2. TOPIC: SURFACE AREA OF A
SOLIDSUBTOPIC: SURFACE AREA OF A CONE /FRUSTRUM
PRIO KNOWLEDGE REQUIRED :-
Length of an arc (circles) form one.
Similarity (triangles) form two (previous topic)
• Pythagoras theorem form two.
Mr. Kibet. Novestus
3. CONE
•Cone is a special solid with circular base
base
base
Mr. Kibet Novestus
5. A
B
Cut out a sector whose arc subtends an angle
of 150o
150o
7cm
Mr. Kibet Novestus
6. 7cm
•Find the length of the remaining arc
AB
1500
2100
A
B
SOLnS
FORMULA:
l =
Ø 2∏ r
360
Where:-
l = length of the arc AB
Ø = 210O
r = 7cm
∏ = 22/7l
Therefore:
210 x 22 x 7 x 2
360 7
AB ( l ) = 25.67cm
= 25.67cm
Mr. Kibet Novestus
7. base
•Fold the sector to form an open cone
A
B
7CM
A B
7CM
Join point A to B as shown to
form a circular base
The circumference of the
circular base is equal to the
length of the arc AB
l
to
C = 25.67 cm
Length of the
arc AB
Mr. Kibet Novestus
8. •NOW WE CAN DEDUCE THE RADIUS OF THE CIRCULAR
BASE R AND THE HEIGHT h
r
A
B
h
• C = ∏ 2 R
• r = C
• 2 ∏
O
p
• OPA makes a right angled triangle as
shown
• OA is the hypotenuse which is the
radius of the initial circle (sector).
• Applying Pythagoras theorem
• h = 72
– 4.082
7cm
r = 25.67
2 ∏
r = 4.08cm
= 5.6 cm
h = 5.6 cmMr. Kibet.
9. • Area of the curved surface
A
B
7cm
210o
• Sector that forms the cone
BA
r
A
B
Area of sector above
210o
x 22 x 72
360 7
= 89.83cm2
The radius of the sector
becomes the slant height
of the cone
7cm
7cm
7cm
p1
p2
p3
The radius of the
circular base is
r = 4.08 cm
Area of sector above
∏ r l
22 x 4.08 x 7
7
= 89.76cm2
Mr. Kibet.
10. •THEREFORE THE AREA OF A CURVED
SURFACE OF A CONE
Is given by:
A = ∏rl
Where : r is the radius of the
base
l is the slant side
l
r
h
Mr. Kibet.
11. •THE TOTAL SURFACE AREA OF A CLOSED CONE
∏ r l + ∏ r2
∏ r2
is the area of the circular base
Mr. Kibet.
12. SURFACE AREA OF A FRUSTRUM
If a cone is cut through a plane parallel to the base and
the top part forms a smaller cone, the bottom part
forms a frustum
frustrum
Small
cone
formed
Mr. Kibet.
13. The surface area steps:
• Complete the cone as follows
Example:
12 cm
10 cm
15 cm
x cm
• let the height be x cm
Mr. Kibet.
14. This will form a smaller cone( 1 )and a larger cone
(2)
12 cm
15 cm
x cm
From the knowledge of similar triangles
There are two triangles A & B
15 cm
(12 – x )cm
A
X cm
10 cm
B
1
2
Mr. Kibet.
15. FROM THE ABOVE DIANGRAMS
x = x + 12
10 15
x = 24 cm
Surface area
of frustrum
=
Area of curved
surface of bigger
cone
-
Area of curved
surface of smaller
cone
Mr. Kibet.
16. Surface area
= ∏ R L - ∏ r l
22
/7 x 15 x 362
+ 152
– 22
/7 x 10 x 242
+ 102
= 1021cm2
Mr. Kibet.