Oasis School Mathematics-10 97 OP is the height. ABCD is its base. Draw OQ ^ BC and join PQ. Base area = 16 × 16 cm2 = 256 cm2 In ∆OPQ right-angled at P, OQ = OP² ² + PQ = 6 8 ² ² + [∵ PQ = 1 2 BC] = 100 = 10 cm. So, total surface area of the given pyramid = Area of the base + Area of 4 triangular faces = 256 cm2 + 4 × ( 1 2 OQ × DC) = 256 cm2 + 4 × 1 2 × 10 × 16 cm2 = 256 cm2 + 320 cm2 = 576 cm2 Again, volume of the pyramid = 1 3 × area of base×height = 1 3 × 256 × 6 cm2 = 512 cm3 example 2 Find the lateral surface area and total surface area of the given square-based pyramid. Solution: Here, O is the vertex and square ABCD is the base of the pyramid. From O, draw OF perpendicular to the base and OE^BC. Join EF. Here, OE = 25 cm, OF = 24 cm Using the Pythagoras Theorem, FE = OE² – OF² = ( ) 25 ² – ( ) 24 ² = 625 – 576 = 49 = 7 cm ∴ DC = 2 FE = 2 × 7cm = 14 cm ∴ DC = BC = 14 cm Now, area of ∆OBC = 1 2 BC × OE = 1 2 × 14 × 25 = 175 cm² We have, lateral surface area of pyramid (LSA) = 4∆OBC = 4 × 175 cm² = 700 cm² 25 cm 24 cm O F A D C E B In the given square base pyramid, a = 16 cm, h = 6cm We have, l² = h² + a² 4 = 6² + 16² 4 = 36 + 64 = 100 l = 100 = 10cm. we have, TSA = a² + 2al =16² + 2 × 16 × 6 = 256 + 320 = 576cm². we have, v = 1 3 a²h = 1 3 16² × 6 = 1 3 × 256 × 6 = 512cm³ Alternative method Here, l = 25cm, h = 24 cm We have, l 2 = 242 + a2 4 or, 252 = 2a2 + a2 4 or, a2 4 = 49 or, a2 = 49 × 4 ∴ a = 14 cm We have, LSA = 2al = 2×14×25cm2 = 700 cm2 TSA = 2al + a2 = 700 + (14)2 = 700 + 196 = 896 cm2 Alternative method
98 Oasis School Mathematics-10 Again, Total surface area = LSA + area of base = 700 cm² + (14)² cm² = (700 + 196) cm² = 896 cm² example 3 In the given square-based pyramid, DC = 18 cm, tSA = 864 cm². Find the height of the pyramid. Solution: In the given square-based pyramid, DC = BC = 18 cm. Draw OF perpendicular to the base and OE ^ BC. Join FE. Area of base = (18)² = 324 cm² Total surface area (TSA) = 864 cm² We have, TSA = LSA + area of base 864 = 4 × ∆OBC + 324 or, 864 – 324 = 4 ∆OBC or, 540 = 4 ∆OBC or, 540 4 = ∆OBC or, 135 = ∆OBC Area of ∆OBC = 135 cm² or, 1 2 BC × OE = 135 or, 1 2 ×18 × OE = 135 or, 9 OE = 135 or, OE = 15 cm Again, using Pythagoras Theorem, OF = OE² – EF² = ( ) 15 ² – ( ) 9 ² = 225 – 81 = 144 = 12 cm ∴ Height of the pyramid = 12 cm. example 4 Slant height of a square-based pyramid is 13 cm and the total surface area is 360 cm². Find the length of the side of the square base. Solution: O F A B D C E Here, a = 18 cm We have, TSA = 2al + a2 or, 864 = 2 × 18l + (18)2 or, 864 = 36l + 324 or, 36l = 864–324 or, l = 540 36 ∴ l = 15 cm Again, l 2 = h2 + a2 4 or, (15)2 = h2 + (18)2 4 or, 225 = h2 + 81 or, h2 = 225–81 or, h2 = 144 or h = 12 cm. Alternative method 13 cm O B E D C A
Oasis School Mathematics-10 99 Let ABCD be the base of the pyramid. O its vertex, OE its slant height. Let, each side of the square base = a Here, area of ∆OBC = 1 2 BC × OE = 1 2 a × 13 LSA = 4 ∆OBC = 4 × 13 2 a cm² = 26 a cm² Again, TSA = LSA + area of base 360 = 26 a + a² or, a² + 26a – 360 = 0 or, a² + (36 – 10) a – 360 = 0 or, a² + 36 a – 10 a – 360 = 0 or, a(a + 36) – 10(a + 36) = 0 or, (a + 36) (a – 10) = 0 Either, a + 36 = 0 ⇒ a = – 36 (not possible) Or, a – 10 = 0 ⇒ a = 10 cm ∴ Length of each side of base = 10 cm example 5 The given figure is a square based pyramid. (a) Which part of the given object represents the base? (b) What do the part OE and OF represent? (c) If the area of ∆OBC = 135cm² and EF = 9cm, find the length of OF and OE. (d) Find the volume of the above pyramid. Solution: (a) Square ABCD is the base of given object. (b) OE represents the height of the pyramid and OF represents slant height of the pyramid or height of the triangular face OE is denoted by ‘h’ and Of is represented by ‘l’. (c) Given, EF = 9cm. then AB = 2EF = 2 × 9cm = 18cm So, a side of the base (a) = BC = AB = 18cm. We have, Area of triangular face = 1 2 al or, 135 = 1 2 × 18 × l Alternative method Here, l = 13 cm, TSA = 360 cm2 We have, TSA = a2 + 2al 360 = a2 + 2a × 13 or, a2 + 26a – 360 = 0 or, a2 + (36–10)a–360 = 0 or, a2 + 36a–10a–360 = 0 or, a(a+36)–10(a+36) = 0 or, (a+36) (a-10) = 0 Either, a + 36 = 0 a = -36 (not possible) or, a–10 = 0 ⇒ a = 10cm. • A D C F B O
100 Oasis School Mathematics-10 or, 135 = 9l or, l = 135 9 = 15cm So, OF = 15cm Now, using the relation, l² = h² + a² 4 or, 15² = h² + 18² 4 or, 225 = h² + 324 4 or, 225 = h² + 81 or, h² = 225 – 81 or, h² = 144 or, h = 144 ∴ h = 12cm (d) We have, Volume of pyramid = 1 3 a²h = 1 3 (18)² × 12 cm³ = 1 3 × 324 × 12 cm³ = 324 × 4cm³ = 1296 cm³ exercise 5.3 1. (a) What is the formula of area of the the base of the square-based pyramid? (b) What is the formula of the area of the base of a pyramid having an equilateral triangle on its base? (c) If the area of the base and height of the pyramid are given, write the formula to find the volume of pyramid. (d) If 'a' be a side of the base of a square-based pyramid and 'h' its height, write the formula for the volume of the pyramid in terms of 'a' and 'h'. (e) If 'a' be a side of the base of square based pyramid and 'l' the slant height, write the formula for the LSA and TSA of the pyramid. 2. (a) Find the volume of a square-based pyramid whose area of the base is 100 cm² and the height is 12 cm. (b) Find the volume of a square-based pyramid whose height is 10 cm and the area of the base 60 cm².
Oasis School Mathematics-10 101 3. (a) Find the lateral surface area of a square-based pyramid in which the area of a triangular face is 36 cm². (b) Find the lateral surface area of a triangular-based pyramid in which the area of a triangular face is 40cm2 . (c) Find the total surface area of a square-based pyramid whose area of the base is 100 cm² and area of a triangular face is 65 cm². 4. Find the volume of the given pyramids. (a) (b) (c) (d) (e) (f) (g) (h) i) 5. Find the lateral surface area and the total surface area of the following pyramids. (a) (b) (c) (d) (e) (f) 10 cm O B C A D18 cm O Q U P S T R 13 cm 12 cm 15 cm O P Q R S 10 cm Q 15 cm E F H G I 12 cm 16 cm 16 cm 10 cm 8 cm 12 cm D C A B O 10 cm E 13 cm 10 cm 25 cm 24 cm 12 cm 10 cm 15 cm 12 cm O D C A B P24 cm 25 cm 6 cm 16 cm 16 cm 10 cm12 cm O D B C A O 17 cm 15 cm
102 Oasis School Mathematics-10 6. (a) In the given figure, the total surface area of the given square based pyramid is 96 cm2 and a side of the square base is 6 cm. Find the slant height of the pyramid. 6 cm 6 cm (g) (h) (i) 24 cm 25 cm 13 cm 10 cm O Q T S R P 10 2 cm 10 cm (b) Total surface area of the given square-based pyramid is 384 cm². Find the slant height and height of the pyramid. (c) The given figure is a square-based pyramid. Total surface area of the pyramid is 144 cm². If the length of each side of the square base is 8 cm, find- (i) its slant height (ii) its height (iii) its volume. 7. (a) In the given figure, the length of each side of the square base is 14 cm. If the volume of the pyramid is 1,568 cm3 , find the (i) height of the pyramid. (ii) TSA of the pyramid. (b) In the given figure, the length of each side of the square base is 10 cm. If its volume is 400 cm3 , find (i) its height. (ii) its slant height. (iii) its lateral surface area. 8. (a) Total surface area of the given square-based pyramid is 96 cm². If its slant height is 5 cm, find its height. 8 cm 14 cm 10 cm O B D C A 5 cm 12 cm
Oasis School Mathematics-10 103 (b) Total surface area of the given square-based pyramid is 864 cm². If its slant is 15 cm, find (i) length of each side of the base. (ii) height of the pyramid. (iii) volume of the pyramid. 9. (a) Volume of the given square-based pyramid is 1568 cm3 . If EF = 7 cm, calculate the area of the triangular surface. (b) Volume of the given square based pyramid is 1,280 cm3 . If MN = 8 cm. Calculate the TSA of the pyramid. 10. (a) In the given figure, the length of each side of the square base is 8 cm. If the area of the triangular faces is 80 cm2 , find the length of OE and the volume of the pyramid. (b) The given figure is a square based pyramid, where UT = 6cm, area of triangular faces is 240 cm2 . Find the length of OT, OU and volume of pyramid. 11. (a) The vertical height and length of base of the square-based pyramid are in the ratio of 2 : 3. If total surface area of the pyramid is 384 sq.cm., find the slant height. (b) In the given pyramid, (i) What do the part OP, BD and OD represent? (ii) Write their relation. (iii) If BD = 12 2 cm and OP = 8cm, find the length of OD. (iv) Find the slant height of the pyramid. (v) Find the LSA and TSA of the pyramid. A O D E B C 8 cm O Q T U R S P O A P D B C O B E F D C A O P Q N S R M 12. On a square field 256 ft², a square based pyramid shaped tent of height 15 ft is to be constructed. (a) Find the length of each side of the base. (b) Find the area of cloths required to prepare the tent. (c) If the tent of similar structure is made with the cloths of 800 ft², find the height of the tent 15 cm
104 Oasis School Mathematics-10 Project Work 1. Prepare a net of square based pyramid. Fold it and form the square base pyramid taking this an a model. Prepare a wooden square pyramid. Measure its height, slant height and its side of the base. Calculate is L.S.A, T.S.A and volume. Present it your class room. 2. In a chart paper, draw a net as shown above. A square and 4 isosceles triangles of same size. • Calculate the area of 4 triangles and a square. • Fold them to form a square based pyramid. • Show its different parts. • Calculate its TSA using the formula. • Compare the area obtained in two different cases. Answers 1. Consult your teacher 2. (a) 400 cm3 (b) 200 cm3 3. (a) 144 cm² (b) 120 cm² (c) 360 cm² 4. (a) 600 cm3 (b) 750 cm3 (c) 400 cm3 (d) 1296 cm3 (e) 512 cm3 (f) 253.92 cm3 (g) 363.62 cm3 (h) 784 cm3 (i) 256cm3 5. (a) 240 cm², 384 cm² (b) 360 cm², 504 cm² ( c) 672 cm², 868 cm² (d) 320 cm², 576 cm² (e) 260 cm², 360 cm² (f) 544 cm², 800 cm² (g) 700 cm², 896 cm² (h) 240 cm², 340 cm² 6. (a) 5 cm (b) 10 cm, 8 cm (c) (i) 5 cm, (ii) 3cm (iii) 64 cm3 7. (a) 24 cm, 896 cm² (b) 12 cm, 13 cm, 260 cm² 8. (a) 4 cm (b) 18 cm, 12 cm, 1296 cm3 9. (a) 700 cm2 (b) 800 cm2 10. (a) 5cm, 64cm3 (b) 10cm, 8cm, 384 cm3 11. (a) 10 cm. (b) (iii) 11.67 cm, (iv) 10 cm, (v) 240 cm2 , 384 cm2 12. (a) 16 ft (b) 544 ft² (c) 23.68 ft 13. (a) 196 ft² (b) 22.96 ft (c) 1499.82 ft³ (d) No, 27ft³ less 13. A square based pyramid has edge 25ft and slant height 24 ft. (a) Find its area of base. (b) Find its height. (c) Find its volume. (d) Is the cloths of area 650 ft² is sufficient to cover the pyramid? How much more or less cloths is required to cover the pyramid.
Oasis School Mathematics-10 105 Review Recall the following points in your class room. • What is the name of given figure? • What do the shaded part of the figure represent? • What is the formula to calculate CSA of the given figure? • What is the formula to calculate TSA and volume of given object? • What are the name of given solid objects? • What is the formula to calculate CSA, TSA and volume of above objects? 6.2 Combined Solids Here we study the volume, curved surface area and total surface area of a combined solid formed by a cylinder and a cone, cylinder and a hemisphere, and a cone and a hemisphere. Combined solid of cylinder and cone: This is the combined solid formed by a cylinder and a cone. 6.1 Warm-up Activities Unit 6 Combined Solids Identify the name of the following objects. • Recall the formula to calculate the area of base of cylinder, cone, square based prism. • If 'a' be the length of a side of the base and 'h' be the height of the prism, what is its LSA? • Recall the formula of CSA, TSA and volume of cylinder, cone, sphere and hemisphere. r h
106 Oasis School Mathematics-10 Volume of this combined solid = Volume of cylinder + Volume of cone Curved surface area = CSA of cylinder + CSA of cone Total surface area = Area of base + CSA of cylinder + CSA of cone Combined solid of cone and hemisphere This is the combined solid of a cone and a hemisphere. Volume of this combined solid = Volume of cone + Volume of hemisphere CSA of this combined solid = CSA of cone + CSA of hemisphere TSA of this combined solid = CSA of cone + CSA of hemisphere Combined solid of cylinder and hemisphere This is the combined solid of a cylinder and a hemisphere. Volume of given combined solid = Volume of cylinder + Volume of hemisphere CSA of given combined solid = CSA of cylinder + CSA of hemisphere TSA of given combined solid = Area of base + CSA of cylinder + CSA of hemisphere example 1 Find the volume of the given combined solid. Solution: This is the combined solid of a cylinder and a cone. For cylinder, Diameter (d) = 14 cm Radius (r) = 14 2 cm = 7cm Height (h) = 50 cm We have, volume of cylinder (V1 ) = π r2 h = 22 7 ×7² × 50 = 7,700 cm3 For Cone, Radius (r) = 14 2 = 7 cm Note Volume of Combined solid = Sum of the volumes of solids CSA of a combined solid = Sum of the areas of visible curved parts TSA of combined solid = Sum of the areas of all visible parts Worked Out examples 50 cm 25 cm 14 cm
Oasis School Mathematics-10 107 Slant height (l) = 25 cm We have, h = ² – r² = ( ) 25 ² – ( ) 7 ² = 625 – 49 = 576 = 24 cm We have, Volume of cone (V2 ) = 1 3 π r2 h = 1 3 × 22 7 (7)² × 24 = 1,232 cm3 Now, Volume of given combined solid = V1 + V2 = 7700 cm3 + 1,232 cm3 = 8,932 cm3 example 2 Find the curved surface area and total surface area of the given solid. Solution: This is the combined figure of a cylinder and a hemisphere. For the hemisphere, Radius (r) = 64 cm – 50 cm = 14 cm We have, CSA of hemisphere = 2 π r2 = 2 × 22 7 × (14)² = 1,232 cm² For the cylinder,radius (r) = 14 cm Height (h) = 50 cm We have, CSA of cylinder = 2πrh = 2 × 22 7 × 14 × 50 = 4,400 cm² Now, CSA of given solid = CSA of hemisphere + CSA of cylinder = 1,232 cm² + 4,400 cm² = 5,632 cm² Again, area of base (circle) = π r2 = 22 7 × 14² = 616 cm² Now, TSA of given solid = CSA of hemisphere + CSA of cylinder + area of circle = 1,232 cm² + 4,400 cm² + 616 cm² = 6,248 cm² 64 cm 50 cm
108 Oasis School Mathematics-10 example 3 A tent in the form of a right circular cylinder of height 5 m is surmounted by a cone of slant height 10 m. If the diameter of the base of the tent is 14 m, find the total surface area of the tent. Solution: Total surface area of the tent is the sum of the CSA of the cone and CSA of the cylinder in this figure. For the cylinder, Radius (r) = 14 2 = 7 m Height (h) = 5 m We have, CSA of cylinder = 2 π rh = 2 × 22 7 × 7 × 5 = 220 m² For Cone, Radius (r) = 7 m Slant height (l) = 10 m We have, CSA of cone = πrl = 22 7 × 7 × 10 = 220 m² Now, TSA of given tent = CSA of cylinder + CSA of cone = 220 m² + 220 m² = 440 m² example: 4 the total surface area of the given combined solid is 858 cm². If the slant height of the cone is 25 cm, find the length of their common radius. Solution: Given figure is the combined figure of a cone and a hemisphere. Here, CSA of hemisphere = 2 π r² CSA of cone = π rl = πr × 25 = 25 πr cm² Now, TSA of given figure = CSA of hemisphere + CSA of cone or, 858 = 2 πr² + 25 πr or, 858 = 2 × 22 7 × r² + 25 × 22 7 × r or, 858 × 7 = 2 × 22r² + 25 × 22r or, 39 × 7 = 2r² + 25r or, 2r² + 25r – 39 × 7 = 0 or, 2r² + (39 – 14)r – 39 × 7 = 0 5m 10 m 14 m A 25 cm
Oasis School Mathematics-10 109 or, 2r² + 39r – 14r – 39 × 7 = 0 or, r(2r + 39) – 7 (2r + 39) = 0 or, (2r + 39) (r – 7) = 0 Either, 2r + 39 = 0 ⇒ r = – 39 2 (not possible) or, r – 7 = 0 ⇒ r = 7 cm ∴ Common radius = 7 cm exercise 6.1 1. Observe the given solid objects and answer the questions given below. (i) Which two solid objects are there in the given combined solid? (ii) What are their separate volume? (iii) What is the volume of whole solid? (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) 2. Observe the given combined solid and answer the questions given below. (i) Which two solid are there in the given figure? (ii) Calculate separately their CSA are the area of flat part. (if there is) (iii) Calculate their CSA and TSA. (a) (b) (c) 54 cm 30 cm 4 cm 25 cm 3 cm 10 cm 14 cm 31 cm 140 cm 13 cm 10 cm A O B A' r = 8 cm 10 cm A A O P A' r = 7 cm 25 cm 14 cm 15 cm 12 cm r= 10 cm 9 cm P O P' A' A 40 cm 54 cm 14 cm 37 cm 12 cm 10 cm 80 cm 10 cm 13 cm 24 cm 140 cm 10cm 152 cm
110 Oasis School Mathematics-10 (d) (e) (f) (g) (h) 3. (a) Find the total surface area of the given tent. (b) A tent is of the shape of a right circular cylinder of height 4 cm and radius 14 cm with a cone of height 6 cm. Find the total surface area and volume of the tent. (c) Height of a cylinder is 3 times its radius. It has hemispheres on both ends. If the radius of the object is 7cm, find its volume. (d) Find the volume of a combined solid containing a cylinder and hemisphere, where the height of the cylinder is 50 cm and the height of whole solid is 64 cm. 4. (a) The base area of the cylinder is 100 cm2 and the height of the cylinder is 3 cm. If the volume of the whole solid is 500 cm3 , find the height of the solid so formed. (b) The base of the cylinder is 125 cm² and the height of the cylinder is 3 cm. If the volume of the whole solid is 625 cm³, find the height of the solid. (c) The volume of the given solid object is 38,808 cm3 . Find the volume of the cone. (d) If the volume of the given combined solid is 792 cm³, find the value of x. 63 cm 140 cm 25 cm 50 cm 74 cm 57 cm 14 cm 30 cm 37 cm 14 cm 31 cm 33m 120m 22 m 3 cm 3 cm 36 cm240 cm x cm 6 cm
Oasis School Mathematics-10 111 (e) The given figure is combined figure of a cone and a hemisphere of diameter 42 cm. The volume of the solid is 38,808 cm3 . Find the height of the cone. 5. (a) Curved surface area of the given solid is 924 cm². (i) Find the CSA of hemisphere (ii) Height of the cylinder (iii) Height of the whole solid (iv) TSA of whole solid (b) Curved surface area of a given combined solid is 880 cm². (i) find its common radius (ii) find the height of the cone (iii) find the volume of combined solid 6. (a) If the curved surface area of the given combined solid is 858 cm², (i) find tis common radius (ii) find the height of the cone (iii) find the volume of combined solid (b) If the total surface area of the given combined solid is 2,024 cm², (i) find its common radius (ii) find the height of the cone (iii) find the volume of the solid (c) The curved surface area of the given combined solid is 836 cm². (i) Find its common radius (ii) Find its TSA. 42 cm 14 cm 10 cm 15 cm 25 cm 25 cm 30 cm 12 cm Answers 1. (a) 138.27 cm3 (b) 11,314.29 cm3 (c) 5,852 cm3 (d) 1,950.67 cm3 (e) 1,950.67 cm3 (f) 1,475.05 cm3 (g) 2,200 cm3 (h) 770 cm3 (i) 30,389.33 cm3 (j) 5,338.67 cm3 2. (a) 3,205.71 cm2 , 3,318.85 cm2 (b) 958.58 cm², 1,037.15 cm² (c) 4,604.28 cm², 4,682.86 cm² (d) 2750 cm², 2904 cm² (e) 55440 cm², 67914 cm² (f) 2,508cm2 , 2,662cm² (g) 985.6 cm², 985.6 cm² (h) 858 cm², 858 cm² 3. (a) 19,799.99 (b) 1,022.18cm2,, 3,696cm3 (c) 4,671.33cm3 (d) 36,549.33cm3 4. (a) 9 cm (b) 9 cm (c) 1,848 cm3 (d) 26 cm (e) 42 cm 5. (a) (i) 308cm² (ii) 21cm (iv) 1078cm² (b)(i) 220r 7 cm² (ii) 660r 7 cm² (iv) 7cm 6. (a)(i) 7cm (ii) 24cm (iii) 1950.67 cm³ (b) (i) 7cm (ii) 24cm (iii)5832 cm³ (c)(i) 7cm, (ii) 990 cm²
112 Oasis School Mathematics-10 6.3 Combined Solids of Prism and Pyramid Square based Prism Observe the given figure properly and answer the questions given below. • What is the shape of the base of given object? • If 'a' be the length of a side of the base, what is its area? • If 'h' be the height of given objects, what is the area of a rectangular face? • What is the area of 4 rectangular faces? • What is the LSA of given object? • What is the TSA of given object? • What is the volume of given object? • The shape of the base of given objects is square. • The area of the base of given object a2 . • If 'h' be the height of given object, then the area of a rectangular face is ah. • Area of 4 rectangular faces is 4ah. • LSA of given object is 4ah. • TSA of given object is LSA + 2 × Area of base. So, TSA of given object is 4ah + 2a2 . • Volume of given object is Area of base × height i.e. a2 h. Remember ! In square based prism: • Area of base = a2 • Area of a rectangular face = ah. • LSA = 4ah. • TSA = 2a2 + 4ah • Volume = a2 h. a h The given figure is a combined solid of a cuboid (square-based prism) and a squarebased pyramid. As we know that the volume of a combined solid = Volume of prism + Volume of pyramid = Area of base × height (h1 ) + 1 3 area of base × height (h2 ) While calculating the total surface area of the combined solid, TSA = LSA of pyramid + LSA of prism + Area of base LSA of combined solid = LSA of prism + LSA of pyramid. Square base Height E a G a E B C D h1 h2 O A
Oasis School Mathematics-10 113 Worked Out examples example 1 1. Find the volume of the given combined solid. Solution: This is the combined solid of a prism (cuboid) and pyramid. For prism Area of base = (10 cm)2 = 100 cm2 Height = 12 cm We have, Volume (V1 ) = Area of base × height = 100 × 12 = 1,200 cm3 For pyramid Area of base = (10 cm)2 = 100 cm2 Height = 21cm – 12 cm = 9 cm. We have, Volume of pyramid (V2 ) = 1 3 area of base × height = 1 3 × 100 × 9 = 900 3 cm3 = 300cm3 Hence, the volume of given solid (V) = V1 + V2 = 1200 cm3 + 300 cm3 = 1500 cm3 Remember ! • LSA of pyramid = 4∆ = 2al • LSA of cuboid (prism) = perimeter of base × height = 4ah • LSA of combined solid = LSA of cuboid (prism) + LSA of pyramid • TSA of combined solid = LSA of pyramid + LSA of cuboid (prism)+area of base. • Volume of Prism = Area of base × height = a2 h. • Volume of Pyramid = 1 3 Area of base × height = 1 3 a2 h. • Volume of combined solid = Volume of Prism + Volume of Pyramid. = a2 h2 + 1 3 a2 h1 12cm 10cm 21cm Alternative method For Prism, h = 12cm, a = 10 cm Volume of Prism (V1 ) = a2 h = 102 × 12 = 1200 cm3 For Pyramid, a = 10 cm, h = 21cm – 12cm = 9cm Volume of Pyramid (V2 ) = 1 3 a2 h = 1 3 102 × 9 = 300 cm2 Volume of combined solid = V1 + V2 = (1200 + 300) cm3 = 1500 cm3
114 Oasis School Mathematics-10 example 2 Find the volume of the given combined solid. Solution: This figure is combined solid of two pyramids whose combined height is 30 cm. Let, h1 = height of upper pyramid h2 = height of lower pyramid i.e. h1 + h2 = 30 cm Area of base = (8)2 = 64 cm2 Volume of upper pyramid (V1 ) = 1 3 area of base × height = 1 3 × 64 × h1 = 64h1 3 cm3 Volume of lower pyramid (V2 ) = 1 3 area of base × height = 1 3 × 64 × h2 = 64h2 3 cm3 Volume of combined solid = V1 + V2 = 64h1 3 + 64h2 3 = 64 3 (h1 + h2 ) = 64 3 × 30 = 640 cm3 example 3 Find the lateral surface area and total surface area of the given solid. Solution: This is the combined solid of a square-based prism (cuboid) and a pyramid. ABCD is the base of the pyramid and EFGH is the base of the prism. Both of them are equal in area. OI is the slant height of the pyramid. For the pyramid, BC = 12 cm, OI = 10 cm Area of ∆OBC = 1 2 BC × OI = 1 2 × 12 × 10 = 60 cm2 30cm 8cm Alternative method Area of base = (8 cm)2 = 64 cm2 Sum of height = 30 cm We have, Volume = 1 3 area of base × sum of the height = 1 3 × 64 × 30 cm3 = 640 cm3
Oasis School Mathematics-10 115 Now, LSA of the pyramid = 4 × 60 cm2 = 240 cm2 For prism (cuboid) Perimeter of the base = 4 × 12 cm = 48 cm. LSA of the prism = Perimeter of the base × height = 48 × 16 cm2 = 768 cm2 Again, Area of base = (12)2 = 144 cm2 Now, LSA of the given figure = LSA of pyramid + LSA of prism. = 240 cm2 + 768 cm2 = 1008 cm2 Again, TSA of the given figure = LSA of pyramid + LSA of prism + Area of base = (240 + 768 + 144) cm2 = 1152 cm2 example 4 Find the lateral surface area and total surface area of the given combined solid. Solution: The given figure is the combined solid of a square-based prism (cuboid) and a square based pyramid. ABCD is the base of the pyramid and EFGH the base of the prism OM is the height of the pyramid. ON is its slant height. For prism Perimeter of the base = 4 × 6 cm = 24 cm Height of the prism = 10cm Lateral surface area of the prism = p × h = 24 cm × 10 cm = 240 cm2 For pyramid Again, height of the pyramid = (14 – 10) cm = 4 cm Alternative method For Prism, Perimeter of the base (P) = 4a = 4 × 12 = 48 cm Height (h) = 16 cm We have, LSA of prism (A1 ) = p × h = 48 × 16cm2 = 768 cm2 For Pyramid, a = 12 cm, l = 10 cm. We have, LSA of pyramid (A2 ) = 2al = 2 × 12 × 10 = 240 cm2 Area of base (A3) = a2 = 122 = 144 cm2 Now, LSA of given figure = A1 + A2 = 768 + 240 = 1008 cm2 TSA of given figure = (1008+144)cm2 = 1,152 cm2 . 10cm G B E E F C N A M H O 6cm 14cm 6cm
116 Oasis School Mathematics-10 ∴ OM = 4 cm, MN = 1 2 DC = 1 2 × 6 cm = 3cm. Using the Pythagoras Theorem in ∆OMN, ON2 = OM2 + MN2 = (4)2 + (3)2 = 16 + 9 = 25. ∴ ON = 5 cm. Area of ∆OBC = 1 2 BC × ON = 1 2 × 6 × 5 cm2 = 15 cm2 ∴ LSA of pyramid = 4∆ = 4 × 15 cm2 = 60 cm2 . ∴ LSA of given solid = LSA of prism + LSA of pyramid = 240 cm2 + 60 cm2 = 300 cm2 TSA of the given solid = LSA + area of base = 300 + (6)2 = 300 + 36 = 336 cm2 example 5 The given figure is the combined solid of square based prism and pyramid, where the slant height of the pyramid is 13 cm and the length of the side of the base is 10cm. If the volume of whole solid is 900 cm3 , find the height of the prism. Solution: Given figure is the combined solid of square based prism and pyramid. Here, to find the volume of prism, Let the height of the prism be h1 Then, area of the base of the prism = (10 cm)2 = 100 cm2 We have, Volume of the prism (V1 ) = Area of base × height = 100 × h1 = 100h1 cm3 . For the pyramid, l 2 = h2 + a2 4 or, l 2 = 42 + 62 4 or, l 2 = 25 or, l = 5 cm LSA of the Pyramid = 2al = 2× 6 × 5 = 60 cm2 LSA of the Prism = 4ah = 4 × 6 × 10cm2 = 240 cm2 LSA of the given figure = 60cm2 +240cm2 = 300 cm2 Area of base = (6)2 cm2 = 36cm2 TSA of the given figure = 300cm2 + 36cm2 = 336cm2 Alternative method 13cm 10cm
Oasis School Mathematics-10 117 Again, for pyramid, Slant height (l) = 13 cm. Length of a side of the base (a) = 10 cm. We have, l 2 = h2 + a2 4 or, 132 = h2 + 102 4 or, 169 = h2 + 25 or, 169 – 25 = h2 or, h2 = 144 ∴ h = 144 = 12 cm. Again, we have Volume of the pyramid (V2 ) = 1 3 a2 h = 1 3 102 × 12 = 400 cm3 . Now, Volume of whole solid (V) = V1 + V2 or, 900 = 100h1 + 400 or, 900 – 400 = 100h1 or, 500 = 100h1 or, h1 = 500 100 = 5 cm. ∴ Height of the prism = 5 cm. exercise 6.3 1. Find the volume of the given combined solids. (a) (b) (c) (d) (e) (f) 15cm 5cm 6cm 6cm 5cm 6cm 8cm 15cm 5cm 8cm 8cm 12cm 13cm 6cm 20cm B 16cm 20cm O D C A 10cm Alternative method Here, aside of the base (a) =16cm Height (h) = 6 cm We have, volume (v) = 1 3 a2 h = 1 3 ×162 ×6 = 512 cm3 Again, l2 = h2 + a2 4 = (6)2 + (16)2 4 = 36 + 64 = 100 ∴ l = 10 cm. We have, TSA = a2 + 2al = 162 + 2×10×16 = 256+576 = 576 cm2
118 Oasis School Mathematics-10 2. The given figure is a combined solid of a square-based prism and a pyramid. From the given data, find - (i) Height of the pyramid. (ii) Slant height of the pyramid. (iii) Area of a triangular face of the pyramid. (iv) LSA of the pyramid. (v) Perimeter of the base of the prism. (vi) LSA of the prism. (vii) Area of base of the prism. (viii) LSA of the whole solid (ix) TSA of the whole solid. 3. Find the lateral surface area and total surface area of the given combined solids. (a) (b) (c) H A B E F I C G 44cm 20cm O H 14cm 4. (a) In the given solid, upper part is a square based pyramid of slant height 5 cm and lower part is a square based prism having each side of base 8 cm. (i) Find the volume of pyramid. (ii) If the volume of whole solid is 448cm³, find the volume of prism. (iii) Find the height of the prism. (b) In the given solid, the lower part is a square based prism and the upper part is square based pyramid. The length of the base of the prism is 10cm and its height is 20 cm. (i) Find the volume of the prism. (ii) Volume of the pyramid. If the volume of whole solid is 2400cm³. (iii) Height of the pyramid (iv) Slant height of the Pyramid. 5 cm 8 cm 20cm 10cm 15cm 10cm 13cm 30cm 24cm 14cm 12cm 10cm 10cm
Oasis School Mathematics-10 119 5. (a) Total surface area of given solid object is 1344cm2 , if the height of the square based prism is 20cm, side of the base is 12cm, find (i) The area of the base (ii) LSA of the prism (iii) LSA of the pyramid in terms of l. (iv) Is the sum of the result of a, b and c equal to TSA? (v) If yes, find the value of l and the height of the Pyramid. (b) Volume of given combined solid is 832 cm3 , find (i) Area of base (ii) Volume of prism (iii) Volume of Pyramid in terms of h. (iv) The value of h. (v) Slant height of the Pyramid (vi) Surface area of the given combined solid. 20cm 12cm 20cm 8cm 8cm Answers 1. (a) 300 cm3 (b) 448 cm3 (c) 832 cm3 (d) 1200 cm3 (e) 240 cm3 (f) 1200 cm3 2. (i) 24 cm (ii) 25 cm (iii) 175 cm2 (iv) 700 cm2 (v) 56 cm (vi) 1120 cm2 (vii) 196 cm2 (viii) 1820 cm2 (ix) 2016 cm2 3. (a) 860 cm2 , 960 cm2 (b) 2,380 cm2 , 2,576 cm2 (c) 480 cm2 , 480 cm2 4. (a) (i)64cm³ (ii) 384cm³ (iii) 6cm (b) (i) 2000 cm³ (ii) 400cm³ (iii) 12cm (iv) 13cm 5. (a) (i) 144cm2 (ii) 960cm2 (iii) 24l (iv) 10cm, 8 cm (b) (i) 64cm2 (ii) 768 cm3 (iii) 64h 3 cm3 (iv) 568 cm2
120 Oasis School Mathematics-10 7.1 Warm up Activities • Pemba is planning to plaster the yard in front of his house. • Chameli has to construct the well using the ring. • A constructor has to construct the square based pillar. • A compound gate has two cylindrical pillars having hemispherical top, its surface it to be painted. How to calculate cost of construction of the objects mentioned above. Recall the formulae to calculate • area of four walls of a room. • volume of a tank Pemba is planning to plaster the yard in front of his home. The cost of plastering per square feet is Rs. 70. He has to calculate the cost of plastering his yard. The structure of his yard is as shown in the figure. A E D B C Let's find the area of the yard. The measurement of each side is given below. AB = 12 ft, BC = 15 ft. CD = 17 ft, ED = 14 ft., AE = 10 ft. A E D B 15ft 17ft 10 ft 12 ft 14 ft C Let's take the measurement of BD and EB. BD = 8 ft and BE = 16 ft. Unit 7 Cost estimation
Oasis School Mathematics-10 121 Now, the area of the yard is the area of ∆ABE + Area of ∆BDE + Area of ∆BDC. For ∆ABE, three sides, a = 10 ft, b = 12 ft, c = 16 ft. s = a+b+c 2 = 10+12+16 2 = 19 ft. Area of ∆ ABE = s(s-a) (s-b) (s-c) = 19(19–10) (19–12) (19–16) = 19 × 9 × 7 × 3 = 3,591 = 59.92 square ft. Again, for ∆ BDE, three sides, a = 14 ft, b = 16ft, c = 8 ft. s = a+b+c 2 = 14+16+8 2 = 19 ft. Area of ∆ ABE = s(s-a) (s-b) (s-c) = 19(19–14) (19–16) (19–18) = 19 × 5 × 3 × 11 = 3,135 = 55.99 square ft. For ∆ BDC, three sides, a = 17 ft, b= 15 ft, c = 8ft. s = a+b+c 2 = 17+15+8 2 = 20 ft. Area of ∆ BDC = s(s-a) (s-b) (s-c) = 20(20-17) (20-15) (20-8) = 20 × 3 × 5 × 12 = 3,600 = 60 square ft. Area of the yard = (59.92 + 55.99 + 60) sq. ft. = 175.91 sq. ft. Here, Cost of plastering 1 sq. ft. = Rs. 70 Cost of plastering 175.91 sq. ft. = Rs. 70 × 175.91 = Rs. 12313.70.
122 Oasis School Mathematics-10 There are many such examples which come in our daily life like, • Estimation of the cost of constructing the well. • Estimation of the cost of constructing the pillar of a gate. • Estimation of the cost of constructing a water tank, etc. Worked Out examples example 1 A water tank having dimensions 3m × 2.5m × 2m is to be constructed. Find (a) how much water the tank can hold. (b) the area of its four walls (c) cost of plastering its four walls at the rate of Rs. 125 per square meter. Solution: Here, the length of the tank (l) = 3 m breadth of the tank (b) = 2.5 m height of the tank (h) = 2 m (a) Volume of the tank (V) = l × b × h = 3 m × 2.5 m × 2 m = 15 m3 = 15 × 1000 liters [∴1m3 = 1,000 l] = 15000 liters (b) Area of 4 walls = 2h(l+b) = 2 × 2(3 + 2.5)m2 = 4 × 5.5 m2 = 22 m2 (c) The cost of plastering 1m2 wall = Rs. 125 The cost of plastering 22m2 wall = Rs. 125 × 22 = Rs. 2,750 example 2 The diameter of a cemented ring of a well is 3 ft. Its height is 1.1 ft. Find (a) the volume of the well if it contains 40 such rings. (b) cost of constructing a well if the cost per ring is Rs. 1,100. (c) volume of water, if the level of water is on the 15th ring from the top. Solution: Given, Diameter of the ring (d) = 3 ft. ∴ Radius (r) = 1.5 ft.
Oasis School Mathematics-10 123 Height of a ring = 1.1 ft. Height of 40 rings = 40 × 1.1 ft. = 44 ft. We have, Volume of the cylinder (well) = πr2 h = 22 7 × (1.5)2 × 44 = 311.14 ft3 (b) Cost per ring = Rs. 1100 Cost of 40 rings = Rs. 1100 × 40 = Rs. 44,000 (c) Here, (40–15) rings are covered by water. ∴ Height of 25 rings = 25 × 1.1 ft. = 27.5 ft. Radius (r) = 1.5 ft. We have, volume of water = πr2 h = 22 7 × (1.5)2 × 27.5 ft3 = 194.46 ft3 example 3 A house contains 12 pillars of dimension 30 cm × 30 cm × 3 m. Find the total cost of constructing the pillars if the construction cost per cubic meter is Rs. 1,450. Also, find the cost of plastering all the pillars at the rate of Rs. 90 per square meter. Solution: Dimension of a pillar = 30 cm × 30 cm × 3 m ∴ Volume of a pillar = 0.3 m × 0.3 m × 3 m = 0.27 m3 Volume of 12 pillars = 12 × 0.27 m3 = 32.4 m3 Construction cost per m3 = Rs. 450 ∴ Total cost of constructing the pillars = 32.4 × 1450 = Rs. 46,980 Again, lateral surface area of a pillar, = perimeter of base × height = 2 (0.3 + 0.3) × 3 m = 3.6 m2 LSA of 12 pillars = 12 × 3.6m2 = 43.2 m2 Cost of plastering per m2 = Rs. 90 Cost of plastering 43.2 m2 = Rs. 90 × 43.2 = Rs. 3,888 30 cm 30 cm 3 m
124 Oasis School Mathematics-10 example 4 A cylindrical shaped tent is made for the accommodation of 77 people in such a way that each person gets the area 2m2 on the ground and 10m3 of air to breathe. Find the height of the tent and cost of canvas required to prepare the tent if the cost per m2 of cloth is Rs. 80 per m2 . Solution: Here, total number of people = 77 Area to be occupied by each people = 2m2 Then, the area of base = 77 × 2 m2 = 154 m2 Again, Since each person needs 10m3 of air to breathe Volume of the tent = 77 × 10 cm3 = 770 cm3 We have, Volume = Area of base × height or, 770 = 154 × h or, h = 770 154 = 5 cm. Again, area of base = 154 m2 πr2 = 154 22 7 r2 = 154 r2 = 154 × 7 22 = 154 ∴ r = 7 cm. Again, area of canvas required to prepare the tent = CSA of tent + area of top = 2πrh + πr2 = 2 × 22 7 × 7 × 5 + 22 7 × 72 = 220 m2 + 154 m2 = 374 m2 Since the cost per m2 cloths = Rs. 80 Total cost = Rs. 374 × 80 = Rs. 29920. example 5 A square based Pyramid shaped tent is made for the accommodation of 100 people. For each person 4 sq. meter ground and 12m3 of air to breathe. Find, (a) height of the ten (b) slant height of the tent (c) LSA of the tent (d) total cost to prepare the tent at the rate of Rs. 120 per m2 .
Oasis School Mathematics-10 125 Solution: Total number of people = 100. Since each person needs 4 square meter area on the grounds. Area of the base (A) = 100 × 4 = 400 m2 Again, each person needs 12m3 of air to breathe Then, volume of tent (V) = 100 × 12 = 1200 m3 (a) We have, Volume of the Pyramid = 1 3 Area of base × height 12000 = 1 3 × 400 × h h = 1200 × 3 400 = 3 m (b) Again, Area of base = 400 m2 a2 = 400 a = 20 m. In square based Pyramid, l 2 = h2 + 92 4 l 2 = 92 + 202 4 l 2 = 81 + 100 l 2 = 181 l = 181 = 13.45 m. (c) Again, LSA of Pyramid = 2al = 2 × 20 × 13.45 = 538 m2 (d) Cost per m2 of cloths = Rs. 120 Total cost = 538 × 120 = Rs. 64,560. example 6 A semi cylindrical tunnel is made on a rectangular field of dimension 50 ft. × 28 ft. The tunnel is covered by plastic. Find (a) the cost of plastic to cover the tunnel at the rate of Rs. 20 per per ft2 (b) the cost to harvest the vegetable on the field at the rate of Rs. 50 per square feet. Solution: In the given figure, ABCD is a rectangular field of dimension 50 ft. × 25 ft.,
126 Oasis School Mathematics-10 A semi cylindrical shaped tunnel is made on it. The area to be covered by plastic = Its CSA + Area of its 2 semi circular face. = πrh + 2 × 1 2 πr2 = πrh + πr2 = 22 7 × 14 × 50 + 22 7 × (14)2 = 2200 + 616 = 2816 square feet. (a) Since the cost to cover the tunnel by plastic at the rate of Rs. 20 per ft2 Total cost = Rs. 2816 × 20 = Rs. 56,320. (b) Again, area of the field = (50 × 28) square feet. = 1400 square feet. Hence, cost to harvest the vegetable at the rate of Rs. 50 per m2 is Rs. 1400 × 50 = Rs. 70,000. example 7 The adjoining figure is the figure of two pillars mounting a square pyramid on the top. Find the cost of tiling both pillars at the rate of Rs. 20 per square feet. Solution: In the given figure, the lower part is a square based Prism and the upper part is squared based pyramid. Taking lower part, Perimeter of the base = 4a = 4 × 6 ft = 24 ft. We have, LSA of the prism (A1 ) = Perimeter of the base × height = 24 × 6 = 144 ft2 For Pyramid, a = 6ft, h = 4 ft. We have, = l2 + a2 4 = 42 + 62 4 = 16 + 9 = 25. ∴ l = 5 ft. We have, LSA of pyramid (A2 ) = 2al = 2 × 6 × 5 A D B C 28 ft. 50 ft. 1.5ft. 10 ft. 6 ft. 4 ft. 10 ft. 6 ft.
Oasis School Mathematics-10 127 = 60 ft2 TSA of a pillar = A1 + A2 = 144 ft2 + 60 ft2 = 204 ft2 Cost of tiling a pillar at the rate of Rs. 20 per square feet is Rs. 204 × 20 = Rs. 4080 Cost of tiling two pillars = 2 × Rs. 4080 = Rs. 8160. example 8 A cylindrical tank having diameter 7cm and depth 5m is constructed to collect the rain water. If has hemispherical top, find (a) volume of in cylinder (b) Volume of the hemispherical (c) Volume of the tank, (d) Cost to plaster the wall, base and the top of the tank at the rate of Rs. 35 per m² and labour cost is Rs. 5 per m2 . (e) Cost of fill the tank by water at the rate of Rs. 2500 per 1000 litre. Solution: Given, figure represent a cylindrical meter tank with hemispherical top. (a) For cylinder Radius (r) = 7m 2 = 3.5m. we have, Volume of the cylinder (V1 ) = πr²h = 22 7 (3.5)² × 5 = 1925m³ = 1925 × 1000 litre = 192500 litre (b) For hemisphere Radius (r) = 7m 2 = 3.5m we have, Volume of the hemisphere (V2 ) = 2 3 πr³ = 2 3 × 22 7 (3.5)³ = 89,833 m³ = 89,833× 100 litre = 89,833 litre Now, (c) Volume of the tank = Volume of the cylinder + Volume of the hemisphere = 192500 litre + 89,833 litre = 2, 82, 333 litre. 7m 5m
128 Oasis School Mathematics-10 (d) Area of the base (A1 ) = πr² = 22 7 × (3.5)² = 38.5m² CSA of the cylinder (A2 ) = 2πrh = 2 × 22 7 × 3.5 × 5 m² = 110 m² CSA of the hemisphere (A3 ) = 2πr² = 2 × 22 7 (3.5)² = 77 m² Hence the area of inner wall, base and the top of the tank = 38.5m² + 110m² + 77m² = 225.50m² Cost to plaster the inner wall, base and the top of the tank = 225.50 × 35 = Rs. 7,892.50. Labour cost = 225.50 × 5 = Rs. 1127.50 Total cost to plaster = Rs. 7,892.50 + 1127.50 = Rs. 9020. (e) Capacity of tank is litre = 2,82,333 litre Cost for 1000 l = Rs. 2500 Cost for 1 l = Rs. 2500 1000 Cost for 2,82,333l = Rs. 2500 1000 × 2,82,333 = Rs. 7,05,832.50 exercise 7.1 1. (a) A cylindrical water tank has height 3m and diameter 7m. If it has a hemispherical top, find how much water can be held by the tank. (b) A community has a cylindrical underground water tank having diameter 14m and depth 5m. It has a hemispherical top above the ground. (i) Find the cost of plastering its inner wall, base and inner surface of the hemisphere at the rate of Rs. 105 per square meter. (ii) Find the capacity of the tank. 2. (a) At a wedding party, a man managed a conical tent which contains 23100 m3 air. If the height of the tent is 50 cm, find the cost of the cloth required to prepare the tent. Given that the cost per square meter of cloth is Rs. 250. (b) In a wedding ceremony, a man has to make arrangements for the accommodation of 150 people. For this purpose, he plans to build a conical tent in such a way that each person has 4 square meters of space on the ground and 20 cubic meter of air. Find the height of the tent.
Oasis School Mathematics-10 129 (c) A conical tent is made for the accommodation of 96 people. For each person there is 6π sq. meter ground and 44 cubic meter space. Find, (i) height of the tent (ii) cost to prepare the tent at the rate of Rs. 42 per square meter. 3. (a) A cylindrical shaped tent is made for the accommodation of 88 people in such a way that each person has 3 m2 space on the ground and 15m3 of air to breathe, find, (i) height of the tent (ii) area of cloths required to make the tent (iii) cost to prepare the tent at the rate of Rs. 50 per square meter and the labour cost is Rs. 2500. (b) A cylindrical shaped tent of height 10 m is made in such a way that it contains 360 m3 of air, find (i) area of the base (ii) radius of the base (iii) total surface area of the tent (iv) cost to prepare to tent at the rate of Rs. 50 per m2 4. (a) A square based Pyramid shaped tent is made in such a way that it contains 192 m3 of air. If the height of the tent is 9m. Find: (i) the area of the base (ii) a side of the base. (iii) slant height of the tent (iv) area of cloths required to make the tent (v) cost to prepare the tent at the rate of Rs. 70 per m2 . (b) A square based Pyramid shaped tent is made for the accommodation of 90 people in such a that each people gets 10 ft2 area on the ground and 30 ft3 of air to breathe, then: (i) Find the height of the tent (ii) Find a side of the base (iii) Find the slant height of the tent (iv) Find the surface area of the tent (v) Find the cost to prepare the tent at the rate of Rs. 40 per ft2 5. (a) A school compound gate has two pillars having a square base. Each pillar has a square based pyramid on the top as shown in the figure. Find (i) total surface area of each pillar. (ii) cost of paving tiles on its wall at the rate of Rs. 70 per square foot, if the labour cost is Rs. 40 per square feet. 1.5ft. 7 ft 1 ft 1 ft 1.5ft.7 ft 1 ft 1 ft
130 Oasis School Mathematics-10 (b) A shopping complex gate has two pillars. Each pillar is square based with dimension 2 ft. × 2ft. On the top of each pillar there is a square based Pyramid of height 2 ft. If the total height of each pillar is 10 ft. Find the cost of tiling both pillars at the rate of Rs. 120 per square feet. (c) An apartment compound gate has two pillars cylindrical in shape. The diameter of the base of each pillar is 2.8 ft. Each pillar has a hemispherical top. The height of both pillars is 13.4 ft. then: (i) Find the total surface area of each pillar. (ii) Find the cost of paving tiles at the rate of Rs. 85 per square foot. 6. (a) A half cylinder tunnel is made on the rectangular field of dimension 20ft × 14 ft. as shown in the figure. Find (i) the cost to cover the tunnel by plastic at the rate of Rs. 20 per square feet. (ii) the cost to harvest the vegetable on the field at the rate of Rs. 15 per square feet. (b) A half cylinder shaped tunnel is made on the rectangular field of dimension 50 ft. × 28 ft. along its length. An aluminium gate of dimension 3 ft × 2 ft is made on one of its semi-circular face. Find: (i) The cost to prepare the aluminium door at the rate of Rs. 200 per square feet. (ii) Cost to cover the tunnel by plastics except door at the rate of Rs. 80 per square feet. (iii) Cost to harvest vegetable if the labour charge is 30 per square feet and other expense is Rs. 405. 7. A square based pyramid shaped tank, equal base with pillar is kept at the top the pillar of the base 240 cm × 240 cm and height 4m. If the capacity of the pyramid shaped tank is 1728 litre, then: (a) Find the lateral surface area of pillar. (b) Find the height of the pyramid. (c) Find the cost to plaster the wall of the pillar at the rate of Rs. 250 per m². (d) Find the cost to colour the tank at the rate of Rs. 500 per m². 8. A tent has height 8m and radius of the base 3m. Its lower part is cylinder of height 4m and upper part is a cone. Find: (a) The surface area of the cylinder (b) The surface area of the cone 20 ft. 14 ft. 13.4 ft 13.4 ft 2.8 ft 2.8 ft
Oasis School Mathematics-10 131 (c) Cost of wrap the tent at the rate of Rs. 125 per m². (d) If you have 125 m² wrapping material to wrap the tent, how much more or wrapping materials is required? 9. A combined solid object is formed by joining the bases of a cone and a cylinder having same radii. The radius of the base of the solid object is 10.5cm and height of the cylindrical part is 40 cm. (a) Find the surface area of the object if the cost of painting at the rate of Rs. 50 per 100 cm² is Rs. 2112. (b) Find the slant height of the cone. (c) Find the height of the cone. 10. A solid object is made with the combination of a cylinder and a cone having same radii. The height of the cylinder is 30 cm and the slant height of the cone is 25cm. (a) Find the surface area of the object if the total cost of painting its surface at the rate of Rs. 150 per 100 cm² is Rs. 3036. (b) Find the common radius (c) Find the height of the cone. 11. A cylindrical water tank has depth 5m and the radius 7m. It has hemispherical top. (a) Find the capacity of the tank in litre. (b) Find the cost to fill the tank by water at the rate of Rs. 1500 per 1000 litre. 12. In Karmabeer farm house the ghee of the cow is collected in the cylindrical jar having diameter 70cm and height 1m. It has hemispherical top. (a) Find the quantity of ghee in kg that can be stored in the jar if the density of ghee is 0.8gm/cm³. (b) If the ghee of the jar is poured into small cylindrical jar of radius 3.5cm and height 5cm, find how many such jars can be filled with ghee. Answers 1. (a) 2,05.333m3 or, 2,05,333 liters (b) (i) Rs. 71610 (ii) 1488666.67 liters 2. (a) Rs. 8,94,811.29 (b) 15m (c) 7m, Rs. 79,200 3. (a) (i) 5m (ii) 552 m2 (iii) Rs. 30,100 (b) (i) 36π m2 (ii) 6m (iii) 156π m2 (iv) Rs. 24514.28 4. (a) (i) 64 m2 (ii) 8m (iii) 9.85 m (iv) 157.58 m2 (v) Rs. 11030.72 (b) (i) 9 ft. (ii) 30 ft (iii) 17.49 ft (iv) 1049.40 ft2 (v) Rs. 41976 5. (a) 31.16 ft2 (ii) Rs. 3427.60 (b) Rs. 17506.56 (c) (i) 117.92 ft2 (ii) Rs. 20,046.40 6. (a) Rs. 11880 (ii) Rs. 4200 (b) (i) Rs. 1200 (ii) Rs. 2,24,800 (iii) Rs. 42,405 7. (i) 38.4m² (ii)0.9m (iii) 7.2m² (iv)Rs.9600 (v) Rs.3600 8. (i)75.43m² (ii)47.14m² (iii) Rs. 15321.25 (iv)2.43 m²less. 9. (i)4224 cm²(ii) 37.5cm (iii) 36cm. 10. (i) 2024cm² (ii) 7cm (iii) 24cm 11. (i) 1488666.67 litre (ii) Rs, 22,33,000 12. (a) 379.87 kg, (b) 2467 (nearly)
132 Oasis School Mathematics-10 Project Work 1. Visit the cement factory nearby your home or city. Observe its shape (cylindrical and conical). Ask the measurement of different dimension. • Estimate the cost of colour its surface at the market rate. • Estimate the cost of make in mixture at the given market rate. 2. A school needs 2,00,000 litre water in a month. School has decided to make a cylindrical tank in the premises. School has separated circular land of area 38.5m². • Which of the depth 4m, 5m, 6m or 7, is suitable? • Present your report in your classroom. 3. Measure length of different dimension of the yard infront of your home. Draw its structure in a chart paper. Divide it into different triangles. • Find the area of the yard in square ft. • Ask the rate of plastering the yard from different sources and find the total cost of plastering in that rate. • Present your result in your classroom. Miscellaneous exercise 1. A solid object of wood in conical shape is cut into two equal parts as shown in the figure . (a) Find the height of object if the diameter of the base is 14cm and length of generator is 25cm. (b) Find the surface area formed by the generator. (c) Find the area of semi-circular and triangular faces of each part (d) Is cone a prism or pyramid? Ans: [(a) 24 cm, (b) 550 cm2 , (c) 245 cm2 (d) Consult your teacher] 2. Study the given questions and answer the questions given below: (a) Develop the surfaces of a right square pyramid. (b) Find the volume of the pyramid if the base area is 144cm² and height is 8cm. (c) Find the total surface area of the pyramid. (d) On which base the name of the pyramid is given? Ans: [(b) 384cm3 , (c) Ans: 384 cm2 ] 14 cm 25 cm
Oasis School Mathematics-10 133 3. Study the given figure and answer the questions given below: (a) Which pyramid is formed by the developed surface in the figure? (b) Find the diameter of the base of the solid object if the circumference of the circle on the figure is 44cm. (c) In the figure OA = 25cm, find the area formed by the generator. Ans: [(A) consult your teacher, (b) 14 cm, (c) 550cm2 ] 4. Two solid cones are placed inside a cylindrical tube. The ratio of their capacities are 4:3, (a) Find the heights of the cones. (b) Find the volume of the remaining portion inside the cube, if the area of base is 154 cm2 . Ans: [(a) 24 cm, 18 cm, (b) 4312 cm3 ] 5. A solid consists of a cone and a hemisphere which share a common base. The cone has a height of 24cm and a base diameter of 14cm. (a) The solid is melted and recast to form a cylinder with height of 42cm. Find the radius of the cylinder. (b) If each tin of paint is enough to paint an area of 0.5m², how many tins of paints are required to paint 100 identical cylinders? Ans: [(a) 3.84 cm (b) 22 tins] 6. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of cone is 2cm and the diameter of the base is 4cm. The toy is packed in a cylindrical packed of same base and height as of toy. (a) Find the volume of the cylindrical packet. (b) Find the volume of the toy. (c) How much more space does the cylindrical packet cover? Ans: [(a) 16πcm³, (b) 8πcm2 , (c) 8πcm2 ] 7. In the figure a square pyramid is given where OP = 4cm and PD = 5cm (a) Find the length of BD. (b) Find the volume of the pyramid. (c) Find the length of the side of the base of the pyramid. Ans: [(a) 6cm, (b) 24cm3 , (c) 3 2cm ] 42 cm 24 cm P D B C A O O A B C D E F G H I J K L M
134 Oasis School Mathematics-10 Attempt all the questions: Full Marks: 30 1. Observe the given figure and answer the questions given below. (a) Which two solid objects are there in this combined solid? (1) (b) Calculate the common radius of given solid. (1) (c) Calculate the curved surface area of both solids. Also find the area of base. (3) (d) Calculate CSA and TSA of given object.? (1) 2. In the given square pyramid, (a) What do BC, OF and OE represent? (1) (b) If BC = 10cm, OF = 12cm, then what is the length of OE. (1) (c) Calculate the total surface area of given solid . (2) (d) Calculate its volume (2) 50cm 25cm 25cm D C B F A O E 8. The bottom of the glass has a hemispherical raised portion. The diameter of base is 7cm and height of the glass is 14cm. (a) What quantity of juice can you drink with the glass? (b) If the cost of 1 litre of juice is Rs. 130, how much should be paid for one glass of juice? Ans: [(a) 449.25 cm3 , (b) Rs. 58.40] 9. The length, breadth and height of a room are 7m, 6m and 5m respectively. (a) Find the total cost of parketing the room at Rs. 250 per m² with Rs. 1500 labour charge. (b) Estimate the cost of coloring the walls and ceiling at Rs. 100 per m². Ans: [(a) Rs. 12000, (b) Rs. 17,200] 10. A rectangular water tank having dimensions on the base 50 m × 44 m is filled with water by a pipe having diameter 14 cm. The water flows the water at the rate of 5 km/hr through the pipe. (a) How much water is flown through pipe in 1 hour? (b) Determine the time in which the level of water in the tank will rise by 7cm. (c) Find the cost of filling water in the tank at the rate of Rs. 10 per m³, if the height of the tank is 10 m. Ans: [(a) 77 m3 , (b) 12 minutes, (c) Rs. 222000] 14 cm
Oasis School Mathematics-10 135 3. A semi cylindrical tunnel in made on the rectangular field of dimension 50 ft × 28 ft. as shown in the figure. (a) What is the area of the field? (1) (b) What is the cost of harvesting the vegetable on the field at the rate of Rs. 2 per square feet. (1) (c) If the field is covered by semi cylindrical tunnel, find the area of plastic required to cover it. (2) (d) Find the cost of cover the tunnel by plastics. (1) 4. A school compound gate has two pillar having square base of dimension 2 ft × 2 ft. the height each pillar is 6 ft. (a) What is the perimeter of its base? (1) (b) What is the LSA of each pillar? (2) (c) If both pillars have square based Pyramid 3 ft. on the top, find the surface area of each pillar. (2) (d) Find the cost of painting both pillar at the rate of Rs. 10 per square feet. (1) 5. A man has to construct a water tank of dimensions 3m × 3m × 2m. (a) find the area of the base. (2) (b) find the capacity of the tank. (2) (c) find the area of the inner walls of the tank. (1) (d) find the cost of plastering its walls at the rate of Rs. 90 per m2 . (1) (e) find the cost of to fill the tank with water at the rate of Rs. 5 per litre. (1) Project Work I. Take the measurement of the dimensions of the yard of your home. Find its area. Take the size and rate of tiles from a tile house. Calculate how many tiles you need to cover it. Calculate the total cost of paving the yard. II. Visit a contractor. Ask him or her about the rate of plastering, coloring the wall, ceiling and floor. Measure the dimensions of the rooms of your home and estimate the cost of - • plastering the wall, ceiling and floor. • coloring the four walls.
136 Oasis School Mathematics-10 AlgebrA Contents • Sequence and series • Quadratic Equation • Algebraic Fraction • Exponential Equation expected learning Outcomes At the end of this unit, students will be able to develop the following competencies: • To insert the Arithmetic mean (A.M’s) and Geometric mean (G.M’s) between two numbers. • To calculate the sum of given number of terms in Arithmetic and Geometric sequence. • To solve the quadratic equation using different method. • To solve the verbal problems related to the quadratic equation. • To simplify the sum or difference of two or more than algebraic fractions. • To solve the exponential equations. Materials required • Graph, A4 size paper, etc. Estimated Teaching Hours 32
Oasis School Mathematics-10 137 8.1 Warm-up Activities Let’s discuss the following points in your classroom. • What are the next 2 terms of the sequence 5,12,19, 26, .................? • What is the nth term of the sequence 19, 16, 13, 10, ........................? • Guess the next 3 terms of the sequence 2,6, 18, 54, ..................? • Find the difference between the features of the sequences 2, 5, 8, 11 ................. and 3, 9, 27, .................. ? • If x, y, z are in AP, what is the relation among x, y and z? Arithmetic mean between two numbers ‘a’ and ‘b’. Let ‘a’ and ‘b’ any two numbers. ‘A’ be their arithmetic mean. Then, a, A, b are in A.P. Now, A - a = b - A or, A + A = a + b or, 2A = a + b or, a = a + b 2 Hence, arithmetic mean (A.M.) = a + b 2 . ‘n’ Arithmetic means between ‘a’ and ‘b’. Let ‘a’ and ‘b’ be any two numbers. m1 , m2 , m3 , m4 , ..... mn be ‘n’ arithmetic means between ‘a’ and ‘b’. By the definition, a, m1 , m2 , m3 , ..... mn b are in A.P. Since, there are ‘n’ arithmetic means then the total number of terms in the series be (n + 2). We have, t n = a + (N - 1)d (N = number of terms) or, t n+2 = a + (n + 2 - 1)d or, b = a + (n + 1)d or, b - a = (n + 1)d or, d = b - a n + 1 Unit 8 Sequence and Series
138 Oasis School Mathematics-10 Worked Out examples example 1 Insert 4 A.M.’s between 5 and 25. Solution: Here, Let, m1 , m2 , m3 , m4 be 4 A.M.’s between 5 and 25. Then, 5, m1 , m2 , m3 , m4 , 25 are in A.P. Hence, a = 5, tn = 25, n = 6 We have, t n = a + (n - 1)d 25 = 5 + (6 - 1)d or, 25 - 5 = 5d or, 20 = 5d or, d = 20 5 = 4. Now, m 1 = t 2 = a + d = 5 + 4 = 9 m 2 = t 3 = a + 2d = 5 + 2 × 4 = 13 m 3 = t 4 = a + 3d + 5 + 3 × 4 = 17 m 4 = t 5 = a + 4d = 5 + 4 × 4 = 21. example 2 There are ‘m’ arithmetic means between 3 and 21. If the third mean is 12, find the value of m. Solution: Here, Since, there are m arithmetic means between 3 and 21, then the number of terms in the series be (m + 2). Then,tm+2 = 21. Again, 3rd mean is 12: The fourth term is 12. [3rd mean = 4th term] We have, t n = a + (n - 1)d t 4 = 3 + (4 - 1)d 12 = 3 + 3d 12 - 3 = 3d 9 = 3d ∴ d = 3 Again, tm+2 = a + (m + 2 - 1)d 21 = 3 + (m + 1) × 3 21 – 3 = 3m + 3
Oasis School Mathematics-10 139 18 = 3m + 3 18 - 3 = 3m 15 = 3m ∴ m = 15 3 = 5. exercise 8.1 1. (a) Find the A.M. between 5 and 12. (b) What is the arithmetic mean of –5 and –7? (c) If the A.M. between 2 and a be 18, find the value of a. 2. (a) Insert 5 A.M’s between -7 and 17. (b) Insert 6 A.M’s between -3 and 32. (c) Insert 3 A.M’s between 5 and 65 (d) Insert 4 A.M’s between 24 and 39. 3. (a) There are m A.M’s between 15 and 45 of the third mean 30, find the value of m. (b) There are ‘x’ A.M’s between 3 and 21. If the third mean is 12, find the value of x. 4. (a) There are ‘n’ arithmetic means between 4 and 24. If the ratio of third mean to the last mean is 4:5, find the value of n. (b) There are n A.M’s between 3 and 39. Find the value of n so that third mean : last mean = 3:7. 5. There are two numbers ‘a’ and ‘b’. (a) Write the formula to calculate the arithmetic mean between 2 and 32. (b) If there are 3 arithmetic means between these numbers, what are the means? (c) If there are ‘n’ arithmetic means between 2 and 32; the ratio of last mean to the second mean is 9:4,find the value of n. 6. Some means are inserted between 30 and 3. (a) If the difference between two successive terms is – 3, find how many can be inserted? (b) If there are only 4 A.M’s between these two numbers, what will be the difference between two successive terms? (c) If the ratio of first mean to the last mean is 7:4, find the number of means. 1. Consult your teacher. 2. (a) -3, 1, 5, 9, 13 (b) 2, 7, 12, 17, 22, 27 (c) 20, 35, 50 (d) 27, 30, 33, 36 3. (a) 5 (b) 5 4. (a) 4, (b) 8. 5. (a) 17 (b) 19 2 , 17, 49 2 (c) 5 6. (a) 8 (b) – 27 5 (c) 2 Answers
140 Oasis School Mathematics-10 Sum of the n terms of the Arithmetic Sequence Let ‘a’ be the first term and ‘d’ be the common difference of an Arithmetic sequence, then First term (t1 ) = a Second term (t2 ) = a + d Third term(t3 ) = a +2d ......................................... ......................................... nth term (tn) = a + (n – 1) d. remember ! nth term tn = a + (n – 1)d Common difference (d) = t2 – t1 . =t3 – t2 = t4 – t3 = ................ . Sum of n terms of an Arithmetic Sequence Let a be the first term and ‘d’ be the common difference of an arithmetic sequence. Then First term (t1 ) = a Second term (t2 ) = a + d Third term (t3 ) = a + 2d and so on. Similarly, we get nth term (tn) = a +(n – 1)d. Let Sn represents sum of n terms of an Arithmetic sequence. Then, Sn = t1 + t2 + t3 + t4 + ............ + tn–2 + tn–1 + tn = a + (a + d) + (a + 2d) + (a + 3d) + ....................+ a + (n – 3)d + a + (n – 2)d + a + (n – 1)d = na + [d + 2d + 3d + .................. + (n – 3)d + (n – 2)d + (n – 1)d] = na + d [1 + 2 + 3 + .............. + (n – 3) + (n – 2) + (n – 1)] = na + d [(1 + n – 1) + (2 + n – 2) + (3 + n – 3) + ............................... to n–1 2 term] = na + d [n + n + n + ........... to n –1 2 term] = 2na + nd (n – 1) 2 = na + d.n(n – 1) 2 = n[2a + (n – 1)d] 2 = n 2 [2a + (n – 1)d] Hence, the sum of n terms of an Arithmetic sequence Sn = n 2 [2a + (n – 1)d] Sum of n terms of an Arithmetic sequence in terms of a and l. Let ‘a’ be the first term and ‘l’ be the last term of the arithmetic sequence. Then sum of n terms of arithmetic sequence Sn = a + (a + d) + (a + 2d) + (a + 3d) + ............. + (l – 3d) + (l – 2d) + (l – d) + l. .............(i)
Oasis School Mathematics-10 141 Again, Sn = l + (l – d) + (l – 2d) + (l – 3d) + ................+ (a + 3d) + (a + 2d) + (a + d) + a. ......................(ii) Adding equation (i) and (ii) 2Sn = (a + l) + [(a + d) + (l – d)] + [(a + 2d) + (l – 2d)] + [(a + 3d) + (l – 3d)] + ............. to n terms or, 2Sn = (a + l) +(a + l) + (a + l) + (a + l) + ................... to n terms or, 2Sn = n (a + l) or, Sn = n 2 (a + l) Hence, the sum of n terms of an A.S is Sn = n 2 (a + l) Again, Sn = n 2 [a + a +(n – 1)d] [l = tn = a + (n – 1)d] = n 2 [2a + (n – l)d] Hence, the sum of n terms of an A.S is Sn = n 2 [2a + (n – l)d] remember ! • If ‘a’, ‘d’ and ‘n’ are given, Sn = n 2 [2a + (n – l)d] • If ‘a’, ‘l’ and ‘n’ are given, Sn = n 2 (a + l) • Three terms of an A.S can be assumed as (a –d), a and (a + d). • Five terms of an A.S can be assumed as (a – 2d), (a – d), (a + d) and (a + 2d). Worked Out examples example 1 Find the first term (a), common difference (d) and the sum of the series 3 = 5 + 8 + 11 + ........ to 8 terms. Solution: Given series, 3 + 5 +8 + 11 + ............. to 8 terms. Here, first terms (a) = 3. common difference (d) = 5 – 3 = 2. number of terms (n) = 8.
142 Oasis School Mathematics-10 We have, Sum of terms Sn = n 2 [2a + (n – l)d] = 8 2 [ 2 × 3 + (8 – 1).2] = 4 [6 + 14] = 4 × 20 = 80. example 2 Find the sum of the series 24 + 20 + 16 + ................. + (–4). Solution: Given series, 24 + 20 + 16+ ............ + (–4). Here, first terms (a) = 24, common difference (d) = 20 – 24 = – 4. nth term (tn) = –4 We have, or, t n = a + (n – 1) d or, –4 = 24+ (n – 1) (–4) or, –4 = 24 – 4n + 4 or, –4 = 28 – 4n or, 4n = 28 + 4 or, 4n = 32 n = 8. Again, Sn = n 2 (a + l) = 8 2 (24 + (–4) = 4 × 20 = 80 example 3 From the series from the given sigma notation and find the sum ∑ 6 n=1 2n – 3 Solution: Given tn = 2n – 3. When n = 1, t 1 = 2 × 1 – 3 = –1 When n = 2, t 2 = 2 × 2 – 3 = 1 When n = 3, t 3 = 2 × 3 – 3 = 3 When n = 4, t 4 = 2 × 4 – 3 = 5 When n = 5, t 5 = 2 × 5 – 3 = 7
Oasis School Mathematics-10 143 When n = 6, t 6 = 2 × 6 – 3 = 9 Hence, ∑ 6 n=1 2n – 3 = –1 + 1 + 3 + 5 + 7 + 9 Here, First term (a) = –1 Common difference (d)= 1 – (–1) = 2. Number of terms (n) = 6. We have, Sn = n 2 [2a + (n – l)d] = 6 2 [2×(–1) +(6 – 1) × 2] = 3 [–2 + 10] = 3 × 8 = 24. example 4 Find the first term of an arithmetic progression whose common difference is 4 and the sum of its first 20 terms is 820. Solution: Given, Common difference (d) = 4 Sn = 820, n = 20, a = ? We have, Sn = n 2 [2a + (n – l)d] or, 820 = 20 2 [2a + (20 – 1)4] or, 820 = 10 (2a +76) or, 820 10 = 2a + 76 or, 82 = 2a +76 or, 82 – 76 = 2a or, 6 = 2a or, a = 6 2 = 3. Hence, the first term (a) = 3. example 5 The 3rd and 11th term of an arithmetic series 8 and –8, find the sum of first 10 terms. Solution:
144 Oasis School Mathematics-10 Given, 3rd term (t3 ) = 8 11th term (t11) = –8 Sum of first 10 terms(s10) = ? We have, t n = a + (n – 1)d or, t 3 = a + (3 – 1)d or, 8 = a + 2d or, a = 8 – 2d .............. (i) Again, t 11 = a + (11 – 1)d or, –8 = a + 10d or, a = – 8 – 10d ....................... (ii) From equations (i) and (ii). 8 – 2d = – 8 – 10d or, –2d + 10d = –8 – 8 or, 8d = –16 ∴ d = –2 Substituting the value of ‘d’ in (i) a = 8 – 2d = 8 –2(–2) = 8 + 4 = 12. Again, Sn = n 2 [2a + (n – l)d] S10 = 10 2 [2×12 +(10 – 1) (–2)] = 5 [24 –18] = 5 × 6 = 30 example 6 The sum of first 9 terms of an arithmetic series is 72 and the sum of first 17 terms is 289. Find the sum of first 25 terms. Solution: Given, Sn = 72, when n = 9 Sn = 289 when n = 17 Sn = ? when n = 25.
Oasis School Mathematics-10 145 We have, Sn = n 2 [2a + (n – l)d] S9 = 9 2 [2a + (9 – 1)d] or, 72 = 9 2 [2a + 8d] or, 72 = 9 2 × 2 (a + 4d) or, 72 = 9 (a + 4d) or, 72 9 = a + 4d or, 8 = a + 4d ∴ a = 8 – 4d ............... (i) Again, S17 = 17 2 [2a +(17 – 1)d] 289 = 17 2 [2a + 16d] 289 = 17 2 × 2 (a + 8d) 289 17 = a + 8d 17 = a + 8d ∴ a = 17 – 8d ............. (ii) From equations (i) and (ii) 8 – 4d = 17 – 8d or, 8d – 4d = 17 – 8 or, 4d = 9 or, d = 9 4 Substituting the value of d in (i) a = 8 – 4d = 8 – 4 × 9 4 = –1. Again, Sn = n 2 [2a + (n – l)d] S25 = 25 2 [2(–1) + (25 – 1). 9 4] = 25 2 [–2 + 24 × 9 4 ] = 25 2 [–2 + 54]
146 Oasis School Mathematics-10 = 25 2 × 52 = 25 × 26 = Rs. 650. Example 7 If the sum of the Arithmetic series 2 + 5 + 8 +......... to n terms is 155, find the value of n. Solution: Given series, 2 + 5 + 8 + ........................ Here, first term (a) = 2. common difference (d) = 5 – 2 = 3. sum of n terms (Sn) = 155. we have, Sn = n 2 [2a + (n – l)d] 155 = n 2 [2 × 2 + (n – 1) × 3] or, 155 = n 2 [4 + 3n – 3] or, 310 = n (3n+ 1) or, 3n² + n – 310 = 0 or, 3n² + (31 – 30)n – 310 = 0 or, 3n² + 31n – 30n – 310 = 0 or, n(3n +31) – 10 (3n +31) = 0 or, (3n + 31) (n – 10) = 0 Either, 3n + 31 = 0 ⇒ 3n = –31 n = –31 3 (not possible). or, n – 10 = 0 ⇒ n = 10 Hence, the given series has 10 terms. example 8 The sum of three numbers of an A.S is 51. The sum of their squares is 939. Find the numbers. Solution: Let, the 3 numbers in A.S be (a – d), a and (a + d) Now, a – d + a + a + d = 51