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Volume of a pyramid 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 Egyptian Pyramid These are pyramids V Vertex Slant edges height D C A B x x x cube Three congruent pyramids x 1 Volume of the pyramid = x3 x 3 1 2 x = x x 3 1 = base area height 3 For any pyramid, 1 Volume of pyramid = base area height 3 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 Solution : V VF2 = (152 - 92 ) cm2 15 cm VF = 15 - 9 cm 2 2 E F = 12 cm D C 9 cm F A Volume of the pyramid E 9 cm B 15 cm 1 = base area height 3 1 =( 192 12) cm3 3 = 768 cm3 B Pyramid B A frustum Pyramid A = - Volume of the frustum = Volume of Pyramid A - Volume of Pyramid B 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 Solution : 1 Volume of VEFGH = ( ( 8 8 ) 6) cm3 6 cm 3 H G = 128 cm3 E F 12 cm 1 Volume of VABCD = ( ( 16 16 ) 12) cm3 3 D C = 1024 cm3 Volume of frustum ABCDEFGH A B = (1024 - 128 ) cm3 = 896 cm3 C V V D C D C V V A B A B V Total surface area of pyramid VABCD = + + + + lateral faces Base Total surface area of a pyramid = Base area + The sum of of the area of all lateral faces 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 Solution : Total surface area of pyramid VABCD = Area of ABCD + (Area VAB + Area VDC + D C Area VBC + Area VAD ) A B = 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 ? l l 2πr r Cut here Curved surface area l Remark : Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l θ r After cutting the cone, Curved surface area = Area of the sector Curved surface area = 1/2 ( l ) ( 2π r ) =πrl Curved surface area = πr l Volume of a cone r h 1 h 3 r Volume of a cone = 1 πr2 h 3 How to calculate total surface area of a cone? l l + r r Total surface area =πr2 + πr l Examples 1 a) If h = 12cm, r= 5 cm, what is the volume? Answer: 1 Volume = 3 πr2h = 1 π (52) ( 12) 3 = 314 cm3 b) what is the total surface area? Based Area = π52 = 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 1 1 = 3 πR3 - π r3 3 1 = 3 π( R3 - r3 ) Start Now Exit 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? A. 2 cm B. 2 3cm Answer C. 6cm Answer is C Help D. 12cm E. 36cm To Q2 Q2 A In the figure, the volumes of the cone AXY and ABC are 16 cm3 and 54 cm3 X Y respectively, AX : XB = A. 2 : 1 B C B. 2 : 3 Answer C. 8 : 19 Answer Help D. 8 :27 is A E. 3 16 : 3 38 To Q3 Q3 V In the figure, VABCD is a right pyramid with a rectangular base. If D AB=18cm, BC=24cm and CV=25cm, find C M A B a) the height (VM) of the pyramid, Answer b) volume of the a) 20cm Help pyramid. b) 2880cm3 To Q4 Q4 A The figures shows a right 50cm circular cone ABC. If AD= 48cm and AC= 50cm, find B C (a) the base radius (r) of the cone, (b) the volume of the cone. (Take = 22 ) 7 Answer Help Let V is the volume of the pyramid and y be the length of a side of base 1 V = 3 base area height what is the length of a 1 96 = 3 y2 8 side of the base? 288 = 8y2 Back to Q1 36 = y2 y=6 Therefore, the length of a To Q2 side of base is 6 cm Hints: Using the concept AX : XB = ? of RATIOS AX 16 A ( AB )3 = 54 ( AX )3 = 278 AB X Y AX 2 AB = 3 B C AB = AX + XB and AX = 2, AB = 3 Back to Q2 3 = 2 + XB XB = 1 Therefore, AX : XB = 2 : 1 To Q3 a) the height (VM) of the b) volume of the pyramid pyramid. AC2 =182 + 242 Volume of the pyramid is: AC2 = 900 1 ×base area ×height AC = 30cm 3 MC = 1 AC =15cm = 1 ×18 ×24 ×20 2 3 252 = VM2 + MC2 = 2880cm3 625 = VM2 + 152 Therefore, the volume of 625 - 225 = VM2 the pyramid is 2880cm3 VM2 = 400 Back to Q3 VM = 20cm Therefore, the height (VM) of the pyramid is 20 cm To Q4 (a) the base radius (r) (b) the volume of the cone The volume (V) of cone is: The radius is r, therefore: 1 502 = 482 + r2 V = 3 r2 h 1 22 2500 = 2304 + r2 = 3 7 142 48 196 = r2 = 704 cm3 r = 14 The volume is 704 cm3 The radius is 14cm. A (Take = 22/7) Back to Q4 50cm B C

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posted: | 4/13/2011 |

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