Title : Usage : Cone and Pyramid Subject : Mathematics Target Audience : Form 3 Usage : Lecturing

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

Pyramid 1 Egyptian Pyramid

These are pyramids

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

Volume of pyramid A x cube Three congruent pyramids

x For any pyramid, Volume of pyramid =  base area  height Volume of the pyramid = x3 x x2 x = =  base area  height For any pyramid, Volume of pyramid =  base area  height

Example 1 cm Volume of the pyramid 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 A C D B 15 cm 9 cm E F V E F 15 cm 9 cm Solution : VF2 = (152 - 92 ) cm2 VF = cm = 12 cm Volume of the pyramid =  base area  height = (  192  12) cm3 = 768 cm3

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

Example 2 Solution : = ( = ( V A E D C B H G F  ( 8  8 )  6) cm3 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 A E D C B H G F 6 cm 12 cm Solution : = (  ( 8  8 )  6) cm3 Volume of VEFGH = 128 cm3 = (  ( 16  16 )  12) cm3 Volume of VABCD = 1024 cm3 Volume of frustum ABCDEFGH = (1024 - 128 ) cm3 = 896 cm3

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

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

Example 3 Solution : V D C B A Total surface area of pyramid VABCD 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. 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

How to generate a cone? …... …...

How to calculate the curved surface area ? Cut here

Curved surface area = πr l θ 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

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

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

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

b) what is the total surface area? = π52 Based Area = 25πcm2 Slant height = 122 + 5 2 = 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)

Volume of Frustum

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

Exercises Start Now Exit

Q1 8cm 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? A. 2 cm B. 2 3cm C. 6cm D. 12cm E. 36cm Answer Answer is C Help To Q2

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

V D C A B M 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 a) 20cm b) 2880cm3 Help To Q4

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

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

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

Answer for Q3 AC2 =182 + 242 Volume of the pyramid is: AC2 = 900 1 a) the height (VM) of the pyramid b) volume of the pyramid. AC2 =182 + 242 Volume of the pyramid is: AC2 = 900 ×base area ×height 3 1 AC = 30cm MC = AC =15cm 2 1 = ×18 ×24 ×20 1 3 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

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: V =  r2 h 3 1 502 = 482 + r2 =   142  48 3 1 22 7 2500 = 2304 + r2 196 = r2 = 704 cm3 r = 14 The volume is 704 cm3 A The radius is 14cm. C B 50cm 48cm (Take  = 22/7) Back to Q4

End of lesson!