FODA DEL PROCESO DE MENTORÍA
PROPUESTA: MANUAL DEL MENTOR
6.7. Acciones y estrategias de mentoría recomendadas
Ch20-H8555.tex 31/7/2007 10: 18 page 163
Volumes and surface areas of common solids 163
Sec tion 2
Now try the following exercise
Exercise 76 Further problems on volumes
Ch20-H8555.tex 31/7/2007 10: 18 page 164
164 Engineering Mathematics
Sec tion 2
12. Find the volume (in cm3) of the die-casting shown in Figure 20.10. The dimensions are in millimetres. [220.7 cm3]
100
60
50 25
30 rad
Figure 20.10
13. The cross-section of part of a circular ventila-tion shaft is shown in Figure 20.11, ends AB and CD being open. Calculate (a) the vol-ume 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 sys-tem, 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.
[(a) 1458 litre (b) 9.77 m2(c) £140.45]
2 m A
B 500 mm
1.5 m
1.5 m
800 mm
C D
Figure 20.11
20.4 Volumes and surface areas of frusta of pyramids and cones
The frustum of a pyramid or cone is the portion remain-ing when a part containremain-ing the vertex is cut off by a plane parallel to the base.
The volume of a frustum of a pyramid or cone is given by the volume of the whole pyramid or cone minus the volume of the small pyramid or cone cut off.
The surface area of the sides of a frustum of a pyra-mid or cone is 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 frus-tum 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 a frustum of a cone. With reference to Fig. 20.12:
r I h
R Figure 20.12
Volume=13πh(R2+Rr+r2) Curved surface area=πl(R+r)
Total surface area=πl(R+r)+πr2+πR2 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. 20.13
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
The height of the large cone=3.6+7.2=10.8 cm.
Ch20-H8555.tex 31/7/2007 10: 18 page 165
Volumes and surface areas of common solids 165
Sec tion 2
4.0 cm 2.0 cm
3.0 cm 6.0 cm
3.6 cm
Q P A
E
C D
B R 1.0 cm
Figure 20.13
Volume of frustum of cone
=volume of large cone
−volume of small cone cut off
=13π(3.0)2(10.8)−13π(2.0)2(7.2)
=101.79−30.16=71.6 cm3 Method 2
From above, volume of the frustum of a cone
=13πh(R2+Rr+r2), where R=3.0 cm,
r=2.0 cm and h=3.6 cm Hence volume of frustum
=13π(3.6)[(3.0)2+(3.0)(2.0)+(2.0)2]
=13π(3.6)(19.0)=71.6 cm3
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. 20.13, 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 164, 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
Ch20-H8555.tex 31/7/2007 10: 18 page 166
166 Engineering Mathematics
Sec tion 2
The frustum is shown shaded in Fig. 20.14(a) as part of a complete pyramid. A section perpendicular to the base through the vertex is shown in Fig. 20.14(b)
By similar triangles: CG BG =BH
AH
Height CG=BG BH
AH
= (2.3)(3.6)
1.7 =4.87 m
4.6 cm 4.6 cm
8.0 m
8.0 m 2.3 m 2.3 m 3.6 m
4.0 m 2.3 m 1.7 m
(a) (b)
C
G D
B
A H
E F
Figure 20.14
Height of complete pyramid =3.6+4.87=8.47 m Volume of large pyramid = 13(8.0)2(8.47)
=180.69 m3 Volume of small pyramid cut off
= 13(4.6)2(4.87)=34.35 m3 Hence volume of storage hopper
=180.69−34.35=146.3 m3
Problem 19. Determine the lateral surface area of the storage hopper in Problem 18
The lateral surface area of the storage hopper consists of four equal trapeziums.
From Fig. 20.15, area of trapezium PRSU
= 12(PR+SU)(QT )
OT=1.7 m (same as AH in Fig. 20.14(b)) and OQ=3.6 m.
By Pythagoras’ theorem,
QT =
OQ2+OT2=
3.62+1.72=3.98 m
4.6 m 4.6 m
8.0 m 0
8.0 m Q
T P
R S
U
Figure 20.15
Area of trapezium PRSU= 12(4.6+8.0)(3.98)
=25.07 m2 Lateral surface area of hopper=4(25.07)
=100.3 m2 Problem 20. A lampshade is in the shape of a frustum of a cone. The vertical height of the shade is 25.0 cm and the diameters of the ends are 20.0 cm and 10.0 cm, respectively. Determine the area of the material needed to form the lampshade, correct to 3 significant figures
The curved surface area of a frustum of a cone=πl(R+r) from page 164.
Since the diameters of the ends of the frustum are 20.0 cm and 10.0 cm, then from Fig. 20.16,
r=5.0 cm, R=10.0 cm and l=
25.02+5.02=25.50 cm, from Pythagoras’ theorem.
r 5.0 cm
R 10.0 cm
h 25.0 cm
I
5.0 cm
Figure 20.16
Hence curved surface area
=π(25.50)(10.0+5.0)=1201.7 cm2 i.e. the area of material needed to form the lampshade is 1200 cm2, correct to 3 significant figures.
Ch20-H8555.tex 31/7/2007 10: 18 page 167
Volumes and surface areas of common solids 167
Sec tion 2
Problem 21. A cooling tower is in the form of a cylinder surmounted by a frustum of a cone as shown in Fig. 20.17. Determine the volume of air space in the tower if 40% of the space is used for pipes and other structures
12.0 m
25.0 m
12.0 m 30.0 m
Figure 20.17
Volume of cylindrical portion
=πr2h=π 25.0
2 2
(12.0)=5890 m3 Volume of frustum of cone
=13πh(R2+Rr+r2) where h=30.0−12.0=18.0 m,
R=25.0/2=12.5 m and r=12.0/2=6.0 m Hence volume of frustum of cone
=13π(18.0)[(12.5)2+(12.5)(6.0)+(6.0)2]
=5038 m3
Total volume of cooling tower=5890+5038
=10 928 m3
If 40% of space is occupied then volume of air space=0.6×10 928=6557 m3
Now try the following exercise
Exercise 77 Further problems on volumes