Basic Engineering Mathematics, Fifth Edition

Volumes of common solids 245

5cm

5cm

D

C

E

A

B

Figure 27.10

The total surface area consists of a square base and 4
equal triangles.

Area of triangleADE

=

1

2

×base×perpendicular height

=

1

2

× 5 ×AC

The lengthAC may be calculated using Pythagoras′
theorem on triangle ABC,whereAB=12cm and
BC=^12 × 5 = 2 .5cm.

AC=

√

AB^2 +BC^2 =

√

122 + 2. 52 = 12 .26cm

Hence,

area of triangleADE=

1

2

× 5 × 12. 26 = 30 .65cm^2

Total surface area of pyramid=( 5 × 5 )+ 4 ( 30. 65 )

=147.6cm^2

Problem 11. A rectangular prism of metal having

dimensions of 5cm by 6cm by 18cm is melted

down and recast into a pyramid having a

rectangular base measuring 6cm by 10cm.

Calculate the perpendicular height of the pyramid,

assuming no waste of metal

Volume of rectangular prism= 5 × 6 × 18 =540cm^3

Volume of pyramid

=

1

3

×area of base×perpendicular height

Hence, 540 =
1
3

×( 6 × 10 )×h

from which, h=

3 × 540

6 × 10

=27cm

i.e. perpendicular height of pyramid=27cm

27.2.5 Cones

A cone is a circular-based pyramid. A cone of base

radiusr and perpendicular height h is shown in

Figure 27.11.

Volume=

1

3

×area of base×perpendicular height

h

r

l

Figure 27.11

i.e. Volume=^1

3

πr^2 h

Curved surface area=πrl

Total surface area=πrl+πr^2

Problem 12. Calculate the volume, in cubic

centimetres, of a cone of radius 30mm and

perpendicular height 80mm

Volume of cone=

1

3

πr^2 h=

1

3

×π× 302 × 80

= 75398. 2236 ...mm^3

1cm=10mm and

1cm^3 =10mm×10mm×10mm= 103 mm^3 ,or

1mm^3 = 10 −^3 cm^3

Hence, 75398. 2236 ...mm^3

= 75398. 2236 ...× 10 −^3 cm^3

i.e.,

volume=75.40cm^3