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Cone and Pyramid - Mathematics Course for Form 3 students

This course teaches students about the concepts of cones and pyramids in mathematics, including their volume, surface area, and related calculations.

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Cone and Pyramid - Mathematics Course for Form 3 students

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  1. Course Title: EDD 5165A Information Technology in Education Instructor: Dr. LEE Fong Lok Group Members: Tiu Chui Yee (99039490) Li Chun Lan (99040120) Ching Yuk Shan (99043270)

  2. Title : Cone and Pyramid Subject : Mathematics Target Audience : Form 3 and Band 2 students Usage : Lecturing

  3. Prerequisite knowledge : 1. Pythagoras’ Theorem. 2. Area of some plane figures e.g. square, rectangle, triangle, circle, sector 3. Ratio and proportion Objectives : Let the students know and apply the mensuration concepts

  4. Distribution of workload Tiu Chui Yee (99039490): Content of Pyramid Li Chun Lan (99040120): Content of Cone Ching Yuk Shan, Maggie (99043270): Quiz

  5. 1 Pyramid Egyptian Pyramid

  6. These are pyramids

  7. Vertex V   height     D  C    A B Slant edges

  8. A Volume of pyramid x x x cube Three congruent pyramids

  9. x Volume of the pyramid= x3 x x2x x  base area  height = Volume of pyramid =  base area  height = For any pyramid,

  10. Example 1 The figure shows a pyramid with a rectangular base ABCD of area 192 cm2, VE = 15 cm and EF = 9 cm, find the volume of the pyramid. V V 15 cm VF= cm E F 9 cm D C F A E B 9 cm 15 cm =  base area  height = (  192  12) cm3 Solution : VF2=(152 - 92 ) cm2 = 12 cm Volume of the pyramid = 768 cm3

  11. B Volume of a Frustum of a Pyramid Pyramid A Volume of the frustum = Volume of Pyramid A - Volume of Pyramid B  Pyramid B  A frustum - =

  12. Example 2 The base ABCD and upper face EFGH of the frustum are squares of side 16 cm and 8 cm respectively. Find the volume of the frustum ABCDEFGH. V = (  ( 8  8 )  6) cm3 6 cm H G E F 12 cm = (  ( 16  16 )  12) cm3 D C A B Solution : Volume of VEFGH = 128 cm3 Volume of VABCD = 1024 cm3 Volume of frustum ABCDEFGH = (1024 - 128 ) cm3 = 896 cm3

  13. C Total surface area of a pyramid V C D V V V D C A B A B V

  14. + + + + Total surface area of a pyramid Base area The sum of of the area of all lateral faces = + Total surface area of pyramid VABCD = lateral faces Base

  15. Example 3 The figure shows a pyramid with a rectangular base ABCD of area 48 cm2. Given that area of  VAB = 40 cm2 , area of  VBC =30 cm2, find the total surface area of the pyramid. V D C B A Solution : Total surface area of pyramid VABCD = Area of ABCD + (Area VAB + Area VDC + Area VBC + Area VAD ) =Area of ABCD + (Area VAB  2) + (Area VBC  2) =48 cm2 + ( (40  2) + (30  2))cm2 = 188 cm2

  16. How to generate a cone? …... …...

  17. l l 2πr r How to calculate the curved surface area ? Cut here

  18. l θ r Curved surface area Remark : Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l After cutting the cone, Curved surface area = Area of the sector Curved surface area = 1/2 ( l ) ( 2π r ) = π r l Curved surface area = πr l

  19. r h h 1 r 3 Volume of a cone = πr2 h Volume of a cone

  20. l r l r How to calculate total surface area of a cone? + Total surface area =πr2 + πr l

  21. If h = 12cm, r= 5 cm, what is the volume? Answer: Volume = πr2h • = π (52) ( 12) • = 314 cm3 1 1 3 3 Examples 1 a)

  22. = 122 + 5 2 b) what is the total surface area? = π52 BasedArea = 25πcm2 Slant height = 13 cm Curved surface area = π(5) ( 13) = 65π cm2 Total surface area = based area + curved surface area = 25π+65π= 90π = 282.6cm2 (corr.to 1 dec.place)

  23. Volume of Frustum

  24. r R = πR3 - π r3  = - = π( R3 - r3 ) Volume of Frustum Volume of frustum = volume of big cone - volume of small cone

  25. Exercises Start Now Exit

  26. 8cm A. 2 cm B. 2 3cm C. 6cm D. 12cm E. 36cm Answer is C Q1 The volume of a pyramid of square base is 96 cm3. If its height is 8 cm, what is the length of a side of the base? Answer Help To Q2

  27. A X Y A. 2 : 1 B. 2 : 3 C. 8 : 19 D. 8 :27 E. 3 16 : 3 38 B C Answer is A Q2 In the figure, the volumes of the cone AXY and ABC are 16 cm3 and 54 cm3 respectively, AX : XB = Answer Help To Q3

  28. V D C M A B a) 20cm b) 2880cm3 Q3 In the figure, VABCD is a right pyramid with a rectangular base. If AB=18cm, BC=24cm and CV=25cm, find a) the height (VM) of the pyramid, b) volume of the pyramid. Answer Help To Q4

  29. A 50cm 48cm B C (a) the base radius (r) of the cone, (b) the volume of the cone. (Take  = ) Answer 22 7 a) 14cm b) 704cm3 Q4 The figures shows a right circular cone ABC. If AD= 48cm and AC= 50cm, find Help

  30. 1 V =  base area  height 3 1 96 =  y2  8 3 8cm Answer for Q1 Let V is the volume of the pyramid and y be the length of a side of base what is the length of a side of the base? 288 = 8y2 Back to Q1 36 = y2 y = 6 To Q2 Therefore, the length of a side of base is 6 cm

  31. AX AX 16 ( )3= AB AB 54 8 ( )3 = 27 X Y AX 2 = AB 3 B C Answer for Q2 Hints: Using the concept of RATIOS AX : XB = ? A AB = AX + XB and AX = 2, AB = 3 Back to Q2 3 = 2 + XB XB = 1 Therefore, AX : XB = 2 : 1 To Q3

  32. ×base area ×height 1 1 MC = AC =15cm = ×18 ×24 ×20 2 3 1 3 a) the height (VM) of the pyramid b) volume of the pyramid. AC2 =182 + 242 Volume of the pyramid is: AC2 = 900 AC = 30cm Answer for Q3 252 = VM2 + MC2 = 2880cm3 625 = VM2 + 152 Therefore, the volume of the pyramid is 2880cm3 625 - 225 = VM2 VM2 = 400 Back to Q3 VM = 20cm Therefore, the height (VM) of the pyramid is 20 cm To Q4

  33. 1 1 V =  r2 h 3 3 22 =   142  48 7 48cm 50cm B C Answer for Q4 (a) the base radius (r) (b) the volume of the cone The volume (V) of cone is: The radius is r, therefore: 502 = 482 + r2 2500 = 2304 + r2 196 = r2 = 704 cm3 r = 14 The volume is 704 cm3 A The radius is 14cm. (Take  = 22/7) Back to Q4

  34. End of lesson!

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