.5 cm and height of cylindrical head=2 mm=0.2 cm Hence, volume of cylindrical head
=πr2h=π(0.5)2(0.2)=0.1571 cm3 Volume of cylindrical shaft
=πr2h=π 0.2
2 2
(1.5)=0.0471 cm3 Total volume of 1 rivet=0.1571+0.0471=0.2042 cm3 Volume of metal in 2000 such rivets=2000×0.2042
=408.4 cm3
Problem 13. A solid metal cylinder of radius 6 cm and height 15 cm is melted down and recast into a shape com- prising 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 12 cm.
Volume of cylinder=πr2h=π×62×15=540πcm3 If 8% of metal is lost then 92% of 540πgives the volume of the new shape (shown in Fig. 24.7).
12 cm r h
Fig. 24.7
Hence the volume of (hemisphere+cone)=0.92×540πcm3, i.e.1
2 4
3πr3
+1
3πr2h=0.92×540π Dividing throughout byπgives:
2 3r3+ 1
3r2h=0.92×540
Since the diameter of the new shape is to be 12 cm, then radius r=6 cm,
hence2 3(6)3+1
3(6)2h=0.92×540 144+12h=496.8
i.e. height of conical portion, h= 496.8−144
12 =29.4 cm
Problem 14. A block of copper having a mass of 50 kg is drawn out to make 500 m of wire of uniform cross-section.
Given that the density of copper is 8.91 g/cm3, calculate (a) the volume of copper, (b) the cross-sectional area of the wire, and (c) the diameter of the cross-section of the wire.
(a) A density of 8.91 g/cm3 means that 8.91 g of copper has a volume of 1 cm3, or 1 g of copper has a volume of (1/8.91) cm3 Hence 50 kg, i.e. 50 000 g, has a volume
50 000
8.91 cm3=5612 cm3 (b) Volume of wire
=area of circular cross-section×length of wire.
Hence 5612 cm3=area×(500×100 cm), from which, area= 5612
500×100cm2=0.1122 cm2 (c) Area of circle=πr2orπd2
4 , hence 0.1122=πd2 4 from which
d=
4×0.1122 π
=0.3780 cm i.e. diameter of cross-section is 3.780 mm.
Problem 15. A boiler consists of a cylindrical section of length 8 m and diameter 6 m, on one end of which is sur- mounted a hemispherical section of diameter 6 m, and on the other end a conical section of height 4 m and base diam- eter 6 m. Calculate the volume of the boiler and the total surface area.
The boiler is shown in Fig. 24.8.
Volume of hemisphere,
P= 23πr3= 23×π×33=18πm3 Volume of cylinder,
Q=πr2h=π×32×8=72πm3 Volume of cone,
R= 13πr2h= 13×π×32×4=12πm3
Volumes of common solids 185
P
Q A B
R C
4 m I
6 m
3 m 8 m
Fig. 24.8
Total volume of boiler=18π+72π+12π
=102π=320.4 m3 Surface area of hemisphere,
P= 12(4πr2)=2×π×32=18πm2 Curved surface area of cylinder,
Q=2πrh=2×π×3×8=48πm2
The slant height of the cone,l, is obtained by Pythagoras’theorem on triangleABC, i.e.l=
(42+32)=5 Curved surface area of cone,
R=πrl=π×3×5=15πm2
Total surface area of boiler=18π+48π+15π
=81π=254.5 m2
Now try the following exercise