novestus kibet
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MATHEMATICS PRESENTATIONMATHEMATICS PRESENTATION
11 3322
4455 66 77 88 99
Mr. Kibet Novestus
TOPIC: SURFACE AREA OF A SOLIDSUB TOPIC: 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
CONE•Cone is a special solid with circular base
base
base
Mr. Kibet Novestus
7cm
CONSTRUCTION OF A CONE.
• Draw a circle of radius r ( 7cm)
Mr. Kibet Novestus
A
B
Cut out a sector whose arc subtends an angle of 150o
150o7cm
Mr. Kibet Novestus
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
base
•Fold the sector to form an open cone
A
B7CM
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
•NOW WE CAN DEDUCE THE RADIUS OF THE CIRCULAR BASE R AND THE HEIGHT h
rA
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.
• Area of the curved surface
A
B
7cm210o
• Sector that forms the cone
BA
r
AB
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.
•THEREFORE THE AREA OF A CURVED SURFACE OF A CONE
Is given by: A = ∏rlWhere : r is the radius of the base
l is the slant side
l
r
h
Mr. Kibet.
•THE TOTAL SURFACE AREA OF A CLOSED CONE
∏ r l + ∏ r2
∏ r2 is the area of the circular base
Mr. Kibet.
SURFACE AREA OF A FRUSTRUMIf 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
frustrumSmall cone
formedMr. Kibet.
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.
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.
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.
Surface area
= ∏ R L - ∏ r l
22/7 x 15 x 362 + 152 – 22/7 x 10 x 242 + 102
= 1021cm2
Mr. Kibet.
THE END
Mr. Kibet.
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