page 280)
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
2π8
2=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
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
=1
3(area of base)(perpendicular height)
=1
3(3.60×5.40)(14.64)=94.87 cm
3
Area of triangleADF(which equals triangleBCF)=1
2(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)
= 1
2(5.40)(14.75)=39.83 cm 2
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 3πr 3=2 3π 5.0 2 3 =32.7 cm2 Total surface area
=curved surface area+area of circle
=1
2(surface area of sphere)+πr 2 =1 2(4πr 2)+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
= 1
3(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.
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 3πr 3 +1 3πr 2h=0.92×540π Dividing throughout byπgives:
2 3r
3+ 1 3r
2h=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 cm
3=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 500×100cm 2=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= 2 3πr 3= 2 3×π×3 3=18πm3 Volume of cylinder, Q=πr2h=π×32×8=72πm3 Volume of cone, R= 13πr2h= 13×π×32×4=12πm3
P Q B A 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
Exercise 88 Further problems on volumes and surface areas of regular solids (Answers on page 280)
1. Determine the mass of a hemispherical copper container whose external and internal radii are 12 cm and 10 cm. Assuming that 1 cm3of copper weighs 8.9 g.
2. If the volume of a sphere is 566 cm3, find its radius. 3. A metal plumb bob comprises a hemisphere surmounted
by a cone. If the diameter of the hemisphere and cone are each 4 cm and the total length is 5 cm, find its total volume.
4. A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 m and the cylindrical portion has a height of 3.5 m, with a diameter of 15 m. Calculate
the surface area of material needed to make the marquee assuming 12% of the material is wasted in the process. 5. Determine (a) the volume and (b) the total surface area
of the following solids:
(i) a cone of radius 8.0 cm and perpendicular height 10 cm
(ii) a sphere of diameter 7.0 cm (iii) a hemisphere of radius 3.0 cm
(iv) a 2.5 cm by 2.5 cm square pyramid of perpendic- ular height 5.0 cm
(v) a 4.0 cm by 6.0 cm rectangular pyramid of perpen- dicular height 12.0 cm
(vi) a 4.2 cm by 4.2 cm square pyramid whose sloping edges are each 15.0 cm
(vii) a pyramid having an octagonal base of side 5.0 cm and perpendicular height 20 cm.
6. The volume of a sphere is 325 cm3. Determine its diameter.
7. A metal sphere weighing 24 kg is melted down and recast into a solid cone of base radius 8.0 cm. If the density of the metal is 8000 kg/m3 determine (a) the diameter of the metal sphere and (b) the perpendicular height of the cone, assuming that 15% of the metal is lost in the process.
8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 16.0 cm and the side of base is 3.0 cm.
9. A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere is 2.5 m and the slant height of the cone is 4.0 m. Determine the volume and surface area of the buoy.
10. A petrol container is in the form of a central cylindrical portion 5.0 m long with a hemispherical section sur- mounted on each end. If the diameters of the hemisphere and cylinder are both 1.2 m determine the capacity of the tank in litres (1 litre=1000 cm3).
11. Figure 24.9 shows a metal rod section. Determine its volume and total surface area.
1.00 cm radius
2.50 cm
1.00 m
12. Find the volume (in cm3) of the die-casting shown in Fig. 24.10. The dimensions are in millimetres.
50 100 25 30 rad 60 Fig. 24.10
13. The cross-section of part of a circular ventilation shaft is shown in Fig. 24.11, ends AB and CD being open. Calculate (a) the volume of the air, correct to the nearest litre, contained in the part of the system shown, neglecting the sheet metal thickness, (given 1 litre=1000 cm3), (b) the cross-sectional area of the sheet metal used to make the system, in square metres, and (c) the cost of the sheet metal if the material costs £11.50 per square metre, assuming that 25% extra metal is required due to wastage.
2 m 1.5 m 1.5 m 800 mm 500 mm A B D C Fig. 24.11
24.4 Volumes and surface areas of frusta
of pyramids and cones
The frustum of a pyramid or cone is the portion remaining when a part containing the vertex is cut off by a plane parallel to the base.
Thevolume of a frustum of a pyramid or coneis given by the volume of the whole pyramid or cone minus the volume of the small pyramid or cone cut off.
Thesurface area of the sides of a frustum of a pyramid or coneis given by the surface area of the whole pyramid or cone minus the surface area of the small pyramid or cone cut off. This gives the lateral surface area of the frustum. If the total surface area of the frustum is required then the surface area of the two parallel ends are added to the lateral surface area.
There is an alternative method for finding the volume and surface area of afrustum of a cone. With reference to Fig. 24.12:
Volume=13πh(R2+Rr+r2)
Curved surface area=πl(R+r)
Total surface area=πl(R+r)+πr2+πR2
r
h I
R
Fig. 24.12
Problem 16. Determine the volume of a frustum of a cone if the diameter of the ends are 6.0 cm and 4.0 cm and its perpendicular height is 3.6 cm.
Method 1
A section through the vertex of a complete cone is shown in Fig. 24.13.
Using similar triangles AP DP= DR BR Hence AP 2.0= 3.6 1.0 from which AP=(2.0)(3.6) 1.0 =7.2 cm
4.0 cm 2.0 cm 1.0 cm 3.0 cm 6.0 cm 3.6 cm Q P A E C D R B Fig. 24.13
Volume of frustum of cone
=volume of large cone−volume of small cone cut off
=1 3π(3.0)2(10.8)− 1 3π(2.0)2(7.2) =101.79−30.16=71.6 cm3 Method 2
From above, volume of the frustum of a cone
=1 3πh(R
2+Rr+r2), where R=3.0 cm, r=2.0 cm and h=3.6 cm
Hence volume of frustum
=1 3π(3.6) (3.0)2+(3.0)(2.0)+(2.0)2 =1 3π(3.6)(19.0)=71.6 cm 3
Problem 17. Find the total surface area of the frustum of the cone in Problem 16.
Method 1
Curved surface area of frustum=curved surface area of large cone−curved surface area of small cone cut off.
From Fig. 24.13, using Pythagoras’ theorem: AB2=AQ2+BQ2, from which AB =[10.82+3.02]=11.21 cm and
AD2=AP2+DP2, from which AD =[7.22+2.02]=7.47 cm Curved surface area of large cone
=πrl=π(BQ)(AB)=π(3.0)(11.21)=105.65 cm2 and curved surface area of small cone
=π(DP)(AD)=π(2.0)(7.47)=46.94 cm2 Hence, curved surface area of frustum=105.65−46.94
=58.71 cm2 Total surface area of frustum
=curved surface area+area of two circular ends
=58.71+π(2.0)2+π(3.0)2
=58.71+12.57+28.27=99.6 cm2
Method 2
From page 186, total surface area of frustum
=πl(R+r)+πr2+πR2,
where l=BD=11.21−7.47=3.74 cm, R=3.0 cm and r= 2.0 cm.
Hence total surface area of frustum
=π(3.74)(3.0+2.0)+π(2.0)2+π(3.0)2=99.6 cm2
Problem 18. A storage hopper is in the shape of a frustum of a pyramid. Determine its volume if the ends of the frustum are squares of sides 8.0 m and 4.6 m, respectively, and the perpendicular height between its ends is 3.6 m.
The frustum is shown shaded in Fig. 24.14(a) as part of a com- plete pyramid. A section perpendicular to the base through the vertex is shown in Fig. 24.14(b).
By similar triangles:CG BG= BH AH HeightCG=BG BH AH =(2.3)(3.6) 1.7 =4.87 m Height of complete pyramid=3.6+4.87=8.47 m Volume of large pyramid=1
3(8.0)