Basic Engineering Mathematics, Fifth Edition

242 Basic Engineering Mathematics

Outer diameter,D=25cm= 0 .25m and inner diame-

ter,d=12cm= 0 .12m.

Area of cross-section of copper

=

πD^2

4

−

πd^2

4

=

π( 0. 25 )^2

4

−

π( 0. 12 )^2

4

= 0. 0491 − 0. 0113 = 0 .0378m^2

Hence,volume of copper

=(cross-sectional area)×length of pipe

= 0. 0378 × 2. 5 =0.0945m^3

27.2.3 More prisms

A right-angled triangular prism is shown in Figure 27.4

with dimensionsb,handl.

I

b

h

Figure 27.4

Volume=

1

2

bhl

and

surface area=area of each end

+area of three sides

Notice that the volume is given by the area of the end

(i.e. area of triangle=^12 bh) multiplied by the lengthl.

In fact, the volume of any shaped prism is given by the

area of an end multiplied by the length.

Problem 6. Determine the volume (in cm^3 )ofthe

shape shown in Figure 27.5

12mm

16mm

40mm

Figure 27.5

The solid shown in Figure 27.5 is a triangular prism.

The volumeVof any prism is given byV=Ah,where

Ais the cross-sectional area andhis the perpendicular

height. Hence,

volume=

1

2

× 16 × 12 × 40 =3840mm^3

=3.840cm^3

(since 1cm^3 =1000mm^3 )

Problem 7. Calculate the volume of the

right-angled triangular prism shown in Figure 27.6.

Also, determine its total surface area

6cm

40cm

A

B C

8cm

Figure 27.6

Volume of right-angled triangular prism

=

1

2

bhl=

1

2

× 8 × 6 × 40

i.e. volume=960cm^3