The Leading Maths - 10

Mensuration 100 Allied The Leading Mathematics-10 Area and Volume 101 Example-1 In the given right pyramid, the vertical height is 12 cm, side length of square base is 10 cm. (a) Define pyramid. (b) Find the slant height of the given pyramid. (c) Find the total surface area and volume of the pyramid. Solution : (a) A monumental structure with a polygonal base and sloping its sides meet at a point at the top is called pyramid. (b) Here, in a pyramid, the square is a base of side (a) = 10 cm and height (h) = 12 cm. ∴ Slant height (l) = 1 2 4h2 + a2 = 1 2 4 × 122 + 102 = 1 2 676= 1 2 × 26 = 13 cm. (c) Lateral surface area (LSA) = 2al = 2×10×13 = 260 cm2 . Area of square base (A) = a2 = 102 = 100 cm2 . ∴ Total surface area (TSA) = LSA + A = 260 + 100 = 360 cm2 . ∴ Volume of pyramid (V) = 1 3 × Base area × height = 1 3 × 100 × 12 = 400 cm3 . Example-2 Observe the given right pyramid. (a) What is a right pyramid ? Define it. (b) What is the vertical height of the given pyramid ? Find. (c) Find the volume of the pyramid. Solution : (a) A pyramid in which its apex is directly above the centroid of its base, is called a right pyramid. (b) Here, in the given figure, the square base side of the pyramid (a) = 10 cm, slant height (l) = 13 cm ∴ Vertical height (h) = l 2 – (a/2)2 = 132 – (10/2)2 = 169 – 25= 144 = 12 cm Now, we have (c) Volume of the given pyramid (V) = 1 3 × Base area (A) × height (h) = 1 3 a2 h = 1 3 × 100 × 12 = 400 cm3 . 10 cm 12 cm 10 cm 10 cm 10 cm 13 cm

Mensuration 102 Allied The Leading Mathematics-10 Area and Volume 103 Example-3 From the given pyramid, (a) Find the length of the side of the square base. (b) What is the slant height of the pyramid ? Find. (c) Find the total surface area of the pyramid. Solution : (a) In the given pyramid, the height (h) = 6cm, the length of edge side (e) = 10 cm. In rt. ∠d ∆POC, CP = e2 – h2 = 102 – 62 = 100 – 36 = 64 = 8 cm ∴ AC = 2 × CP = 2 × 8 = 16 cm Again, in rt.∠d ∆ABC, AB2 + BC2 = AC2 or, a2 + a2 = 162 or, 2a2 = 256 or, a2 = 128 ∴ a = 8 2 cm. (b) Slant height (l) = 1 2 4e2 – a2 = 1 2 4 × 102 – (8 2 ) 2 = 1 2 400 – 128 = 1 2 272 = 1 2 × 4 17= 2 17 cm (c) Now, we have TSA of pyramid (TSA) = 2al + a2 = 2 × 8 2× 2 17 + (8 2 ) 2 = 32 34+ 128 = 314.59 cm2 . PRACTICE 5.1 Read Think Understand Do Keeping Skill Sharp 1. (a) Write the formula to calculate the volume of a pyramid when its height and the base area are given. (b) Write the formula to calculate the volume of a squared pyramid when its height and length of base are given. (c) What is the formula to calculate the lateral surface area of a pyramid when its slant height and length of base are given? (d) What is the total surface area of a squared pyramid having slant height l cm and base length a cm? a a A B C D P O l 10 cm 6 cm

Mensuration 102 Allied The Leading Mathematics-10 Area and Volume 103 (e) Write down the relation among base length ‘a’ vertical height h and slant height l of a square base pyramid. (f) Write down the relation among the vertical height ‘h’ base diagonal; ‘d’ and edge ‘e’ of a square pyramid. 2. Circle ( ) the correct answer. (a) The formula to calculate the volume of a squared base pyramid of the height b cm and base side a cm ............ . (i) V = ab cm2 (ii) V = ab cm3 (iii) V = a2 b cm3 (iv) V = 1 3 a2 b cm3 (b) Which is he correct relation between the base length ‘a’ vertical height 'h' and slant height 'l' of a squared base pyramid ? (i) h = l 2 – (a/2)2 (ii) l = 1 2 4h2 + a2 (iii) a = 2 l 2 – h2 (iv) All of the above (c) The relation between base length 'a’, slant height 'l' and edge ‘e’ of a squared pyramid ......... . (i) l = e2 – (a/2)2 (ii) a = 2 e2 – l 2 (iii) e = l 2 + (a/2)2 (iv) All of the above (d) In the right squared base pyramid, the vertical height is 4 cm, side length of square base is 6 cm. What is its slant height ? (i) 4 cm (ii) 12 cm (iii) 24 cm (iv) 5 cm (e) What is the volume of a pyramid whose base area is 60 cm2 and vertical height, 8 cm ? (i) 160 cm2 (ii) 160 cm3 (iii) 480 cm3 (iv) 480 cm2 (f) What is the lateral surface area of a pyramid whose base length is 10 cm and slant height 10 cm ? (i) 1000 cm2 (ii) 100 cm2 (iii) 200 cm2 (iv) 50 cm2 (g) What is the total surface area of a pyramid whose base length is 5 ft and slant height 10 ft? (i) 120 ft2 (ii) 150 cm2 (iii) 500 ft2 (iv) 125 ft2 Check Your Performance Answer the given questions for each problem. 3. (i) (ii) (iii) (iv) (a) Define a pyramid. (b) Find the vertical height of each pyramid. (c) Find the lateral surface area, total surface area and volume for each pyramid. 12 cm 15 cm 15 cm 18 cm 18 cm 15 cm 16.5 cm 16.5 cm 20.3 cm 8 cm 10 cm

Mensuration 104 Allied The Leading Mathematics-10 Area and Volume 105 4. (i) Observe the given square pyramid. (a) Why is the given pyramid a right pyramid ? (b) Find the vertical height of the given pyramid. (c) Find its slant height. (d) Find its lateral surface area. (ii) Observe the given pyramid. (a) Why is the given pyramid a square pyramid ? (b) Find the slant height of the given pyramid. (c) Find its vertical height. (d) Find its volume. (iii) Observe the given right pyramid. (a) Compute the slant height of the given pyramid. (b) Find the edge of the given pyramid. (c) Find its vertical height. (d) Find its volume. 5. (i) (a) If the base area of a pyramid is 45 cm2 and its height is 9 cm, find its volume. (b) Find the volume of a pyramid inscribed in a cube with base area 100 cm2 . (c) If the volume of a pyramid with squared base area 2.25 m2 is 6.75 m3 , find its height. (ii) (a) If the height of a pyramid with squared base area 64 cm2 is 8 cm, find its slant height. (b) If the height of a square pyramid with volume 384 cm3 is 8 cm, find its slant height. (c) If the base area of a square pyramid of volume 1000cm3 is 100cm2 , find its slant height. (iii) (a) If the volume of a square pyramid with base side 16 cm is 1536 cm3 , find its slant height. (b) If the volume of a square pyramid having height 12 cmis 900 cm3 , find its length of base side. (c) If the volume of a square pyramid having base side 24 cm is 5760 cm3 , find its length of slant height. 6. (i) Find the slant height of the adjoining square pyramid with volume 243 cm3 in cube. (ii) Find the slant height of the adjoining square pyramid with volume 486 cm3 whose height is double of the base side. 10 cm Volume = 400 cm3 12 cm LSA = 240 cm2 14 cm TSA = 476 cm2 (i) (ii)

Mensuration 104 Allied The Leading Mathematics-10 Area and Volume 105 7. (i) The total surface area of a squared base pyramid with base side 6 cm is 96 cm2 . (a) Find its slant height. (b) Find its height. (c) Find its volume. (ii) The lateral surface area of a square pyramid with base side 8 cm is 80 cm2 . (a) Find its slant height. (b) Find its height. (c) Find its volume. 8. (i) A girl plans to construct a temporary tunnel in the shape of square pyramid by using plywood. She needs the height of the tunnel 4 m and the length of side on the ground 6 m. (a) Find the lateral surface area of the tunnel. (b) How much quantity of air is contained in it? Find. (ii) In a marriage ceremony of Dolma's daughter, there was arrangementfor accommodation of 200 persons. For this purpose she plans to build a pyramid shaped tent such that each person must have 4 sq. ft. of the space of ground and 10 cubic ft. of air to breath in average. (a) What should be the height of the tent ? Find it. (b) Find the lateral surface area of the tent. 3. (i) (b) 12 cm (c) 540 cm2 , 864 cm2 , 1296 cm3 (ii) (b) 15 cm (c) 291.72 cm2 , 381.72 cm2 , 360 cm3 (iii) (b) 16.61 cm (c) 612.15 cm2 , 884.4 cm2 , 507.36 cm3 (iv) (b) 5.29 cm (c) 48 cm2 , 192 cm2 , 253.92 cm3 4. (i) (b) 12 cm (c) 13 cm (d) 260 cm2 (ii) (b) 10 cm (c) 8 cm (d) 384 cm2 (iii) (a) 10 cm (b) 12.21 cm (c) 7.14 cm2 (d) 466.48 cm2 5. (i) (a) 135 cm (b) 333.33 cm (c) 9 cm (ii) (a) 135 cm (b) 10 cm (c) 30.41 cm (iii) (a) 18.36 cm (b) 15 cm (c) 32.31 cm 6. (i) 10.06 cm (ii) 18.55 cm 7. (i) (a) 5 cm (b) 4 cm (c) 48 cm3 (ii) (a) 5 cm (b) 3 cm (c) 64 cm3 8. (i) (a) 62 cm2 (b) 48 m3 (ii) (a) 7.5 ft. (b) 16.005 ft2 . Answers

Mensuration 106 Allied The Leading Mathematics-10 Area and Volume 107 5.2 Right Cone At the end of this topic, the students will be able to: ¾ find the CSA, TSA and Volume of cone and solve related problems. Learning Objectives I Introduction to Cone Lets discuss the following activities. Activity 1 Draw a circle with radius 10 cm and cut down a sector of 120°. When we join the radii of the sector, what type of solid object is it? Discuss the solid formed by this sector. Activity 2 Study the pyramids in which the number of sides of bases are increasing from 3 to many number of sides. ....... ....... Icosagonal Pyramid Pyramid (Cone) Hexagonal Pyramid Heptagonal Pyramid Pentagonal Pyramid Square Pyramid Triangular Pyramid Octagonal Pyramid What kind of pyramid do you find at last? It will be formed a circle based pyramid. In general, a cone is a solid formed by joining the points of a circle to a point not in the plane of the circle with straight lines. How many vertices, edges and faces are in that figure ? This is the shape of birth day cap or ice cream cone or topper part of pencil/pen, traffic cone, funnel, tent, top of stupa/ temple/mosque, carrot/radish, tree, ice/sand/pebble stupa, etc. II Parts and Types of Cone Activity 3 When a right triangle ACB is rotated about a vertical stick XY as shown in the fig. (a), what kind of shape do you find ? A cone is a circular base pyramid that is generated by a right triangle. The vertical stick XY is called the axis of the cone. 10 10 Ice stupa made by Engineer, innovator and education reformist Sonam Wangchuk for Preserving water in Ladakh X Y Y X B C l r h B C A A

Mensuration 106 Allied The Leading Mathematics-10 Area and Volume 107 The smooth lateral surface generated by the hypotenuse of the right triangle is called curved surface and its length that is represented by its hypotenuse AC is called slant height (l) of the cone that terminates at the point A, called vertex or apex. The base BC of the right triangle becomes the radius (r) and that of perpendicular AB, vertical height (h). The base of a cone becomes a circle. When the vertical height of the cone is perpendicular to the centre of its circular base is called a right cone. Otherwise,it is called oblique cone. III Surface Area of Cone Activity 4 Let us consider a cone having a circular base of radius r and height h. In the right ∆VOB, VB (l) = h2 + r 2 . The net of the cone is as alongside; The sector ACBV is the lateral surface area or the curved surface area of the cone which is the sector of a circle with radius l, slant height of the cone. Arc length of the sector is equal to the circumference of the circular base of the cone. Let’s cut the sector into 16 equal pieces and arrange them into a rectangular form at where the length of the sector (circumference of circular base = 2πr) divided into two equal parts as πr and πr as shown in the adjoining figure. Here, the length of the rectangle is πr and breadth is l. From this we can say that, Curved Surface Area of Cone (CSA) = Area of Rectangle = l × b = πr × l = πrl. The total surface area of the cone is the sum of area of the sector (curved surface area) and the area of the circular base (A) of the cone. i.e., Total Surface Area of Cone (TSA) = A + CSA = πr2 + πrl = πr(r + l). A l h h C Z W r r Right Cone B" Oblique Cone X V Y l V A B h O r C V A B C r l l l πr l 2πr = circumference Slant height Lateral surface Vertex Height Base r Axis

Mensuration 108 Allied The Leading Mathematics-10 Area and Volume 109 "Alternatively" The sector formed by the given cone is divided into n-triangles with height l and bases b1, b2, b3, b4, ....., bn as shown in the figure. Then, Area of sector (CSA) = Areas of n-triangles = 1 2 b1l + 1 2 b2l + 1 2 b3l + 1 2 b4l + ........ + 1 2 bnl = 1 2 (b1+ b2 + b3 + b4 + ........ + bn) l = 1 2 Circumference of circular base of the cone × l = 1 2 × 2πr × l = πrl In next method, Complete a circle taking slant height l as radius. Then, the area of the sector can be calculated as; Arc length of the sector (c) Circumference of circle with radiusl (c1) = Area of sector Area of circle with radius l or, 2πr 2πl = Area of sector πl 2 ∴ Area of the sector (CSA) = πrl = 1 2 Circumference (c) × slant height Therefore, Surface area of cone = A (base) + 1 2 c (base circumference) × slant height IV Volume of Cone The volume of a cone is the amount of space occupied by a cone in a three-dimensional plane. We can easily derive the formula to find the volume of a cone by the following methods. Activity 5 Let us take a right cone and a right cylinder with the same height and diameter of the base equal to each other. Fill the cone by any substance and pour into the cylinder three times. The cylinder is exactly fulfilled. That means three times of the volume of the cone is equal to the volume of the cylinder. Thus, 3 × Volume of Cone = Volume of Cylinder or, Volume of Cone = 1 3 πr 2 h "Alternatively" Merge the cone into the cylinder filled by water with the same height and height of the cone, which replaces the water and collect it in another same sized cylinder, which onethird of the total water in it and the remaining water is that of two-thirds. l b1 b2 b3 bn b4 ........ ........ l l Sector c = 2 πr c1 = 2 πl h Equal bases

Mensuration 108 Allied The Leading Mathematics-10 Area and Volume 109 ∴ Volume of Cone, V = 1 3 Vol. of Cylinder = 1 3 πr 2 h. In short, Volume of Cone = Volume of prism = 1 3 Ah = 1 3 πr 2 h "Next Method" We can also find the formula of the volume of a cone by using; (i) Two cones and hemi-sphere with equal height and radius. Volume of cone = 1 2 Volume of Hemi-sphere = 1 2 × 2 3 πr 3 = 1 3 πr 2 × r = 1 3 πr 2 h cu. units. (ii) Four cones and sphere with equal height and radius. Volume of cone = 1 4 Volume of sphere = 1 4 × 4 3 πr 3 = 1 3 πr 2 h cu. units. Remarks: (i) If h = d = 2r of the cone, the volume of the cone is 2 3 πr3 , which is the volume of the hemi-sphere. (ii) When the radius and height of cone, hemisphere and cylinder are equal, the ratio of the volume of cone, hemi-sphere and cylinder (1/3πr 3 , 2/3πr 3 , πr 3 ) is in the ratio 1:2:3. (iii) When the diameter and height of cone, sphere and cylinder are equal, the ratio of the area and volume of cone, sphere and cylinder (1/3πr 3 , 2/3πr 3 , πr 3 ) are both in the ratio 1:2:3. (iv) The relation between the volume of cone,sphere and cylinder with equal length of diameter and height is 2 3 πr3 + 4 3 πr3 = 2πr3 . Merging Cone Overflowed water Remaining water h h = r r h h r h r h = r h r h r h r h r d h = d r r = = 1

Mensuration 110 Allied The Leading Mathematics-10 Area and Volume 111 Points to be Remembered 1. A circular base pyramid is called a cone. 2. The generator of a cone is called the axis of the cone. 3. The smooth lateral surface of a cone is called curved surface. 7. The circular flat surface of a cone is called a base. 5. The point opposite the base of a cone is called vertex or apex. 4. The perpendicular length of an apex from a point of circular base of a cone is called slant height (l). 6. The vertical distance of the vertex from the centre of the base of a cone is called vertical height (h). 7. When the vertical height of the cone is perpendicular to the centre of its circular base is called a right cone. Otherwise,it is called oblique cone. 8. Curved Surface Area of Cone (CSA) = πrl sq. units. 9. Total Surface Area of Cone (TSA) = πr(r + l) sq. units. 10. Volume of Cone (V) = 1 3 πr2 h cu. units. Example-1 Observe the given conical tent and answer the following questions. (a) What is a cone ? Define. (b) Find the slant height of the given cone. (c) Find the lateral surface area, total surface area and volume of the cone. (d) Calculate the cost of canvas required for making the tent at Rs. 35 per sq. m. Solution: (a) A pyramid with circular base is called a cone. (b) Here, Height of cone (h) = 48 cm, Radius of base (r) = 28 2 = 14 cm Slant height (l) = h2 + r 2 = 482 + 142 = 2500 = 50 cm (c) Area of the curved surface (CSA) = πrl = 22 7 ×14 × 50 = 2200 cm2 Total surface area of the cone (TSA) = (πrl + πr 2 ) = 22 7 × 14 × 50 + 22 7 × 14 × 14 = 2816 cm2 . Slant height Lateral surface Vertex Height Base r Axis V A B c = 2πr r Sector r r r h h h = d Sphere Cone h = d 48 cm 28 cm

Mensuration 110 Allied The Leading Mathematics-10 Area and Volume 111 Volume of the cone = 1 3 πr 2 h = 1 3 × 22 7 × 14 × 14 × 48 = 9856 cm3 . (d) Cost of canvas = Rs. 35 × 2200 = Rs. 77000. Example-2 A cone of height 20 cm and diameter 16 cm is completely filled with water. The water is poured in a cuboid of dimension 26 cm × 18 cm. (a) Calculate the volume of the water in the cuboid. (b) Find the height of water in the cuboid. Solution: (a) For cone, Diameter (d) = 16 cm, Radius (r) = d 2 = 8 cm, Height (h) = 20 cm We know that, Volume of water = Volume of the cone = 1 3 πr 2 h = 1 3 × 22 7 × 8 × 8 × 20 cm3 . (b) For cuboid; Length (l) = 26 cm, Breadth (b) = 18 cm, Height of water = h = ? Volume of water in a cuboid = 26 × 18 × h Hence, 1 3 × 22 7 × 8 × 8 × 20 = 26 × 18 × h That is, h = 1 3 × 22 7 × 8 × 8 × 20 × 1 26 × 18 = 2.86 cm. Hence, the height of the water in the cuboid is 2.86 cm. Example-3 There are two cones. The curved surface area of the one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii. Solution: Let, h1, r1 and l1 be the height, radius and slant height of the 1st cone. Similarly, h2, r2 and l2 are the height, radius and slant height of the 2nd cone. Also, l2 = 2l1 By question, CSA of 1st cone = 2 × CSA of 2nd cone πr1l1 = 2 × πr2l2 or, r1 l1 = 2 × r2 × 2l1 So, r1 : r2 = 4:1. 16 cm 20 cm 20 cm 18 cm h = ?

Mensuration 112 Allied The Leading Mathematics-10 Area and Volume 113 Example-4 A conical tent is to accommodate 110 persons. Each person must have 1 m2 of the space on the ground and 2 m3 of air to breathe. Find the height of the tent. Solution: Here, No. of persons = 110 Area required for each person = 1m2 ∴ Base area (A) = 110 × 1 = 110 m2 . Again, Air required for each person = 2m3 . ∴ Volume of the cone (V) = 110 × 2 = 220 m3 . Also, V = 1 3 × Base area × Height or, 220 = 1 3 × 110 × h or, h = 6 m Hence, the height of cone is 6m. PRACTICE 5.2 Read Think Understand Do Keeping Skill Sharp 1. (a) Define a right cone. (b) Write the formula to compute the volume of a cone when its height and base area are given. (c) Write the formula to compute the volume of a cone when its height and radius of the base are given. (d) What is the curved surface area of a cone having the slant height ‘l’ cm and radius ‘r’ cm of the base? (e) Write down the relation between the height, slant height and radius of a cone. 2. Circle ( ) the correct answer. (a) The volume of a right cone of the height b cm and base side a cm ............ . (i) V = ab cm2 (ii) V = ab cm3 (iii) V = a2 b cm3 (iv) V = 1 3 a2 b cm3 (b) Which is the correct relation between the radius 'r' vertical height 'h' and slant height 'l' of a squared base pyramid ? (i) l = h2 + r 2 (ii) l = r 2 – h2 (iii) l = h2 – r 2 (iv) l = h2 + r 2 (c) In the right cone, the vertical height is 4 cm and base area is 42 cm2 . What is its volume? (i) 84 cm3 (ii) 168 cm3 (iii) 56 cm3 (iv) 24 cm3 (d) What is the curved surface area of a right cone having radius 7 cm and slant height 13 cm? (i) 386 cm3 (ii) 168 cm2 (iii) 186 cm2 (iv) 286 cm2

Mensuration 112 Allied The Leading Mathematics-10 Area and Volume 113 (e) The volume of a right cone of the base area 25 cm2 is 75 cm3 . What is its height ? (i) 9 cm (ii) 18 cm (iii) 6 cm (iv) 12 cm (f) The CSA of a right cone of the base radius 7 cm is 176 cm2 . What is its height ? (i) 4 cm (ii) 7 cm (iii) 8 cm (iv) 10 cm Check Your Performance Answer the given questions for each problem. 3. Observe the following cones: (i) (ii) (iii) (iv) (a) Write the definition of a cone. c=22 ft. (b) Find the unknown parts of the given cone. (c) Find the curved surface area, total surface area and volume for each cone. 4. (i) (a) The volume of a cone is 1848 cm3 . If the radius of its circular base is 14 cm, find the height of the cone. (b) The height of the right circular cone is 50 cm and its volume is 23100 cm3 . What is the radius of the base of the cone? (ii) (a) The radius and height of a right cone are equal. If its volume is 9702 cm3 , find the circumference of its circular base. (b) The volume of a right cone is 8624 cm3 . If its diameter and height are equal, what will be the whole surface area of the cone? (iii) (a) A cone having radius 70 m has a curved surface area 33000 m2 . Find its volume. (b) A cone of slant height 25 cm has a curved surface area 550 cm2 . Find its volume. 5. (i) The curved surface area of a conical shelter with slant height 50 ft is 220 ft2 . (a) Find the radius of the base. (b) Find the height of the roof. (c) Find its volume. (ii) The curved surface area of a cone with base radius 7 cm is 550 sq. cm. (a) Find the slant height of the cone. (b) Find its height. (c) Find its volume. 6. (i) If the total surface area of a cone whose base circumference 22 m is 148 1 2 sq.m, find the slant height of the cone. (a) Find the slant height of the base. (b) Find the height of the cone. (c) Find its volume. 5 cm 12 cm 3 3 cm 2 cm 5 2 cm 5 cm 9 ft.

Mensuration 114 Allied The Leading Mathematics-10 Area and Volume 115 (ii) The total surface area of a right circular cone with diameter of base 24 cm is 300π cm2 . (a) Find the slant height of the base. (b) Find its height. (c) Find its volume. 7. (i) A tent of height 22 m is in the form of a right circular cone of the radius 60 m. Find the curved surface area and total surface area of the tent. Also, calculate the capacity of air contained in the tent. (ii) If the diameter of a conical circular well is 28 m and slant height of the cone is 21 m, calculate the area of the plastic required to cover its curved area and the area of lid to cover its opened circular mouth. Also, calculate the capacity of the well. (iii) The radius of the base of a right circular cone made by iron is 15 cm. The conical slant height is 39 cm. Find the weight of the cone if one cubic centimeter of iron weighs 7.8 gram. Also, calculate its CSA and TSA. 8. (i) The diameters of two cones are equal. If their slant heights are in the ratio 5:4, find the ratio of their curved surface areas. If the CSA of the first cone is 550 cm2 , find the CSA of the second cone. (ii) Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 2:1. Show that their volumes are in the ratio 4:3. If the diameter and height of the first cone is 28 cm, find TSA of the second cone. 9. (i) Find what length of canvas of 1.5 m in width is required to make a conical tent 9 m in 14 m respectively. Find the cost of constructing it at Rs. 20 per m3 and also find the cost of green paint on its curved surface at Rs. 8 per m2 . 3. (i) (b) 13 cm (c) 204.29 cm2 , 282.86 cm2 , 314.16 cm3 (ii) (b) 3 cm (c) 18.86 cm2 , 28.29 cm2 , 9.42 cm3 (iii) (b) 5 cm (c) 111.12 cm2 , 189.69 cm2 , 130.95 cm3 (iv) (b) 8.29 ft. (c) 99 ft2 , 137.5 ft2 , 106.39 ft3 4. (i) a. 9 cm b. 21cm (ii) a. 132 cm b. 2618.25 cm3 (iii) a. 680988 m3 b. 1232 cm3 5. (i) a. 14 ft b. 48 ft c. 9856 ft3 (ii) a. 25 ft b. 24 ft c. 1232 6. (i) a. 10 m b. 9.37 m c. 120.25 m3 (ii) a. 13 cm b. 5 cm c. 754.29 m3 7. (i) 12051.6m, 23365.89 m, 82971.43 m3 (ii) 1540 m2 , 3207.31 m3 8. (i) 66188.57 gm, 1838.57 cm2 , 2545.71 cm2 (ii) 5:4, 440 cm2 9 (i) 59.40 m, Rs. 11591.91, 21 persons (ii) Rs. 24640, Rs. 4400 Answers

Mensuration 114 Allied The Leading Mathematics-10 Area and Volume 115 5.3 Combined Solids At the end of this topic, the students will be able to: ¾ find the CSA, TSA and Volume of combined two solids. Learning Objectives I Introduction We come across many objects with basic shapes like cubes, cuboids or prisms, cylinder or pyramids, cones or spheres, hemi-spheres and with the known formulae of area and volume that have been derived in the earlier chapter. Now, these basic shapes conjoin and form a shape different from the original. The shapes formed by the combination of different shapes are called composite shapes. The composite solid is a solid made up of common geometric solids. The solids are generally prisms, pyramids, cones, cylinders, spheres and hemi-spheres. In order to find the surface area and volume of a combined solid, we need to know how to find the surface area and volume of prisms, pyramids, cones, cylinders, spheres and hemispheres. How will we derive the measurements of the new shape? The surface area is the area that describes the material that will be used to cover a geometric solid. The surface on sloping form of the solid isthe lateralsurface. When we determine the surface areas of a geometric solid we take the sum of the area for each geometric form within the solid. The volume is a measure of how much a solid can need to make it. In our daily life, we are using combined solids in many fields such as designing of top of roof and wall, tents, capsules, and icecream cones, LP gas or petrol tankers, pencils, etc. Discuss the combined solids as shown in the pictures alongside. a2 4ah 4al + 2 a2 a2 h 1 3 a2 h 2al 2al+a2

Mensuration 116 Allied The Leading Mathematics-10 Area and Volume 117 II Combined Solids Formed by Prism and Pyramid The combined solids formed by the prisms, pyramids or both are given below: 4 cm 6 cm 8 cm 5 cm 13cm 13cm 8 cm 20 in 15 in 15 in 13 in Can you draw the nets of the above combined solids separately ? We can find the lateral surface area (LSA), total surface area (TSA) and volume (V) of the above combined solid. From the above activities we derive the following relations: SN Combined Solid LSA TSA Volume 1. a h l b c Triangular Prism+Cuboid TSA of Solid–3Base LSA of Solid + Base Area Vol. of Prisms 2h(l + b) + 2b(a + c) + 2A 2h(l+b)+ lb+2b(a+c)+2A 'OR' 2(lb+lh+bh) + b(a+l+c) + 2A – 2lb lbh + Ab 2. a l1 l2 h2 h1 h Pyramid+ Pyramid (Square) LSA of Pyramids LSA of Solid Vol. of Pyramids 2a(l1 + l2) 2a(l1 + l2) 3 12 a2 h 'OR' 3 12 a2 (h1 + h2) 3. l h2 a h1 Prism+ Pyramid (Square) LSA (Pyr.+Pri.) LSA of Solid + Base Area Vol. (Pyr.+Prism) 2a(2h1 + l) 2a(2h1 + l) + a2 a2 h1 + 1 3 h2 III Combined Solids Formed by Cylinder with Cone or Hemisphere or Sphere The combined solids formed by the cylinder with cone or hemisphere or sphere are given below. Which separate solids form these combined solids ? 30 cm 50cm 14cm 15 cm 7cm 10cm 10.5cm 4cm 12 cm Can you draw the nets of the above combined solids?

Mensuration 116 Allied The Leading Mathematics-10 Area and Volume 117 Now, find their lateral surface area, total surface area and volume separately. From the above activities we derive the following relations: SN Combined Solid CSA TSA Volume 1. h2 l h1 r Cylinder + Cone CSA (Cylinder + Cone) CSA (Solid + Base Area) Vol.(Cylinder + Cone) πr(l + 2h1) πr(l + 2h1 + r) πr 2 h1 + 1 3 h2 2. l h O r Cylinder – Cone CSA (Cylinder + Cone) CSA (Solid + Base Area) Vol.(Cylinder – Cone) πr(l + 2h) πr(l + 2h + r) πr 2 h – 1 3 πr 2 h = πr 2 h 1 – 1 3 = 2 3 πr 2 h 3. h r Cylinder + Hemisphere CSA (Cylr + Hemisp.) CSA (Solid + Base Area) Vol.(Cylr + Hemisph.) 2πr(r + h) πr(3r + 2h) πr 2 h + 2 3 r 4. r h Cylinder–Hemisphere CSA(Cylinder+Hemisph.) CSA (Solid + Base Area) Vol.(Cylinder–Hemisp.) 2πr(r + h) πr(3r + 2h) πr 2 h – 2 3 r IV Combined Solids Formed by Cone with Cone or Hemisphere or Cylinder or Sphere The combined solids formed by the cone with cone or hemisphere or cylinder are given below: 12 cm 7 cm 10 cm 4 7 cm cm 10 cm Can you draw the nets of the above combined solids?

Mensuration 118 Allied The Leading Mathematics-10 Area and Volume 119 Now, find their lateral surface area, total surface area and volume separately. From the above activities we derive the following relations: SN Combined Solid CSA TSA Volume 1. h2 h1 l1 l2 r Cone + Cone CSA (Cone + Cone) CSA (Solid) Vol.(Cone+Cone) πr(l1 + l2) πr(l1 + l2) 1 3 πr 2 (h1 + h2) 2. r l h Cone + Hemisphere CSA(Cone + Hemisphere) CSA (Solid) Vol.(Cone+Hemisph.) πr(2r + l) πr(2r + l) 1 3 πr 2 (2r + h) 3. r l Hemisphere – Cone CSA (Hemisp.+Cone) CSA (Solid) Vol.( Hemisph.– Cone) πr(2r + l) πr(2r + l) 1 3 πr 3 Points to be Remembered 1. When we have to find the lateral or curved or both surface area (LSA/CSA) of the combined solid, we find the outer surfaces of the combined solids without bases and attaching parts of separate solids and then sum them. We use the formulae to find the area of separate parts. 2. When we have to find the total surface area (TSA) of the combined solid, we find all the visualized surfaces of the combined solids without attaching parts of separate solids and then sum them. We use the formulae to find the area of separate parts. 3. When we have to find the volume (V) of the combined solid, we find the separate solids of the combined solids such as prisms, pyramids, cones, cylinders, spheres and hemi-spheres and then sum them. We use the formulae to find the volume of separate solids.

Mensuration 118 Allied The Leading Mathematics-10 Area and Volume 119 Example-1 Observe the given crystal shown in the adjoining figure: (a) Find the slant heights of both pyramids of the given crystal. (b) Find the surface area of the given crystal. (b) Find the volume of the given crystal. Solution: (a) Given, the height of crystal (h) = 18cm, the height of upper pyramid (h1) = 10cm, ∴ The height of lower pyramid (h2) = 18 – 10 = 8cm, The base side of pyramid (a) = 8cm ∴ The slant height of upper pyramid (l1) = h1 2 + a2 4 = 100 + 64 4 = 116 cm The slant height of upper pyramid (l2) = h2 2 + a2 4 = 64 + 64 4 = 80 cm Now, we know that (b) The surface area of the crystal = 2a(l1 + l2) = 2 × 8( 116+ 80) = 315.43cm2 . (c) The volume of the crystal = 3 12 a2 h = 3 12 × 82 × 18 = 166.28cm3 . Example-2 A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 meters. The height of the cylindrical portion is 14m, while the vertex of the cone is 19 m above the ground. Find the area of the canvas required for the tent. Solution: For the cylindrical portion, Radius = 12 m and Height = 14 m So, the canvas required for cylindrical portion = Area of curved surface of cylindrical portion = 2 × 22 7 × 12 × 14 = 1056 m2 For the conical portion, Radius = 12 m and Height = (19 – 14) = 5 m Slant height = r 2 + h2 = 122 + 52 = 169 = 13 m So, the canvas required for conical portion = area of curved surface of conical portion = 22 7 × 12 × 13 = 490.29 m2 So, the total area of canvas = 1056 + 490.28 = 1546.29 m2 . 10 cm 8 cm 18 cm 13 m 24 m 19 m 14 m

Mensuration 120 Allied The Leading Mathematics-10 Area and Volume 121 “Alternatively” Here, Radius of solid (r) = 12m, Height of Cylinder (h1) = 14m Height of cone (h) = 5m ∴ Slant height (l) = r 2 + h2 = 122 + 52 = 169 = 13m Now, we have The canvas required for the given tent = CSA of the given tent = πr(l + 2h1) = 22 7 × 12(13 + 2 × 14) = 22 × 12 × 41 7 = 1546.29 m2 Example-3 A cylinder having radius 7 cm and height 10 cm is combined with a cone at its one side. If the total surface area of the solid is 783.25 cm2 , find the height of the cone. Solution: Here, in the given solid the radius of base of cylinder and cone (r) = 7 cm the height of the cylinder (h1) = 10 cm the height of the cone (h2) = ? the slant height of the cone (l) = ? the total surface area of the solid (TSA) = 792 cm2 Now, the have TSA of solid = CSA of cone + CSA of cylinder + Area of circle or, 792 = πrl + 2πrh1 + πr2 or, 792 = πr(l + 2h1 + r) or, 792 = 22 7 × 7 (l + 2 × 10 + 7) or, 792 7 = l + 27 or, l = 36 – 27 = 9 cm The height of the cone (h2) = l 2 – r 2 = 92 – 72 = 81 – 49 = 32 = 4 2 cm Hence, the volume of the solid (V) = Volume of cone + Volume of cylinder = 1 3 r 2 h2 + r 2 h1 = r 2 h2 3 + h1 = 22 7 × 72 4 2 3 + 10 = 154(1.886 + 10) = 154 × 11.886 = 1830.44 cm3 TSA = 792 cm2 7 m l h2 10 m

Mensuration 120 Allied The Leading Mathematics-10 Area and Volume 121 PRACTICE 5.3 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the lateral surface area of a square Cristal with length ‘4‘ cm and slant height l1 cm and l2 cm? (b) What is the volume of the given solid? (c) What are the following parts in the given solid? (i) Volume (ii) Curved surface area (iii) Total surface area Check Your Performance Answer the given questions for each problem. 2. Observe the following combined solids: (i) (ii) (iii) (iv) (a) Find the volume of each solid. (b) Find the lateral surface area and total surface area of each pyramid. 3. (i) The total surface area of a solid formed by two cones having the same radius is 2772 cm2 . If the heights of the cones are 10 cm and 8 cm respectively, (a) Find the radius of the cones. (b) Find the volume of the solid. (ii) If the volume of the given solid formed by cone and hemisphere having radius 7 cm is 1232 cm3 , (a) Calculate the height of the cone. (b) Find its surface area. 4. Find the lateral surface area, total surface area and volume of the following solids: (a) (b) (c) (d) (b) (c) r h h1 h r 140 mm 13 mm 10 mm 8 cm 18 cm 6 m 10 m 4 cm 10 cm 7 cm h 8 cm 8 cm 10 cm 13cm 6 cm 6 cm 10 cm 6 cm 4 cm 12 cm 8 cm 4 cm 14 cm 6 cm 14 cm 15 cm

Mensuration 122 Allied The Leading Mathematics-10 Area and Volume 123 5. Find the curved surface area, total surface area and volume of the following solids: (a) (b) (c) (d) (e) (f) (g) (h) 6. Find the lateral curved area, total surface area and volume of the following solids: (a) (b) (c) (d) 7. (a) If the curved surface area of the given solid is 2090 cm2 , find its volume. (b) Find the total surface area of the adjoining solid if its volume is 21149 cm3 . 8. (a) A circular tent is cylindrical up to a height of 15 m and conical above it. If its diameter is 28 m and slant height of the cone is 21 m, calculate the total area of the canvas required to make the tent. (b) Atent of height 33 m is in the form of a right circular cylinder of radius 60 m and height 22 m surmounted by a right circular cone of the same radius. Find the total surface area of the tent. 9. (a) A circular tent is cylindrical to a height of 3 meter and conical shape above it. If its diameter is 105 cm and the slant height of conical portion is 5 m, calculate the length of the canvas of 5 m wide to make the required tent. Also, calculate the cost of canvas at Rs. 2545.75/m. 7 cm 10 cm 5 cm 6.3 cm 15 cm 8 cm 2.1cm 12 cm 4 cm 140 mm 10 12 mm mm 6 m 10 m 12 in. 13 in. 21 cm 10.5 cm 21 cm25 cm 8 cm 13 cm 10 cm 4.2 m 1.8m 4 cm 10 8 cm cm 18 cm 25 cm 35 cm 28 m 28 m 21 m 15 m 60 m 22 m 33 m

Mensuration 122 Allied The Leading Mathematics-10 Area and Volume 123 (b) A Kunyu (pile of straw) in a field is cylindrical in shape with height 1.5 m. and conical above it. If its diameter is 3 m and slant height of the conical portion is 3 m, find the volume of the Kunyu. 10. (a) An iron pillar has some part in the form of a right circular cylinder and remaining part in the form of a right circular cone. The radius of the base of each of cone and cylinder is 49 cm. The cylindrical part is 240 cm high and conical part has 70 cm inclined height. Find the weight of the pillar if one cubic centimeter of iron weighs 0.8 gram. (b) The adjoining parachute has 32 ropes of 32ft each of length to be joined with a man and a semicircular shaped umbrella with circumference 44ft. Find the volume of the parachute with covering ropes and the area of cloth used for umbrella. 2. (i) (a) 11340.48 mm3 (b) 4618.9 mm2 , 4697.47 mm2 (ii) (a) 301.71 mm3 (b) 181.28 mm2 (iii) (a) 234.67cm3 (b) 235.97 cm2 (iv) (a) 733.33 cm3 (b) 345.70 cm2 , 423.70cm2 3. (i) (a) 49 cm, (b) 301.72 cm3 (ii) (a) 10 cm, (b) 576.62cm2 4. (a) 422.40 cm2 , 486.40 cm2 , 746.67 cm3 (b) 799.48 cm2 , 995.48 cm2 , 2156 cm3 (c) 320.52 cm2 , 356.52 cm2 , 432 cm3 (d) 256 cm2 , 272 cm2 , 233.33 cm3 5. (a) 629.25 cm2 , 783.25 cm2 , 1796.67 cm3 (b) 752.40 cm2 , 877.14 cm2 , 2076.11 cm3 (c) 188.22 cm2 , 202.08 cm2 , 184.80 cm3 (d) 4604.29 mm2 , 4682.86 mm2 , 11314.29 mm3 (e) 2160.84 cm2 , 2507.34 cm2 , 4851 cm3 (f) 2544.81 cm2 , 2891.31 cm2 , 5775 cm3 (g) 345.71 cm2 , 424.29 cm2 , 733.33 cm3 (h) 716.67 cm2 , 829.71 cm2 , 1923.43 cm3 6. (a) 247.63 cm2 , 247.63 cm2 , 301.74 cm3 (b) 330 cm2 , 330 cm2 , 477.71 cm3 (c) 235.97 cm2 , 235.97 cm2 , 301.71 cm3 (d) 46.22 cm2 , 46.22 cm2 , 26.48 cm3 7. (a) 6622 cm3 (b) 4048 cm2 8. (a) 2244 cm2 (b) 19800 m3 9. (a) 3.63 m, Rs. 9241.07 (b) 27.77 m3 10. (a) 395.40 kg (b) 2321.29 ft3 Answers Project Work Make a Cylinder and a cone with the same radius and different heights by paper. Now, combine the cone with one end of the cylinder. Find the CSAand TSAand volume of the solid so formed. 240 cm 70 cm 49 cm

Mensuration 124 Allied The Leading Mathematics-10 Area and Volume 125 5.4 Cost Estimation on Construction At the end of this topic, the students will be able to: ¾ estimate the cost of materials and wages on construction by using the properties of geometric shapes or different solid objects. Learning Objectives I Review on Geometric Shapes Geometric shapes are the figures which demonstrate the shape of the objects that we see in our surroundings. The shapes are the forms of objects which have boundary lines, angles and surfaces. Mainly, there are two types of geometric shapes. They are two-dimensional (2D) shapes and three-dimensional (3D) shapes. The 2D shapes are in polygonal shapes of length/height and breadth that are bounded by three or more sides such as different types of triangles, quadrilaterals and so on. For example, land, road-bridge, clothes, court-yard, etc. are the 2D shapes. The 2D shapes are contained on the surfaces of the 3D shapes or solid objects. For example, floor, ceiling and 4 walls of a room, and faces of prism, pyramid, cylinder, cone, etc. A solid object is a three-dimensional object or shape of the measurements length, breadth/width and height. In our day-today life, we come across a number of different types of solids. For example, room, football, birthday hat, water tanks, water pipe, coconut, etc are in single solid shapes. Similarly, the different things that are combination of two or more solid objects of the same or different types, or have a slightly modified shape than the actual solid. For example; water bottle, dome with pipe, pillar with top, Bauddha-stupa, fence, etc. are in the shapes of combined solids. II Cost Estimation on Geometric Shapes In our daily life, we use most man-made-things and making other things that we need. For this, we should estimate the cost of these things so that the shopkeeper does not take more money. A cost estimate is the preliminary stage for any task wherein a reasonable calculation of all the costs of task is done and, therefore, we would take precise judgment. Here, we will discuss the cost estimation of the things that are made from different definite geometric shapes like as triangle, quadrilateral, prism, pyramid, cylinder, cone, sphere and hemisphere. Survey of plot of land Photo Roads with Frame under pass

Mensuration 124 Allied The Leading Mathematics-10 Area and Volume 125 III Cost Estimation of Plane Shapes To estimate the cost of goods that are used on the plane surfaces, at first, we find the area of the plane area of the shape or total number of materials that paste on the plane and then calculate the cost by multiplying the rate of unit cost. If A is the area of the surface of the solid and R is the rate of cost, then we compute the total cost T as given below: Total Cost (T) = Rate of sq. units (R) × Area of plane shape (A) If L is the length of materials used in the plane, then we compute the total cost T as given below: Total Cost (T) = Rate of each unit (R) × Length of material (L) If P is the perimeter of the plane shape, then we compute the total cost T as given below: Total Cost (T) = Rate of each unit (R) × Perimeter of plane shape (P) Similarly, if a is the area of each piece of material used in the surface area and N is the number of pieces of the identical material, then Number of pieces of the same material (N) = Surface area (A) Area of each piece (a) Total cost (T) = Rate of each piece (R) × Number of pieces (N) Activity 1 If we have to need for planting rice on the given quadrilateralshaped field of the plot number 500 as shown in the land map alongside whose dimensions are also given, there are given two options to plant rice on the field; Option (i): When this task is held by using labours, each labour charges Rs. 1250 and the driver of the tractor charges Rs. 700 per hour. To complete the task 7 labours and 2 and half hour of tractor are needed. Option (ii): When this task is given to a contractor, he/she charges Rs. 10000 per Ropani. Which option is beneficial for the land owner ? Discuss. (i) How can we find the total cost using Option (i) ? (ii) Which formula is used to calculate the area of land for Option (ii) ? (iii) How much cost is there for Option (ii) ? For Option (i) For Option (ii) Cost for labour (C1) = Rate of labour × No. of labolur = Rs. 1250 × 7 = Rs. 8750 Cost for tractor (C2) = Rate of tractor per hour ×Time duration for tractor = Rs. 600 × 2.5 = Rs. 1500 ∴Total cost for planting rice (T1) = C1+C2 = Rs. 8750 + Rs. 1500 = Rs. 10250 Semi-perimeter of left triangle (s1) = 60 + 75 + 105 2 = 120 Semi-perimeter of right triangle (s1) = 125 + 105 + 94 2 = 162 Area of quadrilateral land (A) = Ar. of Tf + Ar. of Tr = s1(s1 – a1)(s1 – b1)(s1 – c1)+ s2(s2 – a2)(s2 – b2)(s2 – c2) = 120(120 – 60)(120 – 75)(120 – 105) + 162(162 – 125)(162 – 105)(162 – 94) = 2204.540769 + 4820.035685 = 7024.576454 ft2 = 7024.576454 5476 Ropani = 1.2828 Ropani ∴Total cost for planting rice (T1) = Rs. 10000 × 1.282793 = Rs. 12827.93 Hence, the Option (i) is beneficial than the Option (ii). PN:500 75' 60' 125' 94' 105' Area of qurd. = 1 2 d(h1 + h2) a b c d d a c b

Mensuration 126 Allied The Leading Mathematics-10 Area and Volume 127 IV Cost Estimation on Room Generally, a room has six faces as cuboid or cubic or square base cuboid in shape. Its base face is floor and that of upper is ceiling, and other vertical faces are called walls. There are four walls in the room. If l, b and h are the length, breadth and height of the rectangular room and P is the perimeter of the floor then, Area of floor (Af) = Area of Ceiling (Ac) = lb Area of 4 walls (A4) = 2h(l + b) = Ph Area of 4 walls and ceiling (A4c) = Area of 4 walls and floor (A4f) = 2h(l + b) + lb Area of 4 walls, floor and ceiling (A4fc) = 2(lb + lh + bh) If l is the length, breadth and height of the cubical room then, Area of floor (Af) = Area of Ceiling (Ac) = l 2 Area of 4 walls (A4) = 4l 2 Area of 4 walls and ceiling (A4c) + Area of 4 walls and floor (A4f) = 5l 2 Area of 4 walls, floor and ceiling (A4fc) = 6l 2 If l and h are the length and height of the square based room then, Area of floor (Af) = Area of Ceiling (Ac) = l 2 Area of 4 walls (A4) = 4lh = Ph Area of 4 walls and ceiling (A4c) = Area of 4 walls and floor (A4f) = 4lh + l 2 = l(4h + l) Area of 4 walls, floor and ceiling (A4fc) = 4lh + 2l 2 = 2l(2h + l) If R is the rate of materials used on the floor then, Total cost (T) = Rate (R) × length of the carpet (L). No. of stones (or tiles) required (N) = Area of floor (A) Area of stone (a) Area of floor (A) = Number of items (N) × Area of each item (a) Area of floor (A) = Total cost (T) Rate per item (R) Total cost (T) = Cost per piece (R) × No. of items (N). Note: Area of Floor = Area Ceiling = Area of Carpet/Total Number of tiles or stones or marbles = lb. If the length and breadth of a window are l1 and h1 respectively, the length and breadth of a door are l2 and h2 respectively and n, the number of door or window then, Area of four walls excluding n-windows (Aew) = 2h(l + b) – n×l1h1 Area of four walls excluding n-doors (Aed) = 2h(l + b) – n×l2h2 Area of four walls excluding n-windows and n-doors (Aewd) = 2h(l + b) – n×l1h1 – n × l2h2 Ceiling Wall Wall Floor Room Net of room Wall Floor Wall Wall Wall Ceiling b l h Cubical room l l l Squared base Room h l l b h h2 l1 l2 l h1

Mensuration 126 Allied The Leading Mathematics-10 Area and Volume 127 Activity 2 You want to plaster your new room with dimensions 15 ft by 12 ft by 9 ft having one door of the size 3 ft by 6 ft and two windows of the same size 5 ft by 3.5 ft. Ask your neighbour how much cost was paid for plastering your room. Your neighbour gives two options or quotations as mentioned below: Option (a): Giving responsibility for needed all materials and wages to the contractor Option (b): Buying all materials by yourself and giving responsibility for wages to the contractor For plastering on walls (R1) = Rs. 60/ft2 For plastering on floor (R2) = Rs. 40/ft2 For scotting around room (R3) = Rs. 35/ft2 For plastering on walls (R1) = Rs. 16/ft2 For plastering on floor (R2) = Rs. 14/ft2 For scotting around room (R3) = Rs. 13/ft2 For materials (T4) = Rs. 15,500 Which option or quotation is better ? How can you find ? At first, calculate the area of 4 walls excluding one door and 2 windows, the area of floor and the length of scotting without door. Now, we have Area of 4 walls without 1 door and 2 windows (A4) = 2h(l + b) – l1b1 – 2l2b2 = 2 × 9(15 + 12) – 3 × 6 – 2 × 5 × 3.5 = 379 sq. ft. Area of floor (Af) = l b = 15 × 12 = 180 sq. ft. Length of scotting without door (L) = Perimeter of room (P) – Length of door (l1) = 4(l + b) – l1 = 2(15 + 12) – 3 = 51 ft. Again, calculate the total cost for both options shown below: Total cost for Option (a) Total cost for Option (b) Cost for plastering on walls (T1) = R1 × A4 = Rs. 60 × 379 = Rs. 22,740 For plastering on floor (T2) = R2 × Af = Rs. 40 × 180 = Rs. 7,200 For scotting around room (T3) = R3 × L = Rs. 35 × 51 = Rs. 1,785 Total cost (T) = T1 + T2 + T3 = Rs. 22,740 + Rs. 7,200 + Rs. 1,785 = Rs. 31,725. Cost for plastering on walls (T1) = R1 × A4 = Rs. 16 × 379 = Rs. 6,064 For plastering on floor (T2) = R2 × Af = Rs. 14 × 180 = Rs. 2,520 For scotting around room (T3) = R3 × L = Rs. 13 × 51 = Rs. 663 For materials (T4) = Rs. 15,500 Total cost (T) = T1 + T2 + T3 + T4 = Rs. 6,064 + Rs. 2,520 + Rs. 663 + Rs. 15,500 = Rs. 24,247. For these two options, the total cost amounts are different. The option (a) is more expensive. So, you can work according as the option (b).

Mensuration 128 Allied The Leading Mathematics-10 Area and Volume 129 V Cost Estimation of Solid Objects To estimate the cost of goods that are used on the surfaces of the solid, at first, we find the area of the surface area of the solid or total number of materials that paste on the surfaces of the solid and then calculate the cost by multiplying the rate of unit cost. If A is the area of the surface of the solid and R is the rate of cost, then we compute the total cost T as given below: Total Cost (T) = Rate of sq. units (R) × Area of surface of solid (A) If L is the length of material used in solid and R is the rate of cost, then we compute the total cost T as given below: Total Cost (T) = Rate of each unit (R) × Length of material (L) Similarly, if a is the area of each piece of a material used in the surface area and N is the number of pieces of the identical material, then Number of pieces of the same material (N) = Surface area (A) Area of each piece (a) Total cost (T) = Rate of each piece (R) × Number of pieces (N) If V is the volume of the solid and R is the rate of cost of materials used in the solid, then we compute the total cost T as given below: Total Cost (T) = Rate of cu. units (R) × Volume of solid (V) If V is the volume of the solid, v, the volume of small materials used in the solid and N is the number of pieces of the identical material, then Number of pieces of the same material (N) = Volume of solid (A) Volume of each piece (v) Total cost (T) = Rate of each piece (R) × Number of pieces (N) Activity 3 If you plan to go to far area where there is no house and lodge, let it a tailor to make a conical tent of the diameter 2 m and height 3 m for you as shown in the figure alongside. Discuss on the answers of the following questions: (a) How much clothes is needed to sew the tent with 5% wastage ? (b) How much amount is to be paid to the shopkeeper for the tent clothes of the wide 1.25 m at Rs. 325 per meter ? (c) How much amount is to be paid to the tailor for sewing the tent at Rs. 125.75 per meter ? (d) How much amount is to be paid for painting on the outer side of the tent at Rs. 60 per sq. meter ? (e) How much amount is to be paid for matrix to sleep in the tent at Rs. 2050 per sq. meter ? (i) Which formula is used to find the area of the circular matrix ? (ii) Which formula is used to calculate the tent clothes without 5% wastage? (iii) How can find the cost of materials used on making tent ? l l

Mensuration 128 Allied The Leading Mathematics-10 Area and Volume 129 Activity 4 Suppose, you need to make a solid formed by squared base prism and identical based pyramid as shown in the figure. You need 14 inches base side, 10 inches height of the prism, and 18 inches height of the pyramid and let a carpenter to make it. Discuss on the answers of the following questions: (a) How much wood will the carpenter buy to make the solid of the given size? How much cost for it at Rs. 8,500 per cu. ft ? (b) What is the volume of the solid of the given size made from the buying wood? (c) How much wood is wasted from the buying wood and how much percentage? (d) How much amount does the carpenter pay to the painter for painting its all surfaces of the solid at Rs. 25.50 per sq. ft ? (e) How much amount do you pay to the carpenter at the rate of Rs. 15,500 per cu. ft ? How much percentage does the carpenter gain ? (i) Which formula is used to find the volume of the wooden block that is bought by the carpenter ? (ii) Which formulae are used to make the solid of the given size? (iii) Which formulae are used to find the surface area of the solid of the given size ? Activity 5 You are planning to make an RCC cylindrical pillar of radius 15 cm and height 3 m attaching hemi-sphere of radius 25 cm at the top as shown in the figure and let a carpenter to make it. Discuss on the answers of the following questions: (a) How much concrete do you need to construct the pillar ? (b) How much amount do you need to make the pillar at Rs. 6,250 per cu. cm? (c) How much amount do you need to plaster and paint on the surface of the pillar at Rs. 350 and Rs. 190 per sq. cm respectively ? (i) Which formulae will you need to the volume of the given solid? (ii) Which formulae are used to construct the RCC pillar of the given size? (iii) Which formulae are used to paint on the surfaces of the given solid? 1 3 a2 h 4ah 2al a2 h a2 2pr 2pr2 2prh pr2 h pr 2 2 3 pr 3

Mensuration 130 Allied The Leading Mathematics-10 Area and Volume 131 Example-1 The perimeter of a triangular shape pond is 120 m and its sides are in the ratio 10 : 24 : 26. (a) Find the length of the sides of the pond. (b) How much area in aana is covered by the pond? [Use 1 m = 3.28 ft, 1 aana = 342.25 sq. ft] (c) Calculate the amount to be received on selling it at Rs. 15,25,000 per aana. Solution : (a) Here, the perimeter of the triangular pond (P) = 120 m The ratio of the sides of the triangular pond = 10 : 24 : 26. Suppose, the length of the sides a, b, c of triangle are a = 10x, b = 24x, c = 26x. Now, Perimeter of the triangular pond (P) = 120 m. or, 10x + 24x + 26x = 120 or, 60x = 120 or, x = 120 ÷ 60 = 2 ∴ The length of the sides of triangle are, a = 10 × 2 = 20 m, b = 24 × 2 = 48 m, c = 26 × 2 = 52 m and Semi-perimeter (s) = 120 ÷ 2 = 60 m (b) Area of the pond (A) = s(s – a) (s – b) (s – c) = 60(60 – 20) (60 – 48) (60 – 52) = 60 × 40 × 12 × 8 = 230400 = 480 sq. m = 480 × 3.28 × 3.28 sq. ft = 5164.032 sq. ft = 5164.032 342.25 aana = 15.08848 aana (c) Rate of land (R) = Rs. 15,25,000 per aana Hence, total amount received on selling the land (T) = R × A = Rs. 15,25,000/aana × 15.08847918 aana = Rs. 2,30,09,930.75. Information about land area 1 Bigaha = 20 Kattha 1 Bigaha = 6772.63 m3 1 Bigaha = 72900 sq. ft. 1 Bigaha = 13.31 Ropani 1 Kattha = 20 Dhur 1 Kattha = 338.63 m2 1 Kattha = 3645 sq. ft. 1 Dhur = 16.93 m2 1 Dhur = 182.25 sq. ft. 1 Ropani = 16 aana 1 Ropani = 64 Paisa 1 Ropani = 508.72 m2 1 Ropani = 5476 sq. ft.

Mensuration 130 Allied The Leading Mathematics-10 Area and Volume 131 Example-2 A room is 8 m long, 6 m wide and 4.5 m high. It has one door 2.5 m by 1 m and three windows 1 m by 80 cm. (a) Calculate the area of the walls excluding door and windows. (b) Find the cost of white washing the walls at the rate of Rs. 4.50 per sq. m. Solution: Here, in the room, length (l) = 8 m, breadth (b) = 6 m, height (h) = 4.5 m ∴ Area of four walls (A4) = 2h(l + b) = 2 × 4.5(8 + 6) = 126 m2 Height of door (l1) = 2.5 m, Breadth of door (b1) = 1 m ∴ Area of door (A1) = 2.5 m2 The size of a window is 1 m by 0.8 m ∴ Area of 3 windows (A2) = 3 × 1m × 0.8m = 2.4 m2 ∴ Area of the walls excluding the door and 3 windows = 126 – 2.5 – 2.4 = 121.1 m2 . ∴ Cost of white washing the walls = Rate × area of four walls excluding door and windows = Rs. 4.5 × 121.1 = Rs. 544.95. Example-3 The dimensions of a rectangular prism of wood are 35 cm × 10 cm × 8 cm. A boy overlaps 10 such prisms one upon another vertically using glue. (a) Write the formula to calculate the volume of prism. (b) What is the cost of wood of the new prism formed by 8 small rectangular prisms at the rate of Rs. 8,500 per cu. ft? Find it. (c) What is the cost of panting on the surfaces of the new prism at the rate of Rs. 175.50 per sq. ft ? Solution: (a) The formula to calculate the volume of prism is A = Base area (A) × Height (h). (b) Here, the dimensions of the given prism 35 cm × 10 cm × 8 cm. If 10 such cuboids overlap one upon another vertically, the dimension of new cuboid is given by, Length (L) = 35 cm = 35 × 3.28 100 ft = 1.148 ft. Breadth (B) = 10 cm = 10 × 0.0328 ft = 0.328 ft. Height (H) = 8 × 8 cm = 64 × 0.0328 ft = 2.0992 ft ∴ Base area (A) = L × B = 1.148 × 0.328 = 0.376544 ft2 . 35 cm 8 cm 10 cm 1 m = 100 cm = 3.28 ft ∴ 1 cm = 3.28 100 ft = 0.0328 ft

Mensuration 132 Allied The Leading Mathematics-10 Area and Volume 133 Now, we have Volume of new prism (V) = A × H = 0.376544 × 2.0992 = 0.7904411648 ft3 . Rate of wood per cu. ft (R) = Rs. 8,500 ∴ Cost of wood of new prism (C) = R × V = Rs. 8,500 × 0.790441 = Rs. 6,718.75 (c) Perimeter of the base (P) = 2(L + B) = 2(1.148+0.328) = 2 × 1.476 = 2.952 ft. Painting area on new prism = TSA of new prism = PH + 2A = 2.952 × 2.0992 + 2 × 0.376544 = 6.1968384 + 0.753088 = 6.9499264 ft2 . Rate of painting per cu. ft (R) = Rs. 175.50 ∴ Cost of wood of the new prism (C) = R × TSA = Rs. 175.50 × 6.9499264 = Rs. 1219.71. Example-4 Alisha plans to make a conical tent of a 12 meter in diameter and 6.3 meter in slant height as shown in the figure. (a) Find the height of the tent. (b) Find the length of canvas 2 m width required to make it. (c) Find the cost of the canvas at the rate of Rs. 12.50 per meter. Solution: (a) Here, Diameter of base of the cone (d) = 12m. ∴ Radius of base of the cone (r) = 12 2 = 6m Slant height of the cone (l) = 6.3m ∴ Height of the cone (h) = l 2 – r 2 = 6.32 – 62 = 3.69 (b) Canvas required for cylindrical portion =Area of curved surface of this portion = 2πrh = 2 × 22 7 × 6 × 3.69 = 72.4468 m2 Since the width of the canvas is 2m Alternatively for Volume & TSA Volume of small prism (v) = 35×10×8 = 2800 cm3 Volume of new prism (v) = 8×2800 = 22400 cm3 = 22400×0.03283 ft3 = 0.7900441 ft3 TSA of new prism = 2(LB + BH + LH) = 2(35×10 + 10×64 + 35×80) = 2(350 + 640 + 2800) = 2 × 3790 = 7580 × 0.03282 ft2 = 6.9499264 ft2

Mensuration 132 Allied The Leading Mathematics-10 Area and Volume 133 So, the length of the canvas (l) = CSA b = 72.4468 2 = 36.22m (c) The cost of the canvas at the rate of Rs. 12.50 per meter = Rs. 12.50 × 36.22 = Rs. 452.79. Example-5 The diameter of a spherical copper ball is 42 cm. It is melted and drawn into a cylindrical wire of 3 mm diameter. (a) What is the volume of copper ? Find. (b) Find the length of the wire. (c) Calculate the cost of wire at 65 paisa per meter. Solution: (a) Here, Diameter of the sphere (d) = 42 cm, Radius of the sphere (r) = 42 2 = 21 cm. ∴ Volume of copper (v) = Volume of the spherical copper ball (v) = 4 3 πr3 = 4 3 π × 213 = 38808 cm3 . (b) Diameter of cylindrical wire (D) = 3 mm, ∴ Radius of wire (R) = 3 2 =1.4 mm = 0.14cm. [ ∴10 mm = 1 cm] Let l be the length of the cylindrical wire. Then, the volume of the cylindrical wire (V) = πr2 h = π × 0.142 × l. But, Volume of the cylindrical wire (V) = Volume of the spherical copper ball (v). or, π × 0.14 × 0.14 × l = 38808 or, 22 7 × 0.14 × 0.14 × l = 38808 or, 0.0616 × l = 38808 or, l = 38808 0.0616 = 630000 cm = 6300 m [ ∴100 cm = 1 m] ∴ The length of the wire is 6300 m. (c) Cost of wire per meter (c) = 65 paisa = Rs. 0.65 ∴ The total cost of total wire (T) = c × l = Rs. 0.65 × 6300 = Rs. 4095 Example-6 A man used 150 cylindrical iron rods at one end with a conical point tip to use on the wall of his house compound. The diameter of the cylindrical part is 1 cm. The total height of the rod is 1.26 m and the height of the conical part is 12 cm. (a) Find the quantity of iron to be used for these rods if 1 cm3 = 25 gm.

Mensuration 134 Allied The Leading Mathematics-10 Area and Volume 135 (b) Find the outer total surface area of these 150 rods. (c) If the cost of iron is Rs. 108/kg and the cost of painting enamel on its surface is Rs. 0.15/cm2 , calculate the total cost for ready of these rods. Solution: (a) For the cylindrical portion, Diameter (d) = 1 cm ∴ Radius (r) = 1 2 = 0.5 cm Height of rod (H) = 1.26 m = 126 cm Height of conical part (h1) = 12 cm ∴ Height of cylindrical part (h2) = 126 – 12 = 114 cm Now, we have Volume of iron of 1 rod (v) = Vol. of conical part + Vol. of cylindrical part = 1 3 πr2 h1 + πr2 h2 = 1 3 × 22 7 × 0.52 (12 + 114) = 22 21 × 0.25 × 126 = 33 cm3 ∴ Volume of iron of 150 rods (v) = 33 × 150 = 4950 cm3 If 1 cm3 = 25 gm, the quantity of iron for making 150 rods (v) = 4950 × 25 gm = 123750 gm = 123.750 kg (b) For conical part, slant height (l) = h2 + r 2 = 122 + 0.52 = 144 + 0.25 = 144.25 = 12.01 cm ∴ Outer surface area of 1 rod on the wall = CSA of conical part + CSA of Cylindrical part = πrl + 2πrh2 = 22 7 × 0.5(12.01 + 2 × 114) = 2640.11 7 cm2 ∴ Outer total surface area of 150 rods = 2640.11 7 × 150 = 56573.7857cm2 (c) If the cost of iron is Rs. 108/kg and the cost of painting enamel on its surface is Rs. 0.15/cm2 , then the total cost for ready of these rods is = Rs. 108/kg × 123.750 kg + Rs. 0.15/cm2 × 56573.7857cm2 = Rs. 13365 + Rs. 8486.07 = Rs. 21851.07. 1 cm1.26 m 12 cm

Mensuration 134 Allied The Leading Mathematics-10 Area and Volume 135 PRACTICE 5.4 Read Think Understand Do Keeping Skill Sharp 1. (a) What is the volume of the solid formed by a hemi-sphere and a cylinder of height 'a' cm having the same radius 'b' cm? (b) Write the volume of the solid formed by a cylinder of height 'p' cm and a hemi-sphere having the same radius 'q' cm? (c) What is the volume of the solid formed by a hemi-sphere and a cone of height 'a' cm having the same radius 'b' cm? (d) What is the curved surface area of the solid formed by a hemi-sphere and a cone of height 'p' cm having the same radius 'q' cm? (e) What is the total surface area of the solid combined by a cylinder of height 'a' cm and a cone of height 'b' cm having the same radius 'c' cm ? (f) Write the volume of the solid combined by a cube with base side 'a' cm and a pyramid of height 'b' cm. (g) Write the lateral surface area of the solid combined by a cube with base side 'x' cm and a pyramid of slant height 'y' cm. (h) What is the total surface area of the solid combined by a square prism with base side 'p' cm and a pyramid of slant height 'y' cm ? Check Your Performance 2. (i) A woman bought a triangular land. She measures the boundaries of her triangular open land and finds their measures 90 ft, and 108 ft and 120 ft. She plans to sell it after bound walls around it. [1 aana = 342.25 sq. ft] (a) Calculate the cost of construction wall at Rs. 450 sq. ft. (b) If she sold it at Rs. 25,50,000 per aana, how much amount did she get in total ? (c) If she invests Rs. 2,60,00,000 in total at first, what percent did she gain ? Find. (ii) Abroker bought a triangular land for Rs. 45,00,000 in total in Terai region. He measures the boundaries and finds the perimeter 200 m, shortest boundary 40 m, the ratio of other two boundaries 17:15. He plansto sell it after level maintaining. [1 kattha = 338.63 sq. m] (a) Find the two unknown boundaries of the land. (b) Calculate the cost of maintaining level at Rs. 1420 per kattha. (c) If he sold it at Rs. 15,20,000 per kattha, how much amount did he get in total ? (d) What percent did he gain ? Find.

Mensuration 136 Allied The Leading Mathematics-10 Area and Volume 137 3. (i) The given figure is the newly land buying by a school for playing ground. The length from the corners A to C is 32 meter. The perpendicular lengths from other two corners B and D to AC are 20 meter and 25 meter respectively. (a) Write the formula to find the area of that type of land. (b) Find the area of the ground ABCD. (c) A male labour takes Rs. 1500 per day for sowing the ground and completes the work in 3 days. A female labour takes Rs. 50 per sq. meter for plating grass on it. How much amount should pay for the ground ? Find it. (ii) The given trapezium ABCD is the parking region of a house. (a) What is the width of the parking region? Find. (b) Find the area of the region ABCD. (c) If the size of a cement block paving on the parking region is 6 inches by 6 inches, how many blocks are needed to pave on it? Find. (d) The cost of 1 tip of 2000 blocks is Rs. 70,000. A labour takes Rs. 1400 per day. If 3 labours complete the work in 4 days, how much amount does the house-owner pay at all ? (iii) The given shape PQRS is the open ground. (a) Find the perimeter of the ground. (b) If the ground is around by the fence 5 times, how long fence is needed ? (c) The cost of the fence is Rs. 50 per meter and a labour takes Rs. 1450 per day. If the work is completed in 4 days, find the cost for it. 4. (i) A girl measures her room and finds the length 15 ft , breadth 12 ft and height 9 ft. In this room, there are one door of the size 3 ft by 6 ft and two windows of the size 5 ft by 7 ft each. (a) Write the formula to find the area of 4 walls without door and windows. (b) She calculates the area of the walls excluding door and windows. What does she find ? (c) How much does she expense for wall painting at the rate of Rs. 15.50 per sq. ft ? Find. (ii) A student measures his classroom and finds the length 12 m, breadth 9 m and height 2.75 m. There are two doors of the size 2 m by 1.25 m each and 3 windows of the size 2.25 m by 1.25 m each. (a) Which formula does he use to find the area of 4 walls and floor without doors and windows? Write. (b) He calculates the area of the walls excluding doors and windows. What does he find ? D C Q P A B D 45 ft 35 ft 26 ft C P A B P S Q R 25 24 25

Mensuration 136 Allied The Leading Mathematics-10 Area and Volume 137 (c) How much does he expense for plastering on 4 walls and floor at the rate of Rs. 115.25 per sq. m ? Find. 5. (i) Your math book is in the size of 25 cm × 18.5 cm × 1.9 cm. A girl vertically overlaps 1 dozen of such books one upon another. (a) Which formula do you use to calculate the volume of prism formed by such 1 dozen of books ? (b) What is the cost of paper of the new prism formed by the 1 dozen of books at the rate of 160 paisa per cu. cm3 with printing? Find it. (c) What is the cost of paper packing it at the rate of Rs. 0.25 per sq. cm ? Find. (ii) The diameter of a circular coin with a thickness of 4 mm is 2.2 cm. A conductor collects vertically 20 such coins one upon another as a form of cylinder. (a) Which formula do you use to calculate the volume of the cylinder formed by such 20 coins ? (b) What is the cost of the metal to make such 20 coins at the rate of Rs. 1.05 per cu. cm3 ? Find it. (c) What is the cost of plastic packing it at the rate of Rs. 0.35 per sq. cm ? Find. 6. (i) Amar plans to make a conical tent 20 ft in diameter and 26 ft in slant height for the fire camp. (a) What is the height of the tent ? Find. (b) Find the length of canvas 5.5 ft width required to make it. (c) Calculate the cost of the canvas at the rate of Rs. 10.75 per ft2 . (ii) A woodcutter makes a pyramid shelter of a square base with sides 14 ft each and 17 ft in edge by using a wooden board. (a) Find the vertical height and slant height of the shelter. (b) If he did not use wood on the ground, how much wood did he use for it ? Find. (c) Find the cost of the wood at the rate of Rs. 75.50 per ft. 7. (i) The circumference of a great circle of a spherical iron ball is 44 cm. It is melted and drawn into a cylindrical wire of 4 mm diameter. (a) What is the volume of the iron ? Find. (b) Find the length of the iron wire made by the same iron ball. (c) Calculate the cost of wire at 28 paisa per meter. (d) Calculate the cost of copper coated at 8 paisa per meter. (ii) The base side and slant height of a square pyramid of wax are 20 cm and 25 cm respectively. It is melted and drawn into 10 small conical candles of height 4 cm. (a) What is the volume of the pyramid made by the wax ? Find. (b) Find the base radius of the small candles. (c) Calculate the cost of papering on all candles at 6 paisa per sq. cm.

Mensuration 138 Allied The Leading Mathematics-10 Area and Volume 139 8. (i) A man makes 1000 cylindrical iron bullets at one end with a conical pointed tip. The diameter of the bullet is 0.7 cm. The total height of the bullet is 1.4 cm and the height of the conical part is 0.8 cm. (a) Find the quantity of iron to be used for these bullets if 1 cm3 = 7.8 gm. (b) Find the outer total surface area of these 1000 bullets. (c) If the cost of iron is Rs. 108/kg and the cost of brass galvanised on its surface is Rs. 0.2/cm2 , calculate the total cost for ready of these bullets. (ii) Aman used 220 squared rods at the upper end with the pointed tip of the pyramid on the wall for a fence. The base side of the rod is 0.8 cm. The total height of the rod is 1.75 m and the height of the pyramid part is 1.5 cm. (a) Find the quantity of iron to be used for these rods if 1 cm3 = 7.8 gm. (b) Find the total surface area of these rods on the wall. (c) If the cost of iron is Rs. 105/kg and the cost of painting enamel on its surface is Rs. 0.15/cm2 , how much amount did he pay for the ready of these rods? Find. 2. (i) (a) Rs. 6142 (b) Rs. 3,84,07,500 (c) 33.85% (ii) (a) 75 m (b) Rs. 6290.60 (c) Rs. 67,33,600 (d) 49.43% 3. (i) (b) 720 m2 (c) Rs. 8100 (ii) (a) 24 ft (b) 960 ft2 (c) 3840 (d) Rs. 151800 (iii) (a) 106 m (b) 530 m (c) Rs. 8450 4. (i) (b) 398 ft2 (c) Rs. 6169 (ii) (b) Rs. 102.5625 (c) Rs. 11820.33 5. (i) (b) Rs. 16870 (c) Rs. 272.58 (ii) (b) Rs. 117.13 (c) Rs. 10.65 6. (i) (a) 24 ft (b) 148.57 ft (c) Rs. 1597.13 (ii) (a) 13.82 ft (b) 333.66 ft2 (c) Rs. 25191.33 7. (i) (a) 1437.33 cm3 (b) 114.33 m (c) Rs. 32.01 (ii) (a) 3054.67 cm3 (b) 8.54 cm (c) Rs. 289.39 8. (i) (a) 10.374 kg (b) 5038 cm2 (c) Rs. 2127.99 8. (ii) (a) 1910.92 kg (b) 122830.4 cm2 (c) Rs. 219071.16 Answers

Mensuration 138 Allied The Leading Mathematics-10 Area and Volume 139 CONFIDENES LEVEL TEST - III Unit III : Mensuration Class: 10, The Leading Maths Time: 45 mins. FM: 23 Attempt all questions. 1. In the given right pyramid made by plywood, the vertical height is 71 ft, side length of square base is 112 ft. (a) Which is the correct formula for finding lateral surface area (LSA) of the square pyramid with side length a, vertical height h and slant height l? [1] (i) 2h(a + l) (ii) 2a(a + l) (iii) a(a + l) (iv) a(a + h) (b) Find the area of triangular part of the pyramid. [2] (c) Find the cost of coloring in the triangular part of the pyramid when the cost of per square ft is 3.20 paisa. [1] (d) How many pieces of matrices of the size 7 ft × 4 ft can be kept on the ground of pyramid without doubling or overlapping? [1] 2. Two square based pillars of height 8 ft each with four faces are shown at the gate of a stadium have one and one pyramid of height 4 feet, each having the same base on their tops. The base of each pillar is 6 ft × 6 ft. (a) Write the formula to find the volume of the given pillars. [1] (b) Calculate the area of total surface area of the pyramid. [2] (c) If the pillars with pyramid are plastered at the rate of Rs. 110 per sq. ft, what will be the total cost? Find it. [1] 3. A cone has its height 12 cm and slant height 13 cm. If it is filled up with icecream in such a way that its upper part is hemi-sphere as shown in the figure. (a) Find the curved surface area of conical part of icecream. [2] (b) How much quantity of the icecream is given? [2] (c) If the cost of icecream is Rs. 0.75 per cc and the cost of a cone is Rs. 5, find the total cost of whole icecream. [1] 4. Given pencil as shown in the figure is made up of a cylinder and a cone. The diameter of the cylinder and cone are equal with length 3 4 inches. If the height of cylinder is 2.5 inch, height of cone is 3 4 inches. (a) Write the formula for finding the volume of the given combined solid. [1] (b) Find the total surface area of the given pencil. [3] 5. The diameter of a spherical iron ball is 70 cm. It is melted and drawn into a cylindrical wire of 2 mm diameter. (a) What is the volume of iron ? Find. [2] (b) Find the length of the wire. [2] (b) Calculate the cost of wire at 25 paisa per meter. [1] Best of Luck 71 fr 112 fr 112 fr 2.5 In In In 3 4 3 4

Mensuration 140 Allied The Leading Mathematics-10 Area and Volume 141 Additional Practice – III Right Pyramid 1. What is the volume of a squared pyramid having vertical height 5cm and base length 4 cm? 2. Calculate the area of total surface of the solid to be made by the following nets. (a) (b) (c) 3. Find the volume and total surface area of a squared base pyramid with vertical height 20 cm and base side 30 cm. 4. The length of base side of a squared base pyramid is 10 cm. If the total surface area of the pyramid is 240 cm2 , find its vertical height. 5. The vertical height of a pyramid is 12 cm. If its volume is 400 cu. cm, find the area of its total surface. Also, calculate the cost of paining its all surfaces at the rate of Rs. 1.75 per sq. cm. Right Cone 6. (a) What is the volume of a cone with height 10 cm and base area 15 cm2 ? (b) What is the total surface area of a cone having the slant height ‘l ‘m and radius ‘r’ m of the base? 7. Find the curved surface area of the cone to be made by the following nets: (a) (b) (c) 8. Calculate the surface area of a closed cone having 12 cm height with a diameter of 10 cm. 9. If the circumference of a cone is 44 cm and the sum of its radius and slant height is 32 cm, find its TSA. 6 cm 12 cm 6 cm 10 cm 13 cm 10 cm 10 cm 15 cm 14 cm 7cm 42 cm 42 cm 21cm 21cm

Mensuration 140 Allied The Leading Mathematics-10 Area and Volume 141 10. The curved surface area and the total surface area of a cone are 550 cm2 and 704 cm2 respectively. Find the slant height of the cone. 11. A cone of height 30 cm and diameter 26 cm is completely filled with water. The water is poured in a cuboid of dimension 20 cm × 18 cm × 16 cm. Find the height of the water in the cuboid. 12. (a) What is the length of canvas of the width 1.75 m required to make a conical tent of the diameter 7 m and slant height 5 m? Also, find the cost of canvas at the rate of Rs. 175.50 per meter and the persons can sit inside the tent if each person covers the space of 2 m × 2 m. (b) The slant height and diameter of a cone are 30 m and 10.5 m respectively. Find the cost of a constructing it at Rs. 45 per m3 and also the cost of white paint in its curved surface at Rs. 14 per m2 . Combined Solids 13. Find the lateral surface area, total surface area and volume of the following solids: (a) (b) 14. A cylindrical pencil is 17.6 cm long and has a radius of 7 mm. It has 1.5 cm of its length sharpened to a point. Calculate its volume and surface area. 15. The diagram shows the cross section of a circular pipe. The bore of the pipe is 1.4 cm and the wall of the pipe is 2 mm thick. Find; (i) the area of cross section of the bore. (ii) the cross section area of the metal used to make the pipe. (iii) the volume of the metal pipe of length 7 m. 26 cm 30 cm 20 cm 18 cm 16 cm 14 m 12 m 12 m 8 m 6 cm 6 cm 6 cm 4 cm 1.4 cm

Mensuration 142 Allied The Leading Mathematics-10 Area and Volume PB 16. Find the volume and surface area of the following objects: (a) (b) (c) (d) 1. 26.67 cm3 2. (a) 180 cm2 (b) 273.2 cm2 (c) 340 cm2 3. 6000 cm3 , 2400 cm2 4. 7 cm 5. 360 cm2 , Rs. 630 6. (a) 50 cm3 7. (a) 308 cm2 (b) 882 cm2 (c) 231 cm2 8. 346.5 cm2 9. 704 cm2 10. 25 cm 11. 14.75 cm3 12. (a) Rs. 5515.97, 9 person (b) Rs. 22275, Rs. 6930 13. (a) 204 cm2 , 240 cm2 , 264 cm3 (b) 587.65 cm2 , 731.65 cm2 , 1440 cm3 14. 26.80 cm3 , 78.01 cm2 15. (i) 1.54 cm2 , (ii) 1.0057 cm2 , (iii) 703.99 cm3 16. (a) 24.10 cm3 , 52.89 cm2 (b) 217.20 cm3 , 251.43 cm2 (c) 386.57 cm3 , 300.12 cm2 (d) 1702.38 cm3 , 785.71 cm2 Answers 1cm 2 cm 9 cm 10 cm 25 cm 12 cm 12 cm 2 cm 6 cm 18 cm 5 cm

Algebra PB Allied The Leading Mathematics-10 Sequence and Series 143 Apply the algebraic knowledge, skill and relationship on the solution of the behaviour problems. Marks Weightage : 15 Estimated Working Hours : Competency Learning Outcomes 1. To solve the problems related to the means of arithmetic and geometric sequences. 2. To solve the verbal problems related to arithmetic and geometric series. 3. To solve the algebraic problems related to quadratic equation. 4. To simplify the algebraic fractions. 5. To solve the problems related to exponential equation. 6. Sequence and Series 6.1 Mean of Arithmetic Sequence 6.2 Mean of Geometric Sequence 6.3 Sum of Arithmetic Series 6.4 Sum of Geometric Series 7. Quadratic Equations 7.1 Solution of Quadratic Equation 7.2 Verbal Problems on Quadratic Equation 8. Algebraic Fractions 8.1 Simplification of Algebraic Fractions 9. Indices 9.1 Solving Exponential Equations Chapters / Lessons ALGEBRA IV UNIT 32 Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of Items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks IV Algebra 32 2 2 2 4 3 7 1 2 8 3 15 HSS Way: Mathematics begins at Home, grows in the Surroundings and takes shape in School.

Algebra 144 Allied The Leading Mathematics-10 Sequence and Series 145 WARM-UP Arithmetic Sequence and Geometric Sequence 1) Sequence: A group of things in particular order in which related things follow each other, is called sequence. e.g. 2, 6, 10, 14, .............. 2) Series: The sum (operation of adding) of a list of numbers or things that are generating according to some pattern or rule is called a series. e.g. 2 + 6 + 10 + 14 + .............. 3) Progression: A progression is a pattern of numbers definite rule. e.g. 3, 6, 9, 12, .............. 3) Types of Sequence or Series: (i) Finite Sequence or Series: A sequence or series with countable number of terms, is called finite sequence or series. e.g. 5, 10, 15, 20, ......., 70 (ii) Infinite Sequence or Series: A sequence or series with uncountable number of terms, is called infinite sequence or series. e.g. 5, 10, 15, 20, ........... 4) Summation or Sigma Notation: Summation or sigma notation allows to write a long sum in a single expression. It is short form of series. e.g. S10 = 15 + 21 + 27 + 33 + .... to 10 terms = (6n + 9) ∑ 10 5) Partial Sum: for the finite sequence t n = 1 1, t2, t3, ....... tn–1, tn, tn.= Sn – Sn–1 6) Arithmetic Sequence: A sequence of numbers or things in which every term (except the first term) is obtained by adding or subtracting a constant number to the previous term is called an arithmetic sequence. A difference between any term and its preceding term in an arithmetic sequence is called common difference (d). e.g. 1, 4, 7, 10, ..... 7) If t1, t2, ... , tn - 1, tn is an arithmetic sequence, (i) Common difference, d = tn – tn – 1 (ii) General or nth term/rule, tn = a + (n – 1)d, where, a = first term (t1), n = numbers of terms 8) General form of Arithmetic sequence or progression (AS/AP) is; a, a + d, a + 2d, a + 3d, ......., a + (n – 1)d 9) Geometric Sequence: A sequence of numbers of things where each term after the first is found by multiplying or dividing a constant number to the previous term is called a geometric sequence. The number multiplied (or divided) at each stage of a geometric sequence is called common ratio (r). eg. 2, 6, 18, 54, ...... 10) If t1, t2, ... , tn - 1, tn is an geometric sequence, (i) Common ratio, r = tn tn – 1 (ii) General or nth term/rule, tn = arn – 1, where a = first term (t1), n = numbers of terms 11) General form of Arithmetic sequence or progression (GS/GP) is; a, ar, ar2 , ar3 , ......., arn – 1 Algebraic Formulae SN Expression Factorization Form Expanded Form 1. (a + b)2 (a + b) (a + b) a2 + 2ab + b2 ; (a – b)2 + 4ab 2. (a – b)2 (a – b) (a – b) a2 – 2ab + b2 ; (a + b)2 – 4ab 3. a2 + b2 ---- (a + b)2 – 2ab; (a – b)2 + 2ab R E V I S I T I N G A L G E B R A

Algebra 144 Allied The Leading Mathematics-10 Sequence and Series 145 4. a2 – b2 (a + b) (a – b) ---- 5. (a + b)3 (a + b) (a + b) (a + b) a3 + 3a2 b + 3ab2 + b3 ; a3 + b3 + 3ab(a + b) 6. (a – b)3 (a – b) (a – b) (a – b) a3 – 3a2 b + 3ab2 – b3 ; a3 – b3 – 3ab(a – b) 7. a3 + b3 (a + b) (a2 – ab + b2 ) (a + b)3 – 3ab(a + b) 8. a3 – b3 (a – b) (a2 + ab + b2 ) (a – b)3 + 3ab(a + b) 9. a4 – b4 (a2 + b2 ) (a + b) (a – b) ---- 10. a4 + a2 b2 + b4 (a2 + ab + b2 )(a2 – ab + b2 ) ---- 11. (a + b + c)2 (a + b + c) (a + b + c) a2 + b2 + c2 + 2ab + 2bc + 2ca 12. (a – b – c)2 (a – b – c) (a – b – c) a2 + b2 + c2 – 2ab + 2bc – 2ca 13. (x + a) (x + b) (x + a) (x + b) x2 + (a + b)x + ab, x2 + px + q, where a+ b = p and ab = q HCF and LCM 1. HCF: A factor taken from the given algebraic expressions that exactly divides to the given expressions, is called Highest Common Factors (HCF). i.e., HCF = Product of Common Factors 2. LCM: A lowest multiple taken from the given algebraic expressions that is exactly divided by the given expressions, is called Lowest Common Multiples (LCM). i.e., LCM = Common Factors × Remaining Factors = HCF × Remaining Factors 3. For Two algebraic expressions, HCF × LCM = 1st Expression × 2nd Expression Solving Simultaneous Linear Equations 1. Simultaneous Linear Equations: Two or more linear equations that all contain the same unknown variables are called a system of simultaneous linear equations. Simultaneous linear equations are solved by the following methods: (a) Substitution Method (b) Eliminating Method (c) Graphical Method Indices 1. An index number is a number which is raised to a power. The power is also known as the index. It is the number how many times to multiply the same number. For example, 25 means the multiplication of 2 by itself five times = 2 × 2 × 2 × 2 × 2 = 32. 2. Laws of Indices: SN Laws/Rules SN Laws/Rules SN Laws/Rules 1. am × an = am + n 2. am ÷ an = am – n 3. (am) n = amn = (an ) m 4. (ab)m = ambm 5. (a ÷ b)m = am ÷ bm 6. a–m = 1 am, am = 1 a – m 7. (a ÷ b)–m = bm ÷ am 8. 1m = 1, a0 = 1 9. a m n = (am) 1 n = a n m, a = a1/2 10. a n n = a, a mn pn = a m p 11. ab n = a n × b n 12. a ÷ b n = a n ÷ b n 13. a m n = a mn = a n m 14. If am = an , then m = n, if a ≠ 0. 15. If xn = yn , then x = may or may not be equal to y, if n ≠ 0. R E V I S I T I N G A L G E B R A

Algebra 146 Allied The Leading Mathematics-10 Sequence and Series 147 6.1 Mean/s of Arithmetic Sequence At the end of this topic, the students will be able to: ¾ solve the problems related to the means of arithmetic sequence. Learning Objectives I Introduction Activity 1 Take three sticks of equal length 24 cm each and give each stick to your three friends A, B and C. You say to your friends A, B and C respectively. Then "If you divide the stick into two, four and eight equal parts, on which measures of the stick do you mark? Mark it." Observe the above arithmetic sequences (AS) and discuss the arithmetic mean and means. The term/s between the first and last terms in an AS is/are called an arithmetic mean/s (AM/s). This is one of the oldest methods used to combine observations in order to give a unique approximate value by Babylonian astronomers in the third century BC. It was used by the astronomers to determine the positions of the sun, the moon and the planets. Nowadays this is used in economics, anthropology, weather forecasting, construction, etc. The concept of the arithmetic mean was originated from the Greek mathematician and astronomer Hipparchus of Rhodes (190-120BC). II Arithmetic Means (AM) What is a number exactly between two numbers 0 and 24 ? A number between 0 and 24 is 12. But, how would it be? The missing number = 0 + 24 2 = 24 2 = 12. This exactly middle number is the average value of the given two numbers. Similarly, what are only one exactly middle numbers between the following sets of two numbers: 2, ......, 6; 5, ......, 15; 23, ......, 27; 106, ......, 384. Thus, if a and b are two given numbers then the average value between them is a + b 2 . 24 cm ? ? B 24 cm ? 0, ........., ........., ........., 24 ? A 24 cm 0, ........., 24 C 24 cm ?? 0, ..., ..., ..., ..., ..., ..., ..., 24 ? ? ? ? ? Hipparchus CHAPTER 6 SEQUENCE AND SERIES

Algebra 146 Allied The Leading Mathematics-10 Sequence and Series 147 Now, consider the three terms a, m, b are in AP. Then their common differences are equal. i.e., m – a = b – m or, m + m = a + b or, 2m = a + b or, m = a + b 2 Hence, the AM between a and b is a + b 2 . i.e., AM = a + b 2 Note : If tn – 2, tn – 1, tn are in AP then tn – 1 = tn + tn – 2 2 III n-Arithmetic Means (AMs) Consider the following two numbers as the first and last terms in an AP 0, ___, ___, ___, 24 and 0, ___, ___, ___, ___, ___, ___, ___, 24. How many arithmetic means are there between 0 and 24? What are they? Can you easily find them? What is the common difference between any two successive terms ? But, how many arithmetic means are there between 2 and 2999 with common difference 3 ? What are they? Do you write all means easily ? For removing these types of difficulties, we learn the rules that help to find the arithmetic means as follows. Suppose the AP has n arithmetic means as m1, m2, m3, m4,……., mn between a and b, where a is the first term and b, the last term. Then the AP a, m1, m2, m3, m4,……., mn, b has the total number of terms, N = n + 2. Now, we have the general term of AP, tn b = a + (N – 1)d or, b – a = (N – 1)d or, d = b – a N – 1, where N = total number of terms. or, d = b – a n + 1, where n = total number of means. The n arithmetic means are in the table below: Terms Arithmetic Means Formulae t1 - a t2 m1 a + d = a + b – a n + 1 = an + b n + 1 t3 m2 a + 2d = …… = a(n – 1) + 2b n + 1 t4 m3 a + 3d = …… = a(n – 2) + 3b n + 1 …. …. …………… tn-3 mn-2 a + (n – 2)d = …… = 3a + (n – 2)b n + 1 tn-2 mn-1 a + (n – 1)d = …… = 2a + (n – 1)b n + 1 tn-1 mn a + nd = …… = a + nb n + 1 tn - b = a + (N – 1)d The height of the 1st floor is 8 ft 9 inches and will be made 15 stairs of ladder. In which heights shall these stairs construct ? n and d are fixed in sequence.

Algebra 148 Allied The Leading Mathematics-10 Sequence and Series 149 Alert to Arithmetic Mean/s (i) AM has no difference from the first and last terms. (ii) The AMs between any two numbers, all the differences between any two consecutive terms are not different. Points to be Remembered 1. The term/s between the given two terms in an AP is/are called an arithmetic mean/s (AM/s). 2. AM between two numbers a and b is a + b 2 . i.e., AM = a + b 2 . 3. The common difference of any two terms when inserting n-arithmetic means between a and b is d = b – a N – 1, where N = total number of terms OR d = b – a n + 1 , where n = total number of means. 4. The values of n-arithmetic means between a and b is a + nb, n = Numbers of means and n = 1, 2, 3, ... Example-1 It is given that the arithmetic sequence is 5, ...., 13, ......, ......, ....... (a) Define arithmetic mean (AM) between two numbers. (b) Find the second term of the given AS. (c) What is its 10th term ? Find it. Solution: (a) The average of the two terms is called an arithmetic mean (AM). It is computed by the formula AM = a + b 2 . (b) The second term of the given AS is a + b 2 = 5 + 13 2 = 18 2 = 9. (c) Now, in the given AS, first term (a) = 5, common difference (d) = 9 – 5 = 4 No. of terms up to 10 terms (n) = 10, 10th term (t10) = ? We know that General term (tn) = a + (n – 1)d or, t10 = 5 + (9 – 1) × 4 = 5 + 8 × 4 = 5 + 32 = 37. Example-2 2x, 45, 13x, are the first three terms of the arithmetic series. (a) Find the value of x. (b) Add next two terms in the AS. (c) How much percent increased in each term in the AS ? Solution: (a) Since 45 is the AM between 2x and 13x, so, 45 = 2x + 13x 2 [ m = a + b 2 ] or, 90 = 15x => x = 6. (b) Now, 1st term = 2x = 2 × 6 = 12, 3rd term = 13x = 13 × 6 = 78

Algebra 148 Allied The Leading Mathematics-10 Sequence and Series 149 ∴ Common difference (d) = t2 – t1 = 45 – 12 = 33 ∴ The next two terms are 4th term = t3 + d = 78 + 33 = 111 and 5th term = t4 + d = 111 + 33 = 144. (c) The increased percent in each term in the AS = d a × 100% = 33 12 × 100% = 275%. Example-3 7, p, 13 and q are in an AP. (a) What is p in the AP ? (b) Find the values of p and q. (c) Write the relation between p and q. Solution: (a) p is a arithmetic mean between 7 and 13 of the given AP. (b) Given: 7, p, 13, q are in AP. So, p is the AM between 7 and 13, then p = 7 + 13 2 = 20 2 = 10. Again, 13 is the AM between p and q. So, 13 = p + q 2 or, 26 = 10 + q or, q = 26 – 10 = 16 ∴ p = 10 and q = 16. (c) The relation between p and q is q = p + 2d, where d = 10 – 7 = 3. “Alternatively for p and q” Since 7, p, 13, q are in AP. So, p – 7 = 13 – p = q – 13 Taking the first and second parts, we get p – 7 = 13 – p or, p + p = 13 + 7 or, 2p = 20 ∴ p = 10. Example-4 There are n-arithmetic means between two terms p and q of an AP. (a) How many terms are there in the AP ? (b) What is the common difference of the AP? Find it. (c) If the first and last terms are 8 and 22 respectively, find 5 means between them. Solution: (a) There are n + 2 terms in the AP having n-arithmetic means between p and q. (b) Let the AP be a, m1, m2, m3, m4,……., mn, b, then the AP has the total number of terms, N = n + 2. Now, we have the general term of AP, tn b = a + (N – 1)d or, b – a = (N – 1)d or, d = b – a N – 1, where N = total number of terms. Again, taking the second and third parts, we get 13 – p = q – 13 or, 13 – 10 = q – 13 or, 3 + 13 = q or, q = 16 ∴ p = 10 and q = 16.

Algebra 150 Allied The Leading Mathematics-10 Sequence and Series 151 or, d = b – a n + 1, where n = total number of means. (c) Given, the first term (a) = 8, the last term (b) = 22, the number of AM’s (n) = 5 The total number of terms (N) = 5 + 2 = 7 ∴ The common difference (d) = b – a N – 1 = 22 – 8 7 – 1 = 14 6 = 7 3 = 2 1 3 . Now, we have The first AM (m1) = a + d = 8 + 21 3 = 101 3. The second AM (m2) = a + 2d = 8 + 2 × 2 1 3 = 121 3. The third AM (m3) = a + 3d = 8 + 3 × 2 1 3 = 15. The fourth AM (m4) = a + 4d = 8 + 4 × 2 1 3 = 171 3. The fifth AM (m5) = a + 5d = 8 + 5 × 21 3 = 191 3. Example-5 There aren-arithmetic means between 20 and – 20. If the5thmeanis0, (a) Findthe value of n. (b) How many total terms are there in the AP ? (c) Find the other means. Solution: (a) Given, the first term (a) = 20, The last term (b) = – 20, the number of means = n The common difference (d) = b – a n + 1 = – 20 – 20 n + 1 = – 40 n + 1 By question, the 5th means (m5) = 0 or, a + 5d = 0 or, 20 + 5 – 40 n + 1 = 0 or, 20n + 20 – 200 n + 1 = 0 or, 20n = 180 or, n = 9 Hence, there are 9 arithmetic means between 20 and – 20. (b) There are 9 + 2 = 11 terms in total in the AP. (c) For other means, d = – 40 9 + 1 = – 40 10 = – 4 1st mean (m1) = a + d = 20 – 4 = 16; 2nd mean (m2) = a + 2d = 20 – 2 × 4 = 12 3rd mean (m3) = a + 3d = 20 – 3×4 = 8; 4th mean (m4) = a + 4d = 20 – 4×4 = 4 “Alternatively for d” The common difference (d) = b – a n + 1 = 22 – 8 5 + 1 = 14 6 = 7 3 = 21 3.