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Novestus kibet

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VOLUME AND SURFACE AREA OF A FRUSTRUM

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Novestus kibet

  1. 1. MATHEMATICS PRESENTATIONMATHEMATICS PRESENTATION 11 33 22 4455 66 77 88 99 Mr. Kibet Novestus
  2. 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. 3. CONE •Cone is a special solid with circular base base base Mr. Kibet Novestus
  4. 4. 7cm CONSTRUCTION OF A CONE. • Draw a circle of radius r ( 7cm) Mr. Kibet Novestus
  5. 5. A B Cut out a sector whose arc subtends an angle of 150o 150o 7cm Mr. Kibet Novestus
  6. 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. 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. 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. 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. 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. 11. •THE TOTAL SURFACE AREA OF A CLOSED CONE ∏ r l + ∏ r2 ∏ r2 is the area of the circular base Mr. Kibet.
  12. 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. 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. 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. 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. 16. Surface area = ∏ R L - ∏ r l 22 /7 x 15 x 362 + 152 – 22 /7 x 10 x 242 + 102 = 1021cm2 Mr. Kibet.
  17. 17. THE END Mr. Kibet.

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