Basic Engineering Mathematics, Fifth Edition

254 Basic Engineering Mathematics

Problem 26. 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 8.0mand

4 .6m, respectively, and the perpendicular height

between its ends is 3.6m

The frustum is shown shaded in Figure 27.23(a) as part

of a complete pyramid. A section perpendicular to the

base through the vertex is shown in Figure 27.23(b).

8.0m

4.6cm

8.0m

(a)

B

2.3m

1.7m2.3m

(b)

4.0m

2.3m 3.6m

C

A

E

G D

H F

4.6cm

Figure 27.23

By similar triangles

CG

BG

=

BH

AH

From which, height

CG=BG

(

BH

AH

)

=

( 2. 3 )( 3. 6 )

1. 7

= 4 .87m

Height of complete pyramid= 3. 6 + 4. 87 = 8 .47m

Volume of large pyramid=

1

3

( 8. 0 )^2 ( 8. 47 )

= 180 .69m^3

Volume of small pyramid cut off=

1

3

( 4. 6 )^2 ( 4. 87 )

= 34 .35m^3

Hence,volume of storage hopper= 180. 69 − 34. 35

=146.3m^3

Problem 27. Determine the lateral surface area of

the storage hopper in Problem 26

The lateral surface area of the storage hopper consists

of four equal trapeziums. From Figure 27.24,

Area of trapeziumPR S U=

1

2

(PR+SU)(QT)

4.6 m

8.0 m

P

U

T

S

O

8.0 m

Q

R

4.6 m

Figure 27.24

OT= 1 .7m (same as AH in Figure 27.23(b) and

OQ= 3 .6m. By Pythagoras’ theorem,

QT=

√(

OQ^2 +OT^2

)

=

√[

3. 62 + 1. 72

]

= 3 .98m

Area of trapeziumPRSU

=

1

2

( 4. 6 + 8. 0 )( 3. 98 )= 25 .07m^2

Lateral surface area of hopper= 4 ( 25. 07 )

=100.3m^2

Problem 28. A lampshade is in the shape of a

frustum of a cone. The vertical height of the shade

is 25.0cm and the diameters of the ends are 20.0cm

and 10.0cm, 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 252. Since the diameters of

the ends of the frustum are 20.0cm and 10.0cm, from

Figure 27.25,

r 5 5.0cm

I

h^5

25.0cm

R 5 10.0cm

5.0cm

Figure 27.25