Lesson Notes By Weeks and Term - Senior Secondary School 1
MENSURATION OF SOLID SHAPES
MENSURATION OF SOLID SHAPES
SUBJECT: MATHEMATICS
CLASS: SS 1
DATE:
TERM: 3rd TERM
REFERENCE BOOKS
TOPIC: MENSURATION OF SOLID SHAPES Mensuration: The concept of B – D a shape cube, cuboids, cylinder, triangular prism, cone, rectangular based pyramid, total surface area of cone, cylinder and their volumes. MENSURATION OF SOLID SHAPES Properties of solid shapes A cube has the following properties. A cuboid has the following properties. A triangular prism has the following properties: Properties: A cone has the following properties: A rectangular based pyramid has the following properties: EVALUATION (b) Write down the properties of each of the solid shapes you mentioned in 1a above (c) List one real object for each of the solid shape mentioned in (1a) above Surface Area and Volume of Common Solid shapes A prism is a solid which has uniform cross section. Cubes, cuboids, and cylinders are examples of prisms. In general, Volume of prism = area of uniform cross section X perpendicular height =area of base x height Cube Cuboids Cylinder Triangular prism Cube Volume = l3 Surface area = 6l2 Cuboid Volume =lbh Surface area = 2 (lb + lh + bh) Cylinder Volume = πr2 h Curved surface area = 2πrh Total surface area = 2πrh + 2π r2 = 2πr ( h + r) Examples a) s In the figure above, PQRS is a trapezium b) Solutions 1a.) Volume of prisms = area of uniform cross section X perpendicular height = area of base X length of the prism Area of PQRS = ½ ( 7 + 4) X /QR/ cm2 4cm Since /QR/ = / X S/ Consider triangle P X S / PX /2 + /XS/2= 52 3 2+ /XS/2 = 25 9 + / XS/2 = 25 /XS/2 = 25 – 9 /XS/2 = 16 /XS/ = √16cm = 4cm Thus /XS/ = /QR/ = 4cm Area of PQRS = ½ x ( 7 + 4) x /QR/ cm2 = ½ x 11 x 4 cm2 = 22cm2 Hence, Volume of Prism = area of uniform cross section X length of prism = 22cm2 x 12cm = 264cm3 (b) volume of given cylinder = πr2h from the given cylinder, r = d/2 = 14/2 cm = 7cm h = 4cm volume of given cylinder = π x (7) 2 x 4cm3 22/7 x 49 x 4cm = 22 x 28cm3 = 616cm3 2a) To calculate the total surface area of the solid shapes in 1a and b above. 2b) Total surface area of the given cylinder = 2πrh + 2πr2 = 2πr ( h + r) = 2 x 22/7 x 7 ( 4+ 7 ) cm2 = 44 x 11cm2 = 484 cm2 EVALUATION 1a. A rectangular tank is 76cm long, 50cm wide and 40 cm high. How many litres of water can it hold? Surface area of a Cone A sector of a circle can be bent to form the curved surface of an open cone. In the figure below, the sector OA x B is of radius l and arc A X B subtends angle θ at O. This sector is bent to form a cone of base radius r and slant height o The following points should be noted Curved surface area of cone =θ x πl2 …………..0 360 Also, θ x 2πl = 2 πr 360 Divide both sides by 2π θ x 2πl = 2 πr 360 2π 2π θ x l =r 360 divide both sides by l θ = r 360 l substitute r/l for θ in equation i) above: 360 Curve surface area of cone =r x πl2 l = Πrl Hence, Total surface area = curved surface area of a cone + area of circular base = πr l +π r2 = πr ( l + r) Examples A paper cone has a diameter of 8cm and a height of 3cm a). Make a sketch of the cone and hence use Pythagoras theorem to calculate its slant height. b). Calculate the curved surface area of the cone in terms of π c ) If the cone is cut and opened out into the sector of a circle. What is the angle of the sector? Solutions. From the given information about the paper cone, Diameter = 8cm :. Radius = diameter 2 = 8cm = 4cm 2 using Pythagoras theorem in the right angled triangle OBC l2 = /OB/2 + /BC/ 2 l2 = 32 + 42 l2 = 9 + 16 l2 = 25 Take square root of both sides √ l2 =√ 25 l = 5cm :.the slant height of the paper cone is 5cm = π x 4 x 5 cm = 20 πcm2 c) If the paper cone is cut and opened out into the sector of a circle as shown in the figure above, then area of sector of circle = curved surface area of the cone i.eθx π x (5) 2 = 20 x π 360 5 θx π x 25 = 20 x π 360 12 5 θ = 72 x 20 Divide both sides by 5 5 θ =72 x 20 5 5 θ = 72 x 4 θ = 288o EVALUATION Volume of Pyramids and Volume of cone In general, Volume = 1/3 x base area x height Square based pyramid rectangular based pyramid Cone :. Volume of square based pyramid = 1/3 x b2 x h volume of rectangular based pyramid = 1/3 x l x b x h volume of cone = 1/3 x Πr2 x h Examples 1.A pyramid 8cm high stands on a rectangular base 6cm by 4cm.Calculate the volume of the pyramid. Solutions 1) Volume of a rectangular based pyramid = 1/3 x l x b x h = 1/3 x 6 x 4 x 8 cm3 = 8 x8 cm3 = 64cm3 2) Considering the square base ABCD /DB/ 2= /DC/ 2 + /CB/2 Pythagoras rule: /DB/2 = 42 + 42 /B/2 = 16 + 16. :. √/DB/ = √ 32 /DB/ = 4 √2 cm but / EB/ = ½ /DB/ Since t is the midpoint of / DB/ Then /EB/ = ½ X 4 X √ 2 = 2 √2 cm. Now Consider right angle OEB OE 2 + EB 2 = ( OB)2 OE 2+ ( 2√2) 2 = ( 6) 2 OE 2 + 4 x 2 = 36 OE 2 + 8 = 36 OE 2 = 36 – 8 OE2 = 28 OE = √28 OE = √4 x 7 OE = 2 x √ 7 cm OE = 2 √7cm But OE =height of the pyramid = 2√7 :.volume of square of based pyramid = 1/3 x b2 x h 1/3 x 42 x 2 x √7 cm3 1/3 x 16 x 2 x √7 cm3 = 32 x √7 cm3 3 32 x 2.646cmm3 3 = 32 x.0.882cm3 = 28. 224cm3 = 28.2cm3 to 1 d.p. 3) Since Diameter = 14cm Radius = diameter 2 = 14 cm.=7cm 2 :. Volume of cone = 1/3 πr2 h = 1/3 x 22/7 x ( 7 ) 2 x 24 = 1/3 x 22/7 x 49 x 24 cm3 = 22 x 56cm3 = 1232 cm3 EVALUATION GENERAL EVALUATION READING ASSIGNMENT NGM SS Bk 1 pg 166- 170 Ex 15a Nos 1 (d), 1(f), 2(b) and 29c) pages 168 -169. WEEKEND ASSIGNMENT THEORY
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