Name of Lecturer: Mr. J.Agius
Course: HVAC 2
Lesson 27
Chapter 2. More Areas & Volume
27.1 The Circumference and Area of an ellipse
Major axis
Area = ab
Minor axis
exactly
a
Circumference = (a + b) approximately
b
Example 1
Find the area and the approximate perimeter of an ellipse with a major axis of 200mm and a
minor axis of 150mm
Answer
Area = ab
=   100  75 = 23 600 mm2
Perimeter
= (a + b) =   (100 + 75) = 550 mm approximately
27.2 Frustum of a Cone
A frustum is the portion of a cone or pyramid between the base and a horizontal slice which removes
the pointed portion.
Volume =

1
h R 2  Rr  r 2
3

(h is the vertical height)
Curved surface area = l(R + r)
Total surface area = l(R + r) + R2 + r2
r
l
h
R
(l is the slant height)
2. More Areas & Volume
Page 1
Name of Lecturer: Mr. J.Agius
Course: HVAC 2
Example 2
The bowl shown in the following figure is made from sheet and has an open top. Calculate
the total cost of painting the vessel (inside and outside) at a cost of 1p per 10000 mm2.
 210
R 105
35
50 cm
50
l
 140
r 70
Answer
The second figure shows a half section of the bowl. Using Pythagoras’ theorem on the rightangled triangle
l 2 = 502 + 352
l = 61.0 mm
Now the required total surface area, i.e. inside and outside
= 2{Curved surface area} + 2(Base area)
= 2{l(R + r)} + 2(r2)
= 2{(61)(105 + 70)} + 2(702) = 97800 mm2
97800
 9.78 p
10000
At 1p per 10000 mm2 total cost =
27.3 Frustum of a Pyramid
Area of Top = a
Volume =

1
h A  Aa  a
3

Surface area = Sum of the areas of the
trapeziums forming the
sides plus the areas of the
top and base of the
frustum
h
Area of base = A
2. More Areas & Volume
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Name of Lecturer: Mr. J.Agius
Course: HVAC 2
Example 3
A casting has a length of 2m and its cross-section is a regular hexagon. The casting tapers
uniformly along its length, the hexagon having a side of 200 mm at one end and 100 mm at
the other. Calculate the volume of the casting.
Answer
In the following figure, the area of the hexagon of 200mm side
200
= 6  Area ABO
A
In AOC, AOC = 30o, AC = 100 mm
AC
and
tan AOC 
OC
C
B
60o
O
AC
100
= 173 mm

tan AOC tan 30 o
Therefore
OC 
Therefore
Area of ABO = ½  200  173
Therefore
Area of hexagon = 6  ½  200  173 = 103 800mm2
The area of the hexagon of 100 mm side can be found in the same way. It is 25950mm2.
The casting is a frustum of a pyramid with A = 103800, a = 25950 and h = 2000
Therefore


Volume = 13 h A  Aa  a  13  2000  103800  103800  25950  25950
= 1.21  10 mm
8
2. More Areas & Volume
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