Consultante •Recibe la intervención
RESULTADOS: ANÁLISIS Y DISCUSIÓN
8. Al final o cada cierta unidad de información,
3.4. Valoración de la mentoría 1 Interacción y comunicación
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Volumes and surface areas of common solids 159
Sec tion 2
Total surface area of pyramid=(5×5)+4(30.65)
=147.6 cm2 Problem 6. Determine the volume and total surface area of a cone of radius 5 cm and perpendicular height 12 cm
The cone is shown in Fig. 20.4.
h l 12 cm
r 5 cm Figure 20.4
Volume of cone= 13πr2h=13×π×52×12
=314.2 cm3 Total surface area=curved surface area
+area of base
=πrl+πr2
From Fig. 20.4, slant height l may be calculated using Pythagoras’ theorem
l=
122+52=13 cm
Hence total surface area=(π×5×13)+(π×52)
=282.7 cm2
Problem 7. Find the volume and surface area of a sphere of diameter 8 cm
Since diameter=8 cm, then radius, r=4 cm.
Volume of sphere= 4
3πr3=4
3 ×π×43
=268.1 cm3 Surface area of sphere=4πr2=4×π×42
=201.1 cm2
Now try the following exercise
Exercise 75 Further problems on volumes
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160 Engineering Mathematics
Sec tion 2
11. Find the total surface area of a hemisphere of diameter 50 mm. [5890 mm2or 58.90 cm2] 12. How long will it take a tap dripping at a rate
of 800 mm3/s to fill a 3-litre can?
[62.5 minutes]
20.3 Further worked problems on volumes and surface areas of regular solids
Problem 8. A wooden section is shown in Fig.
20.5. Find (a) its volume (in m3), and (b) its total surface area.
3 m
12 cm r 8 mm
r
Figure 20.5
The section of wood is a prism whose end comprises a rectangle 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)+12π82=292.5 cm2 Volume of wooden section
=area of end×perpendicular height
=292.5×300=87 750 cm3=87 750 m3 106
=0.08775 m3 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)+12(2π×8×300)
=585+7200+4800+2400π
=20 125 cm2 or 2.0125 m2 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. 20.6. To calculate the volume of the pyramid the perpendicular height EF is required. Diagonal BD is 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
Figure 20.6
Hence EB= 1
2BD= 6.490
2 =3.245 cm
Using Pythagoras’ theorem on triangle BEF gives 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 triangle ADF (which equals triangle BCF)
=12(AD)(FG), where G is the midpoint of AD. Using Pythagoras’ theorem on triangle FGA gives:
FG=
15.02−1.802=14.89 cm Hence area of triangle ADF= 12(3.60)(14.89)
=26.80 cm2 Similarly, if H is the mid-point of AB, then
FH=
15.02−2.702=14.75 cm,
hence area of triangle ABF (which equals triangle CDF)
= 12(5.40)(14.75)=39.83 cm2
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Volumes and surface areas of common solids 161
Sec tion 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=21(volume of sphere)
=2
3πr3= 2 3π
5.0 2
3
=32.7 cm3 Total surface area
=curved surface area+area of circle
= 12(surface area of sphere)+πr2
= 12(4πr2)+πr2
=2πr2+πr2=3πr2=3π 5.0
2
2
=58.9 cm2
Problem 11. A rectangular piece of metal having dimensions 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=13(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 diameter 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=12cm=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 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 12 cm
Volume of cylinder=πr2h=π×62×15
=540πcm3
If 8% of metal is lost then 92% of 540π gives the volume of the new shape (shown in Fig. 20.7).
12 cm r h
Figure 20.7
Hence the volume of (hemisphere+cone)
=0.92×540πcm3, i.e. 12!4
3πr3
+13πr2h=0.92×540π Dividing throughout byπgives:
2
3r3+13r2h=0.92×540
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162 Engineering Mathematics
Sec tion 2
Since the diameter of the new shape is to be 12 cm, then radius r=6 cm,
hence 23(6)3+13(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/cm3means that 8.91 g of
cop-per 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=πr2or πd2
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 surmounted a hemispherical section of diameter 6 m, and on the other end a conical section of height 4 m and base diameter 6 m. Calculate the volume of the boiler and the total surface area
The boiler is shown in Fig. 20.8
P
Q
A B R
C
4 m I
3 m 8 m
6 m
Figure 20.8
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
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 Pythago-ras’ theorem on triangle ABC, 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
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Volumes and surface areas of common solids 163
Sec tion 2
Now try the following exercise
Exercise 76 Further problems on volumes