248 Basic Engineering Mathematics

3m r

r (^5) 8cm
12cm
Figure 27.14
(a) The section of wood is a prism whose end com-
prises a rectangle and a semicircle. Since the
radius of the semicircle is 8cm, the diameter is
16cm. Hence, the rectangle has dimensions 12cm
by 16cm.
Area of end=( 12 × 16 )+
1
2
π 82 = 292 .5cm^2
Volume of wooden section
=area of end×perpendicular height
= 292. 5 × 300 =87750cm^3

87750
106
m^3 ,since 1m^3 = 106 cm^3
=0.08775m^3
(b) The total surface area comprises the two ends
(each of area 292.5cm^2 ), three rectangles and a
curved surface (which is half a cylinder). Hence,
total surface area
=( 2 × 292. 5 )+ 2 ( 12 × 300 )
+( 16 × 300 )+
1
2
( 2 π× 8 × 300 )
= 585 + 7200 + 4800 + 2400 π
=20125cm^2 or 2.0125m^2
Problem 17. A pyramid has a rectangular base
3 .60cm by 5.40cm. Determine the volume and total
surface area of the pyramid if each of its sloping
edges is 15.0cm
The pyramid is shown in Figure 27.15. To calculate the
volume of the pyramid, the perpendicular heightEFis
required. DiagonalBDis calculated using Pythagoras’
theorem,
i.e. BD=
√[
3. 602 + 5. 402
]
= 6 .490cm
Hence, EB=
1
2
BD=
6. 490
2
= 3 .245cm
C
F
D
G
A
E
H
B
15.0cm15.0cm
15.0cm15.0cm
5.40cm
3.60cm
Figure 27.15
Using Pythagoras’ theorem on triangleBEFgives
BF^2 =EB^2 +EF^2
from which EF=
√(
BF^2 −EB^2
)

√
15. 02 − 3. 2452 = 14 .64cm
Volume of pyramid

1
3
(area of base)(perpendicular height)

1
3
( 3. 60 × 5. 40 )( 14. 64 )=94.87cm^3
Area of triangle ADF (which equals triangleBCF)
=^12 (AD)(FG),whereG is the midpoint of AD.
Using Pythagoras’ theorem on triangleFGAgives
FG=
√[
15. 02 − 1. 802
]
= 14 .89cm
Hence, area of triangleADF=
1
2
( 3. 60 )( 14. 89 )
= 26 .80cm^2
Similarly, ifHis the mid-point ofAB,
FH=
√
15. 02 − 2. 702 = 14 .75cm
Hence, area of triangle ABF (which equals triangle
CDF)=
1
2
( 5. 40 )( 14. 75 )= 39 .83cm^2