The Leading Maths - 9

Mensuration Mensuration in standard form. But, in our locality, we measure the area of big surfaces like as land, ground, road, building, pond/lake, river, etc. in Ropani-Aana-Paisa-Dam and Bigahaa-Kattha-Dhur. In Tarai region, we use Bigahaa-Kattha-Dhur to measure area of land and Ropani-Aana-Paisa-Dam in hilly and mountain regions. The relations among these units are shown below; Conversion for Terai Customary Units Conversion for Mountainous Customary Units Unit Devisions In Square Metres In Sqare Feet Unit Devisions In Square Metres In Sqare Feet Dhur 16.93 182.25 Daam 1.99 21.39 Kattha 20 Dhur 338.62 3,645 Paisa 4 Daam 7.65 85.56 Bigaha 20 Kattha 6772.41 72,900 Aana 4 Paisa 31.84 342.25 Ropani 16 Aana 509.44 5476 The converting charts of ropani, aana, paisa and dam into bigaha, dhur and kattha and vice versa are given below. Land Measurement System in Nepal 1 hectare = 19 ropani 10 aana 2 paisa 0.99 daam 1 bigaha = 13 ropani 5 aana 0 paisa 0.66 daam 1 acre = 7 ropani 15 aana 1 paisa 0.82 daam 1 ropani = 15 aana 3 paisa 3.95 daam ≈ 16 aana 1 kattha = 10 aana 2 paisa 2.4 daam 1 aana = 1.8 dhur 1 aana = 4 paisa 1 paisa = 4.00012 daam 1 bigaha = 13.31 ropani Cost Estimation of Materials used on Plane Surfaces Plane surfaces such as triangles, squares, rectangles, parallelograms, quadrilaterals, trapeziums, pentagons, hexagons, etc. can be covered by suitable triangles of one or more shapes. Tessellation is the name given to the pattern obtained when a surface is covered in this way and without any gap. In everyday life, we come across various problems such as finding the total land area, size of a room, carpeting a room, painting or plastering of a wall, paving a path, ground and floor and so on. In all such cases, areas are found by measurements and calculations and then estimated the cost to be expensed for different materials. If A is the area of the plane surface and R is the rate of cost, then we compute the total cost T as given below: Total Cost (T) = R × A. 150 Allied The Leading Mathematics-9 Area 151

Mensuration Mensuration Similarly, if a is the area of each piece of a material used in the plane surface area and N is the number of pieces of the same material, then Number of pieces of the same material (N) = Plane surface area (A) Area of each piece (a) and Total cost (T) = Rate of each piece (R) × Number of pieces (N) Example-1 Observe the given small triangular traffic island ABC on the road whose edges are 3 m, 4 m and 5 m. (a) What is right-angled triangle ? define it. (b) Calculate the area of ∆ABC. (c) How much cost requires to plant the grasses on it at the rate of Rs. 125 per square meter? Find it. (d) Is ∆ABC a right-angled triangle? Justify. Solution : (a) A triangle having one right angle is called a right-angled triangle. (b) Here, in the given small triangular traffic island, the edges BC (a) = 4 m, AC (b) = 5 m and AB (c) = 4 m ∴ Its semi-perimeter (s) = a + b + c 2 = 3 + 4 + 5 2 = 6 , Now, using the formula, Area of triangle (∆) = s(s – a) (s – b) (s – c), we get Area of ∆ABC = 6(6 – 4) (6 – 5) (6 – 3) = 6 × 2 × 1 × 3 = 36 = 6 sq. m. (c) Rate of planting grasses on the island (R) = Rs. 125 per square meter ∴ Cost of planting grasses on the island (C) = C × A = Rs. 125 × 6 = Rs. 750. (d) In ∆ABC, 42 = 32 = 16 + 9 = 25 = 52 This satisfy the Pythagoras' theorem. So, ∆ABC is a right-angled triangle. Example-2 A plot of land in the form of a quadrilateral ABDC has the following dimensions: AB = 30 m, BD = 40 m, CD = 20 m, CA = 60 m, and ∠ABD = 90o . (a) Which formula do you use to find the area of a scalene triangle having the sides a, b and c? (b) What is the length of the diagonal AD of the land? Find it. (c) Calculate the area of the quadrilateral land ABDC in ropani system. Use 1 aana = 31.84 m2 . (d) If the land owner wants to sell it at Rs. 3550000 per aana, how much does he/she get from it after giving 5% commission to an agent? Find it. 3 m B C A 4 m 5 m A B D C 30 m 40 m 60 m 20 m 152 Allied The Leading Mathematics-9 Area 153

Mensuration Mensuration Solution : (a) The area of a scalene triangle having the sides a,b and c is s(s – a) (s – b) (s – c). (b) The given quadrilateral ABDC is covered by two triangles ACD and DAB formed by joining the diagonal AD. Since ∠ABD = 90o , AB = 30 m, BD = 40 m, we have, using the Pythagoras' theorem, AD2 = BD2 – AB2 = (40)2 + (30)2 = 1600 + 900 = 2500 = (50)2 and so, AD = 50 m (c) Now, area of rt. triangle ABD (A1) = 1 2 × AB × BD = 1 2 × 30 × 40 = 600 m2 . Again, in ∆CAD, CD (a) = 20 m, AC (b) = 60 m and AD (c) = 50 m Semi-perimeter (s) = a + b + c 2 = 20 + 60 + 50 2 = 130 2 = 65 ∴ Area of isosceles ∆CAD (A2) = s(s – a) (s – b) (s – c) = 65(65 – 20) (65 – 60) (65 – 50) = 65 × 45 × 5 × 15 = 658125m2 . ∴ Area of the land ABDC (A) = Area of ∆ABD + Area of ∆CAD = 600 + 658125 = 600 + 811.25 = 1411.249 sq. m = 1411.249 31.84 aana [∴ 1 m2 = 1 31.84 aana] = 44.3231 aana = 32 aana + 12 aana + 0.3231 aana = 2 ropani + 12 aana 0.3231 × 4 paisa [ ∴ 1 ropani = 16 aana, so 32 aana = 2 ropani] = 2 ropani + 12 aana 1.2926 paisa [ ∴ 1 aana = 4 paisa] = 2 ropani + 12 aana 1 paisa 0.2926 × 4 daam [ ∴ 1 paisa = 4 daam] = 2 ropani + 12 aana 1 paisa 1.17 daam (d) At last, cost of 1 aana land (R) = Rs. 3550000 ∴ Cost of the land (C) = C × A = Rs. 3550000 × 44.3231 = Rs. 157347005. Commission amount to an agent = 5% of Rs. 157347005 = 5 100 × Rs. 157347005 = Rs. 7867350.25 Hence, the amount got by the land owner = Rs. 157347005 – Rs. 7867350.25 = Rs. 149479654.80 152 Allied The Leading Mathematics-9 Area 153

Mensuration Mensuration Example-3 The length of the compound fence of the given triangular pond is 120 m and its sides are in the ratio 10 : 24 : 26. (a) Why is the perimeter of a triangle different from its area ? Give reason. (b) Find the length of sides of the pond. (c) Find the total cost of covering net on its surface at the rate of Rs. 450/m2 . (d) If its water level is 5 m up from its base, how much litre of water does it contain? Calculate it. Solution : (a) The perimeter of a triangle is different from its area because the perimeter is the total length of the boundaries of the plane shape but the area is the covering space by the plane shape. (b) Here, the length of he compound wall of the triangular pond, Perimeter of the pond (P) = 120 m The total number of sides = 10 + 24 + 26 = 60. The length of one part = 120 ÷ 60 = 2 ∴ The length of the sides of triangle are, First side (a) = 10 × 2 = 20 m, Second side (b) = 24 × 2 = 48 m Third side (c) = 26 × 2 = 52 m and (c) Now, semi-perimeter (s) = 120 60 = 60 m ∴ Area of triangle = s(s – a) (s – b) (s – c) = 60(60 – 20) (60 – 48) (60 – 52) = 60 × 40 × 12 × 8 = 230400 = 480 sq. m Again, the rate of covering net on its water surface (R) = Rs. 450/m2 Hence, the total cost of covering net on its water surface (T) = R × A = Rs. 450 × 480 = Rs. 216000. (d) Area of water surface (A) = 480 sq. m, height of water level (h) = 5 m ∴ Volume of water in the pond (V) = Ah = 480 × 5 = 2400 m3 = 24,00,000 litres. [ ∴ 1 m3 = 1000 l] 154 Allied The Leading Mathematics-9 Area 155

Mensuration Mensuration Example-4 The area of an isosceles triangle is 336 cm2 and the length of base is 14 cm. Calculate the length of equal sides. Solution: Given, The length of base of an isosceles triangle (b) = 14cm, The area of the isosceles triangle (A) = 336cm2 , The length of equal sides (a) = ? ∴ s = a + a + b 2 = 2a + 14 2 = a + 7 Now, we have A = s(s – a) (s – b) (s – c) or, 336 = (a + 7) (a + 7 – a) (a + 7 – a) (a + 7 – 14) or, 336 = (a + 7) . 7 . 7 . (a – 7) or, 336 = 7 a2 – 49 or, 48 = a2 – 49 Squaring on both sides, we get 2304 = a2 – 49 or, 2353 = a2 ∴ a = 48.51 cm. Hence, the length of the equal side is 48.51cm. PRACTICE 5.1 Keeping Skill Sharp 1. (a) Write the definition of scalene triangle. (b) Write the formula to compute the area of a scalene triangle with the sides a, b and c. (c) What is the formula to compute the area of a right-angled triangle with hypotenuse h, perpendicular p and base b? (d) What is the area of a parallelogram with base 14 cm and height 8 cm? (e) Write down the area of a rhombus having diagonals 10 cm and 16 cm. (f) Write down the area of a kite having diagonals 12 cm and 15 cm. 2. Circle the correct answer. (a) Which is the correct formula to calculate the area (A) of a scalene triangle with sides a, b and c? (i) A = s(s + a) (s + b) (s + c) (ii) A = s(s – a + s – b + s – c) (iii) A = s(s + a + s + b + s + c) (iv) A = s(s – a) (s – b) (s – c) Alternatively, you can use the formula A = b 4 4a2 – b2 154 Allied The Leading Mathematics-9 Area 155

Mensuration Mensuration (b) Which is the perimeter (P) of a right-angled triangle with legs p and b? (i) P = p + b + h (ii) P = a + b + c (iii) P = p + b + h2 (iv) P = p + b + p2 + b2 (c) What is the area of the given ∆PQR below? (i) 30 cm (ii) 30 cm2 (iii) 43.30 cm (iv) 43.30 cm2 (d) What is the area of the given ∆ABC below? (i) 30.05 cm2 (ii) 0.3977 cm2 (iii) 0.1503 cm2 (iv) 10.04 cm2 (e) What is the area of the given ∆LMN below? (i) 14 cm2 (ii) 24 cm2 (iii) 24 cm (iv) 34 cm2 (f) What is the area of the given ∆CDE below? (i) 26.43 cm2 (ii) 26.83 cm3 (iii) 26.83 cm (iv) 26.83 cm2 10 cm 8 cm N 6 cm L M fig (e) 8 cm 7 cm 9 cm E D C fig (f) 10 cm P Q R fig (c) 10 cm 10 cm B A C fig (d) 5 cm 4.5 cm 5 cm Check Your Performance 3. Observe the following polygons: (i) (ii) (a) Write the definition of polygon. (b) Find the length of the diagonals in each polygon. (c) Find the area of all triangles in each polygon. (d) Find the area of the polygons. 4. (i) Observe a triangular land ABC with sides 5 m, 12 m and 13 m and answer the following questions. (a) Define a right-angled triangle. (b) Find the area of ∆ABC. (c) How much cost is required to plough the land at the rate of Rs. 25 per square meter? Find it. (d) Is ∆ABC a right-angled triangle? Justify. A2 A1 4 cm 4 cm 3 cm 3 cm 8 cm 7 cm 6 cm 8 cm 3 cm A3 A2 A1 156 Allied The Leading Mathematics-9 Area 157

Mensuration Mensuration (ii) Observe a triangular pond with sides 42 ft, 34 ft and 20 ft whose bottom is congruent to its surface and answer the following questions. (a) Define a scalene triangle. (b) Find the area of the pond. (c) How much cost is needed to pave the tiles on its bottom at the rate of Rs. 85 per square ft? Find it. (d) If the height of the water level from the bottom is 15 ft, find the quantity of water in litre. 5. (i) The perimeter of a triangular tarpaulin is 900 cm and its edges are in the ratio 5:12:13. (a) Write the formula to find the area of a scalene triangle with sides a, b and c. (b) Find the edges of the tarpaulin. (c) Find the area of the tarpaulin. (d) Find the cost for waterproof painting on it at Rs. 5.25 per cm2 . (ii) The perimeter of a triangular pond is 540 ft and its sides are in the ratio 25:17:12. (a) Write the formula to find the area of a scalene triangle with base b and height h. (b) Find the length of sides of the pond. (c) Find the area of the pond. (d) Calculate the total cost for paving slates on its floor at the rate of Rs. 325/ft2 . (e) If the pond is empty and you fill it with water of a height of 7 ft, find the cost of water at Rs. 2.15 per liter. 6. Observe the mentioned triangular shapes and answer the following question. (i) An isosceles triangular garden has an area 72 m2 and one of the equal sides is 12 m. (a) Find the length of unknown side of each triangular garden. (b) Find the perimeter of the garden. (c) Calculate the cost for fencing 5 times around its boundaries at the rate of Rs. 276/m. (ii) The area of an isosceles triangular island of the road is 192 m2 and the length of base is 24 m. (a) Calculate the length of equal sides. (b) Find the perimeter of each triangular shape. (c) Calculate the total cost for colouring on its boarders around it at the rate of Rs. 25.25/m. (iii) The length of the sum of three sides of an equilateral triangular paper is 18 cm. (a) Calculate the length of its sides. (b) Find the area of the paper. (c) Calculate the total cost for colouring on its surface at the rate of 60 paisa/cm2 . 156 Allied The Leading Mathematics-9 Area 157

Mensuration Mensuration 7. (i) Observe the plot of land in the form of a quadrilateral PQRS with dimensions; PQ = 45 m, QR = 30 m, RS = 45 m, PS = 21 m, and QS = 44 m (a) Find the area of ∆PQS and ∆QRS. (b) Calculate the area of the quadrilateral land PQRS in bighaha system. Use 1 kattha = 338.62 m2 . (c) If an agent takes 4% commission on selling the land, how much does she/he take commission when it sells on Rs. 1250000 per kattha? Find it. (d) How much money does the land owner get from it after commission ? Find it. (ii) Observe the plot of land in the form of a quadrilateral ABCD alongside. (a) State the Pythagoras' theorem. (b) If, in the land, ∠ADC = 90o , how can you find the area of the land ? Write. (c) Find the area of the land in ropani system. Use 1 aana = 342.25 ft2 . (d) If an agent takes 5% commission on selling the land, how much does she/he take commission when it sells on Rs. 4525000 per aana ? Find it. (e) How much money does the land owner get from it after commission ? Find it. 8. (i) If each side of an equilateral triangular design on a garden is increased by 2 cm, its area increases by 5 3 cm2 . (a) Find the length of each side of the triangular design. (b) Find the area of the triangular design. (ii) The perimeter of a right-angled triangle is 60 cm and its hypotenuse is 26 cm. (a) Find the length of other two sides of the triangle. (b) Find the area of the triangle. 9. (i) An umbrella is made by stitching 8 triangular pieces of canvas of four different colors, each piece measuring 15 cm, 45 cm and 45 cm. (a) How much canvas is required for stitching the umbrella? (b) If the cost of 1 cm2 is Rs. 0.65, find the total cost of the canvas. (ii) A regular octagonal court-yard has a side 12 ft and a main diagonal 20 ft. (a) What is the area of the court-yard? (b) If the cost of 1 sq. ft is Rs. 265 for paving stones, find the total cost of the stones. 10. (i) If a, b and c are the length of three sides and s is the semi-perimeter of a triangle, find the area of the triangle in terms of a, b, c and s. (ii) If a and b are the lengths of two adjacent sides and d is the length of a diagonal of a parallelogram, find the area of the parallelogram in terms of a, b and d. A B C D 74 ft 46.30 ft 33.45 ft 62.30 ft Road 158 Allied The Leading Mathematics-9 Area 159

Mensuration Mensuration 3. (i) (b) 5 cm (c) 6 cm2 (d) 12 m2 (ii) (b) 10 cm, 6 cm (c) 24 cm2 , 24 cm2 , 8.94 cm2 (d) 56.94 cm2 4. (i) (b) 30 m2 (c) Rs. 750 (ii)(b) 336 ft2 (c) Rs. 28560 (d) 142,717.68 litre 5. (i) (b) 150 cm, 360 cm, 390 cm (c) 27,000 cm2 (d) Rs. 141, 750 (ii) (b) 250 ft, 170 ft, 120 ft (c) 9000 sq.ft. (d) Rs. 29,25,000 (e) Rs. 38,35,537.65 6. (i) (a) 16.97 m (b) 40.97 m (c) Rs. 56,538.60 (ii) (a) 20 m (b) 64 m (c) Rs. 1616 (iii) (a) 6 cm (b) 15.59 cm2 (c) Rs. 9.35 7. (i) (a) 453.54 m2 , 628.08 m2 (b) 3 Kattha 3.88 Dhur (c) Rs. 159,500 (d) Rs. 38,28,000 (ii) (c) 7.81 Aana (d) Rs. 17,67,012.50 (e) Rs. 3,35,73,237.50 8. (i) (a) 4 cm (b) 4 3 cm2 (ii) (a) 10 cm, 24 cm (b) 120 cm2 9. (i) (a) 2662.24 cm2 (b) Rs. 1730.45 (ii) (a) 384 sq.ft. (b) Rs. 101,760 Answers Project Work Go to visit a filed and measure sides and diagonal of a plot of irregular land. Find the area of the land. Calculate the total cost of ploughting the land @ Rs. 25 per sq. units and planting rice @ Rs. 45 sq. units. If you want to fence around the land at 5 times, calculate the cost of the fence at the rate of Rs. 135 per unit. 158 Allied The Leading Mathematics-9 Area 159

Mensuration Mensuration 5.2 Surface Area of Room and Cost Estimation At the end of this topic, the students will be able to: ¾ find the area of plane surface of cuboid and cube. ¾ estimate the cost of carpeting, painting, etc. in the room. Learning Objectives Uses of Surface Area of Cuboid and Cube in Our Daily Activities Let's discuss the pictures below; What is a man doing in the picture? How much carpet isrequired to cover the floor of the room? How much amount is needed to cover the carpet on the floor? How can we find ? Discuss. What is a man doing in the picture? How many pieces of tiles are required to pave the walls of the bath room? How much amount will need for it? How can we find? Discuss. What are labourers doing in the picture? How can we estimate colours to be required to paint on the walls? How much amount will need for it? How can we find ? Discuss. What is a girl doing in the picture? How many areas does she paint? How much amount will need to cover the carpet on the floor? Discuss. Remember the formulae of the area and volume of the cuboid, cube and prism that are studied in the class 8, below: SN Name Definition Area (A) Lateral Surface Area (LSA) Total Surface Area (TSA) Volume (V) 1. Cuboid Solid contained six rectangular faces at right angles to each other. A = l × b LSA = 2(l + b) h TSA = 2(lb + bh + lh) V = lbh 2. Cube Solid contained by six equal squares at right angles to each other. A = l2 LSA = 4l2 TSA = 6l2 V = l3 160 Allied The Leading Mathematics-9 Area 161

Mensuration Mensuration 3. Prism Solid with two congruent polygonal faces lying in parallel planes and with the other faces parallelograms or rectangles. A = Area of Base (shape of base) LSA = Ph, P = Perimeter TSA = Ph + 2A V = Ah Cube and Cuboid A cube is a six-faced, three-dimensional figure composed of square-shaped faces of the same size that meet at 90-degree angles, whereas a cuboid is a box-shaped object made of six faces that all meet at 90-degree angles. A cuboid shape can also be a cube if all sides are of the same length, but not all cuboids are cubes. They have 6 faces, 8 vertices and 12 edges. They are like as a room of a house or building. The bottom plane surface is called floor of the room (cube or cuboid), that of top plane surface is called ceiling and the vertical plane surfaces are called the walls. It has one floor, one ceiling and four walls. When we open the cube and cuboid, their nets becomes as follows: Ceiling Wall Floor Cube Cuboid Wall Floor Wall Wall Wall Ceiling Net of cube Net of cuboid Wall Floor Wall Wall Wall Ceiling Areas of Floor, Ceiling and Four Walls In the annexed figure, the rectangle ABCD represents the floor and the rectangle EFGH the ceiling of a room. Its four walls are represented by the rectangles ABFE, BCGF, CDHG and DAEH. In the figure, length, AB = l, breadth, AD = b, height, AE = h. Here, area of the floor ABCD (Af ) = lb, area of the ceiling EFGH (Ac) = lb, and area of the four walls (A4) = Area of ABFE + Area of BCGF + Area of CDHG + Area of DAEH. = lh + bh + lh + bh = 2h(l + b) = Ph, where P = Perimeter = 2(l + b). ∴ The area of four walls, floor and ceiling of the room is given by A4fc = 2h(l + b) + lb + lb = 2h(l + b) + 2lb = 2(lb + bh + lh) D A b B E F G H h l C Rectangular Room 160 Allied The Leading Mathematics-9 Area 161

Mensuration Mensuration Note : (i) A squared base room means a room having equal length, breadth and height. Therefore, in the squared room, Area of floor (Af ) = l2 , Area of ceiling (Ac) = l2 and Area of four walls (A4) = 4lh = Ph, where P = Perimeter = 4l. ∴ The area of four walls, floor and ceiling of the room is given by, A4fc = 4l2 + l2 + l2 = 6l2 (ii) A squared or cubical room means a room having equal length, breadth and height. Therefore, in the squared room, Area of floor (Af ) = l2 , Area of ceiling (Ac) = l2 and Area of four walls (A4) = 4l2 . ∴ The area of four walls, floor and ceiling of the room is given by, A4fc = 4l2 + l2 + l2 = 6l2 (i) Floor Area Problems Floors are generally rectangular or square in shape. Calculation of the area of a floor is important for carpeting, and paving stones, tiles and marbles in a room and also for finding the corresponding costs. Sometimes we have to find the length of the carpet or the like if the breadth is not the same as the breadth of the room. We illustrate various cases with actual examples. It will be useful to remember that (a) Total cost (T) = Rate (R) × length of the carpet (L). (b) No. of stones (or tiles) required (N) = Area of floor (A) Area of stone (a) (c) Area of floor (A) = Number of items (N) × Area of each item (a) (d) Area of floor (A) = Total cost (T) Rate per item (R) (e) 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 = l × b. (ii) Four Walls Area Problems Painting or papering of walls and finding their costs are some of the common household problems. In such problems, we need to consider not only the length, breadth and height of a room but also the sizes of the windows and doors. 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, then Squared base Room h l l Cubical Room l l l b h l h2 1 l2 l h1 162 Allied The Leading Mathematics-9 Area 163

Mensuration Mensuration (a) Area of four walls excluding n-windows (Aew) = 2h(l + b) – n × l1h1 (b) Area of four walls excluding n-doors (Aed) = 2h(l + b) – n × l2h2 (c) Area of four walls excluding n-windows and n-doors (Aewd) = 2h(l + b) – n × l1h1 – n × l2h2 Example-1 The floor of a bathroom is 8 ft × 6 ft in size. (a) Write the relation between the area of floor and ceiling of the bathroom. (b) What length of the linoleum of the wide 3 ft is required for paving it ? (c) What would be the cost of linoleum if its cost is Rs. 250 per feet ? Solution: (a) The area of floor and ceiling of the room are equal. (b) Here, Size of the room = 8 ft × 6 ft = 48 ft2 . Breadth of the linoleum = 3 ft Suppose, the length of the linoleum required = l (say). Since, area of linoleum (A) = area of the floor, we have Length × breadth = 48 ft2 or, l × 3 = 48. ∴ Length of the linoleum (l) = 16 ft. (c) Since, cost per metre or rate (R) = Rs. 250 ∴ Total cost of linoleum = Rate per length (R) × Length of linoleum (l) = Rs. 250 × 16 = Rs. 4000. Example-2 A room is 15 m long and 10 m broad. (a) Write the relation between the area of floor of the room and the area of carpet paving on it. (b) Find the cost of carpeting the floor at Rs. 300/m2 with a carpet of length 25 m. (c) Find the breadth of the carpet. Solution: (a) The area of floor of the room and the area of carpet paving on it are equal. (b) Here, length of the room (l) = 15 m, breadth of the room (b) = 10 m ∴ Area of the room = l × b = 15 × 10 = 150 m2 . But, Area of carpet = Area of the room = 150 m2 . ∴ Cost of carpeting the floor = Rate × Area of carpet = Rs. 300/m2 × 150 m2 = Rs. 45000. (c) Again, area of the carpet = (length × breadth) of carpet or, 150 = 25 × b = (18 m)2 ∴ The required breadth of carpet is 6 m. 162 Allied The Leading Mathematics-9 Area 163

Mensuration Mensuration Example-3 The cost of carpeting the floor of a square room is Rs. 4050 at the rate of Rs. 12.50/m2 . (a) Find the area of the room. (b) Find the length of the room. (c) What is the perimeter of the room ? Find it. (d) Calculate the cost of scotting around the room at Rs. 225 per m. Solution: (a) Given, cost of carpeting the floor = Rs. 4050. Rate of carpeting = Rs 12.50/m2 . ∴ Area of carpet = Cost of carpeting (A) Rate of cost (a) = 4050 12.50 = 324 m2 (b) Area of floor = Area of carpet or, (l) 2 = 324 m2 = (18 m)2 ∴ The required length of the room is 18 m. (c) Perimeter of the room (P) = 4l = 4 × 18 = 72 m. (d) Rate of scotting around the room (R) = Rs. 225 per m ∴ Cost of scotting around the room(C) = R × P = Rs. 225 × 72 = Rs. 16200. Example-4 The cost of paving a room at Rs. 5 per stone is Rs. 8000. (a) Find the number of stones required to pave the room. (b) Find the area of the floor if the length of a room is 12 m and breadth 8 m. (c) Find the length of stone of width 20 cm. Solution: (a) Here, cost of paving stones (T) = Rs. 8000, Rate (R) = Rs. 5/stone. ∴ No. of stones required = Total cost of paving stones (T) Rate of cost per stone (R) = 8000 5 = 1600. (b) Area of floor (A) = 12 m × 8 m = 96 m2 (c) Area of each stone (a) = l × 0.2 m But, no. of stones required = Area of floor (A) Area of each stone (a) or, 1600 = 12 × 8 l × 0.2 or, l = 12 × 8 1600 × 0.2 = 0.3 ∴ The length of stone is 0.3 m = 30 cm. Example-5 A room is 20 m long, 12 m wide and 8 m high. (a) Find the area of four walls of the room. (b) Find the length of the wall-paper 2.5 m wide paving on its four walls. (c) Find the cost of papering on the four walls at Rs. 12.50 per metre. Solution: (a) Here, length of the room (l) = 20 m, breadth of the room (b) = 12 m, Height of the room (h) = 8 m, Width of paper (d) = 2.5 m, rate (R) = Rs. 12.50/m So, the area of the four walls (A4) = 2h(l + b) = 2 × 8(20 + 12) = 512 m2 . 164 Allied The Leading Mathematics-9 Area 165

Mensuration Mensuration (b) Also, the area of the wall paper (a) = length × width = length × 2.5 m Since, the area of wall paper (a) = area of four walls (A4), we have length × 2.5 = 512 or, length = 512 2.5 = 204.80 Hence, the length of the wall paper is 204.8 m. (c) Total cost of papering the four walls (T) = Rate × length of paper = Rs. 12.50 × 204.8 = Rs. 2560. Example-6 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) Find the area of the walls excluding door and windows. (b) The cost of white washing the walls at the rate of Rs. 4.50 per sq. m. Solution: (a) Here, length of the room (l) = 8 m, breadth of the room (b) = 6 m, height of the room (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 . (b) Cost of white washing the walls = Rate × area of four walls excluding door and windows = Rs. 4.5 × 121.1 = Rs. 544.95. Example-7 The cost of papering the four walls of a room at 75 paisa per square metre is Rs. 240. The height of the room is 5 m. (a) Find the area of four walls of the room. (b) Find the length and breadth of the room if their ratio is 5 : 3. (c) Find also the number of paper-sheets of size 40 cm × 20 cm required for papering the four walls. (d) Calculate the cost of papering the four walls at Rs. 3 per sheet. Solution: (a) Here, Cost of papering 4 walls (T) = Rs. 240, Rate (R) = Rs. 0.75/m2 . ∴ Area of 4 walls (A4) = Total cost (T) Rate (R) = Rs. 240 Rs. 0.75/m2 = 320 m2 (b) Since the length and breadth of the room are in the ratio 5:3, we may take length of the room (l) = 5x and its breadth (b) = 3x. Also, height of the room (h) = 5 m. Since, area of 4 walls = 2h(l + b), we have 320 = 10(5x + 3x) ∴ x = 4 164 Allied The Leading Mathematics-9 Area 165

Mensuration Mensuration ∴ Length of the room (l) = 5 × 4 = 20 m and breadth of the room (b) = 3 × 4 = 12 m. (c) Also, the area of the paper-sheet (a) = 40 cm × 20 cm = 0.08 m2 . ∴ Number of paper-sheets required = Area of four walls (A4) Area of each paper (a) = 320 0.08 = 4000. (d) Rate of paper per sheet (R) = Rs. 3 So, total cost of papering the four walls (T) = Rate (R) × No. of paper sheets (N) = Rs. 3 × 4000 = Rs. 12000. Example-8 The cost of carpeting a squared hall at Rs. 12 per m2 is Rs. 768 and the cost of plastering its walls at the rate of Rs. 6 per square metre is Rs. 576. (a) Find the area of the carpet. (b) Find the length of the room. (c) Find the area of 4 walls of the room. (d) Find the height of the room. Solution: (a) Here, Rate of carpeting (R) = Rs. 12/m2 , Total cost of carpeting (T) = Rs 768. ∴ Area of carpet (A) = Total cost (T) Rate of carpeting (R) = 768 12 = 64 m2 (b) But, Area of floor square room = Area of carpet or, l2 = 64 m2 ∴ l = 8 m. ∴ The length of the room is 8 m. (c) Rate of plastering (R1) = Rs. 6/m2 , Total cost of plastering (T1) = Rs. 576. ∴ Area of the 4 walls (A4) = Total cost of plastering (T1) Rate of plastering (R1) = 576 6 = 96 m2 (d) Again, Area of the 4 walls (A4) = 96 or, 2h(l + b) = 576 6 or, 2 × h(8 + 8) = 96 or, 32h = 96 or, h = 96 32 = 3 m Hence, the height of the room is 3 m. PRACTICE 5.2 Keeping Skill Sharp 1. (a) What is the formula to calculate the area of the floor of a rectangular room having length l, breadth b and height h ? (b) What is the area of the four walls of a rectangular room having length a cm, breadth b cm and height c cm ? (c) Write down the formula to find the area of the four walls and ceiling of a rectangular room having the length l, breadth b and height h. 166 Allied The Leading Mathematics-9 Area 167

Mensuration Mensuration (d) What is the formula to calculate the area of the four walls, ceiling and floor of a rectangular room having the length l, breadth b and height h ? 2. (a) N tiles are needed to pave the floor of a room. If the area of each tile is a sq. units, what is the area of the room? (b) What is the cost of N paving bricks at the rate of Rs. R to pave on the floor of the room ? (c) The total cost of paving stones on the floor of a room at Rs. R is Rs. T. What is the area of the floor? 3. (a) What is the area of the ceiling of a rectangular room having length 4 m, breadth 3 m and height 2.5 m? (b) What is the area of the four walls of a cubical room having the length 4 m ? Check Your Performance Observe the given situation and answer the following questions for each situation. 4. (i) A man wants to paste the carpet of 80 cm wide on the floor of a room 10 m long and 6 m broad. (a) Write the relation between the area of room and carpet. (b) Calculate the area of the room. (c) What is the area of the carpet to be pasted on the room. (d) Find the length of the carpet to be needed to paste on the room. (e) If the cost of 1 meter carpet is Rs. 645, calculate the cost of carpet used in the room? (ii) When a girl pasted the carpet on the floor of a squared room of the length 15 ft, a 15 m long carpet used. (a) What formula is used to find the area of the square floor ? (b) Calculate the area of the room. (c) What is the area of the carpet to be pasted on the room. (d) Find the breadth of the carpet. 5. (i) The stones of each 30 cm long and 25 cm broad are required to pave the floor of a rectangular room 20 m by 18 m. (ii) The carpets of each 5 m long and 3 m wide will be required to cover the floor of a room measuring 15 m by 12 m. (a) Write the formula to find the number of materials paving on plane floor. (b) What are the area of the room and each material? (c) How many materials are needed to pave the floor of the above mentioned rooms? (d) If the cost of each material is Rs. 15, calculate the cost of all required materials. 6. (i) The cost of carpeting on the floor of a room at Rs. 430/m2 was Rs. 17200. If the length of the room had been 2 m more, the expense would have been Rs. 19780. (ii) The cost of carpeting a room at Rs. 4.75 per sq. metre is Rs. 646. If the length of the room had been 3 m less, the cost would have been Rs. 532. 166 Allied The Leading Mathematics-9 Area 167

Mensuration Mensuration (a) What is the cost of paving marbles on the floor of the area 'a' sq. m at Rs. 'b' per sq. m? (b) Find the area of the carpet used in the room. (c) What is the area of the room. (d) What is the length and breadth of the room? Find it. 7. (i) A rectangular room has the length 8 m, breadth 6 m and height 4 m. (ii) A square room has the length 10 m and height 3 m. (a) Write the formula to find the area of 4 walls of a rectangular room. (b) Find the area of 4 walls of the above room. (c) Find the area of the 4 walls and ceiling of the room. (d) What percent of the area of the 4 walls and ceiling is more than its 4 walls? Find it. 8. (i) A rectangular room is 12 m long, 6 m broad and 4 m high. (a) How many metres of wall-paper of 80 cm wide will be required to paste on its walls? Find it. (b) Calculate the cost of paper at the rate of Rs. 50 per sq. m. (ii) A rectangular room is 6.5 m long, 5.5 m broad and 3.75 m high. (a) How many pieces of tiles 0.25 sq. meters each will be required to paste on the four walls and ceiling of the room? Find it. (b) Calculate the cost of tiles at the rate of Rs. 245 per piece. 9. (i) The area of four walls of a room is 77 m2 . The length and breadth of the room are 7.5 m and 3.5 m respectively. (ii) A room 6 m long and 5 m wide and the area of whose floor and ceiling together is equal to the area of the four walls. (a) Find the height of the room. (b) Find the four walls and floor of the room. (c) Calculate the cost of plastering on its four walls and floor at Rs. 125/m2 . 10. (i) A room is 12 m long, 8 m broad and 4 m high. It has two doors each 2 m high and 1.5 m wide and 4 windows each 2 m wide and 2.5 m high. Calculate the total cost of papering the walls and ceiling at the rate of Rs. 5.50 /m2 . (ii) A room is 9 m long, 8 m broad and 6.5 m high. It has one door 3 m × 1.5 m and three windows each of dimensions 1.5 m and 1 m. Find the cost of white washing the walls at Rs. 3.75 per m2 . 11. (i) The length of a room is three times its breadth. The cost of carpeting its floor at Rs. 175 per sq. metre is Rs. 13125. (a) Find the area of the floor of the room. (b) Find the length and breadth of the room. (ii) The length of a room is twice its breadth and the area of its 4 walls is 120 m2 . The height of the room is 4 m. (a) Find the length and breadth of the room. 168 Allied The Leading Mathematics-9 Area 169

Mensuration Mensuration (b) Compute the area of the ceiling of the room. (c) Calculate the cost of carving on the ceiling at Rs. 550 per m2 . 12. (i) A hall is 36 m long and 24 m broad. Allowing 40 sq. metres for doors and windows the cost of papering the walls and ceiling at Rs. 8.40 per square metre is Rs 9441.60. Find the height of the hall. (ii) A room is 7 m long and 5 m broad. It has one door measuring 2 m by 1.5 m and two windows, each measuring 1.5 m by 1 m. The cost of painting the walls at Rs. 7.50/m2 is Rs. 495. Find the height of the room. 13. (i) The cost of carpeting a square room at Rs. 160 per sq. metre is Rs. 5760. The cost of painting the four walls at Rs. 10 per sq. metre is Rs. 840. Find the height of the room. (ii) The total cost of painting the four walls of a room at the rate of Rs. 6 per m2 is Rs. 432. Find the cost of carpeting the room at Rs. 850 per sq. metre if the room is 4 m wide and 3 m high. 4. (i) (b) 60 m2 (c) 60 m2 (d) 75 m (e) Rs. 48,375 (ii) (a) l 2 (b) 225 ft2 (c) 225 ft2 (d) 1.4 m 5. (i) (b)360 m2 , 750 m2 (c) 4800 (d) Rs. 72,000 (ii) (b) 180 m2 , 15 m2 (c) 12 (d) Rs. 180 6. (i) (a) b × a (b) 40 m2 (c) 40 m2 (d) 3 m, 13.33 m (ii) (b) 136 m2 (c) 136 m2 (d) 17 m, 8 m 7. (i) (a) 2h(l + b) (b) 112 m2 (c) 160 m2 (d) 42.85% (ii) (a) 4lh (b) 120 m2 (c) 220 m2 (d) 83.33% 8. (i) (a) 180 m (b) Rs. 7200 (ii) (a) 503 (b) Rs. 123,235 9. (i) (a) 3.5 m (b) 103.25 m2 (c) Rs. 12,906.25 (ii) (a) 2.72 m (b) 89.84 m2 (c) Rs. 11,230 10. (i) Rs. 1265 (ii) Rs. 795 11. (i) (a) 75 m2 (b) 15 m, 5 m (ii) (a) 10 m, 5 m (b) 50 m2 (c) Rs. 27,500 12. (i) 2.5 m (ii) 3 m 13. (i) 3.5 m (ii) 27200 Answers 168 Allied The Leading Mathematics-9 Area 169

Introduction Solid is one of the four fundamental states of matter (the others being liquid, gas, and plasma). It is characterized by structural rigidity and resistance to changes of shape or volume. Unlike a liquid, a solid object does not flow to take on the shape of its container, nor does it expand to fill the entire volume available to it like a gas does. The atoms in a solid are tightly bound to each other. There are some solid objects as given below: How can we find out the surface area and volume of the above solid objects? These all solid objects do not have definite shape and size. Stone and pine crone or nut are irregular objects. But bricks, Rubik's cube, ball and book are in specific shapes. We can easily find out the area, surface area and volume of these specific shaped objects. How? A solid shape is a three-dimensional figure that has width, depth and height. Examples of specific solid shapes include cuboids, cubes, polyhedrons, pyramids, spheres, cone, etc. They have faces, edges and vertices. Faces are flat sides on a solid shape which are the plane regions. Edges are the straight locations where two faces come together. Vertex is a point where more than two faces meet. Some solid shapes only have curves and irregularities instead of faces, edges and vertices, such as sphere and egg. Some shapes have a combination of faces and curved parts, such as cylinder and cone. A vertex Edges Faces Vertices Apex Edges Vertices Faces Cube Pyramid Edge Edge Axis Base r Altitude Slant Height Vertex Top Bottom (base) Sphere Cylinder Cone Circular plane surface Curved surface Circular plane surface Many of the everyday objects with which children are familiar are solid shapes. For example, building blocks are often cubes or cuboid. They have six faces, or flat surfaces. Other familiar solid shapes are spheres, which children might recognize as being shaped like balls; cones, like icecream cones or traffic cones; and cylinders, which are shaped like cans. One shape that children might not immediately recognize is a pyramid, which has one rectangular face and four triangular faces. CHAPTER 6 PRISM AND CYLINDER 170 Allied The Leading Mathematics-9 Prism and Cylinder 171 Mensuration Mensuration

6.1 Surface Area and Volume of Prism At the end of this topic, the students will be able to: ¾ find surface area and volume of prism. Learning Objectives Uses of Volume of Cuboid and Cube in Our Daily Activities Let's discuss about the pictures below; What shows the picture above? Do you know? How is object the interesting? Can you see other patterns like it? Rainbow is alike. What is the picture shown? How would it be constructed? How much amount does need for it? How can we find ? Discuss. What is the combination of the above picture 'Nepalese House'? How is the prisms used in it? Discuss. What is the name of this instrument? How does it work? Do you know how is prism used in it? Discuss in short. Cross Section of Prism A prism is a solid object with two identical ends and parallel congruent flat faces or surfaces called bases connected by the lateral faces that are rectangles or parallelograms. When we cut down a solid parallel to bases, the surface obtained is called cross-section. If the cross section of the solid is parallel and congruent to its bases, it is called a prism, otherwise it is not. The area of cross section of the prism is equal to the area of its base. So, Area of Cross Section (A) = Area of Polygonal Base of Prism The given figure below shows the prism. Prism should cut down these ways: Prism should not cut down these ways: Plane section 170 Allied The Leading Mathematics-9 Prism and Cylinder 171 Mensuration Mensuration

Show the following solids which are prism or not in the table below: Examples of prisms Examples of non-prism Cross section Aright prismis a prismin which the joining edges and faces are perpendicularto the base faces.This applies if the joining faces are rectangular. If the joining edges and faces are not perpendicular to the base faces, it is called an oblique prism. Surface Area of Prism The surface area is the area that describes the material that will be used to cover a geometric solid. When we determine the surface areas of a geometric solid we take the sum of the area for each geometric form within the solid. There are two kinds of surface area in the prism: (i) Lateral Surface Area (LSA) of the prism having bases as the polygons (e.g., triangular prism, rectangular prism, pentagonal prism, etc.) or Curved Surface Area (CSA) of the prism having the bases as circles (e.g., cylinder, etc.) and (ii) Total Surface Area (TSA) of the prism. In the adjoining figure, the faces 1 and 2 in both prisms are bases and the faces 3, 4, 5, and 6 in the rectangular prism are lateral surfaces and 3 in the circular prism is curved surface. Here, the surface area occupied by the faces 3, 4, 5 and 6 is called lateral surface area of the rectangular prism. The surface area occupied by the face 3 is called curved surface area of the circular prism. The sum of all faces including bases and lateral faces is called total surface area. To find the surface area of a prism (or any other geometric solid) we open the solid like a carton box and flatten it out to find all included geometric forms. Triangular prism 1 2 5 4 3 Net of Triangular Prism 1 4 2 h 3 5 L-shaped prism h 4 8 5 6 7 3 Net of L-shaped prism 3 4 2 5 6 7 8 1 1 2 Right triangular prism Right rectangular prism Oblique prism 5 3 2 4 6 LSA CSA Bases Rectangular Prism Circular Prism 1 3 2 1 172 Allied The Leading Mathematics-9 Prism and Cylinder 173 Mensuration Mensuration

In the net of triangular prism shown in the above figure, the lateral surface area (LSA) is the sum of the faces 2, 3 and 5, which is in rectangle form with length as h and breadth as the sum of a, b and c, i.e., perimeter of base. Similarly, in the given L-shaped prism, the lateral surface area (LSA) is the sum of the faces 3, 4, 5, 6, 7 and 8, which is in the rectangle form with length as h and breadth as the sum of a, b and c, i.e., perimeter of base. ∴ LSA = Area of rectangle form by the height and the perimeter of the base = P × h The total surface area of the prism is the sum of LSA and two bases of the prism. i.e., TSA = LSA + 2 × Area of Base = Ph + 2A Volume of Prism The volume is a measure of how much a figure can hold and is measured in cubic units. The volume tells us something about the capacity of a figure. To find the volume of a prism (or any other geometric solid) we count the number of the base area in the prism. In the adjoining right triangular prism, the perpendicular and base sides of the base right-angled triangle are p and b respectively. Then the area of the base is A = ½ pb. There are 15 such right triangles in the triangular prism. Therefore, the volume of the triangular prism is 15 times of the base. i.e., Volume of the prism (V) = 15 × A Here, A is the base area of the prism and 15 is the height of the prism. Therefore, the volume of the prism is the product of the base area and height. i.e., Volume of the prism (V) = Base Area (A) × Height (h) Example-1 Observe the given solid and answer the following questions. (a) Is the given solid a prism ? (b) Find the area of the cross-section of the solid. (c) Find the ratio of its lateral surface area and total surfacee area. (d) Calculate its volume Solution: (a) Yes, the given solid is a prism because of the cross section of the solid is congruent and parallel to its bases. (b) Here, in the given prism, AB = 10 cm, BC = 8 cm AF = CD = 3 cm, DE = BC – AF = 8 – 3 = 5 cm, EF = AB – CD = 10 – 3 = 7 cm. Height (h) = 12 cm b A p h A F 3 cm 10 cm B 8 cm C E D 3 cm 12 cm 172 Allied The Leading Mathematics-9 Prism and Cylinder 173 Mensuration Mensuration

Now, we know that Area of cross-section (A) = Base Area (A) = AB × AF + DE × CD = 10 × 3 + 5 × 3 = 30 + 15 = 45 cm2 . Perimeter of the cross-section (P) = AB + BC + CD + DE + EF + FA = 10 + 8 + 3 + 5 + 7 + 3 = 36 cm (c) Lateral Surface Area (LSA) = P × h = 36 × 12 = 432 cm2 , Total Surface Area (TSA) = LSA + 2A = 432 + 2 × 45 = 522 cm2 ∴ The ratio of LSA and TSA = 432 cm2 : 522 cm2 = 24 : 29. (d) Volume (V) = A × h = 45 × 12 = 540 cm3 . Example-2 The area and perimeter of the cross-section of a cuboid having the height 9 cm are 48 cm2 and 28 cm respectively. (a) Find the length and breadth of the cuboid. (b) Find its total surface area. Solution: (a) Here, Area of a cuboid (A) = 48 or, l × b = 48 .............. (i) Perimeter of the cuboid (P) = 28 cm or 2(l + b) = 28 or, l + b = 14 ............... (ii) or, (l + b)2 = 142 or, (l – b)2 + 4lb = 196 or, (l – b)2 + 4 × 48 = 196 or, (l – b)2 = 196 – 192 or, (l – b)2 = 4 ∴ l – b = 2 ................ (iii) Adding the eqn . (ii) and (iii), we get 2l = 16 = l = 8 cm From (ii), 8 + b = 14 ⇒ b = 6 cm (b) Height of the cuboid (h) = 9 cm ∴ TSA of the cuboid (TSA) = 2(lb + bh + lh) = 2(8 × 6 + 6 × 9 + 8 × 9) = 2 × 174 = 348 cm2 Example-3 The ratio of the length, breadth and height of a cuboid of the volume 1944 cm3 is 6:4:3. (a) Find the length, breadth and height of the cuboid. (b) Find the LSA and TSA of the cuboid. V = 1944 cm3 6x 4x 3x 174 Allied The Leading Mathematics-9 Prism and Cylinder 175 Mensuration Mensuration

Solution: (a) Here, the ratio of the length, breadth and height of a cuboid is 6:4:3. Suppose the length (l) = 6x breadth (b) = 4x and height (h) = 3x. Also, the volume of the cuboid (V) = 1944 cm3 or, l × b × h = 1944 or, 6x × 4x × 3x = 1944 or, 72x3 = 1944 or, x3 = 1944 72 = 27 = 33 ∴ x = 3 cm Hence, the length (l) = 6x = 6 × 3 cm = 18 cm, breadth (b) = 4x = 4×3 = 12 cm and height (h) = 3x = 3 × 3 cm = 9 cm. (b) The perimeter of the base of the cuboid (P) = 2(l + b) = 2(18 + 12) = 2 × 30 = 60. ∴ The area of the base of the cuboid (A) = l × b = 18 × 12 = 216 cm2 . ∴ The LSA of the cuboid (LSA) = Ph = 60 × 9 = 540 cm2 . ∴ The TSA of the cuboid (TSA) = Ph + 2A = 540 + 2 × 216 = 540 + 432 = 972 cm2 . Example-4 Each edge of a cuboid is increased by 50%. Find the percentage of increased area in the surface area of the cuboid. Solution: Let, in the cuboid, Length (l) = x, breadth (b) = y and height (h) = z ∴LSA = 2h(l + b) = 2z (x + y). After increasing 50% in each side, then the new length (L) = x + 50% of x = 150x 100 the new breadth (B) = y + 50% of y = 150y 100 the new height (H) = z + 50% of z = 150z 100 ∴ The new LSA = 2H(L + B) = 2 × 150z 100 150x 100 + 150y 100 = 150 100 × 150 100 × 2z(x + y) = 9 4 LSA ∴ Increased area = 9 4 LSA – LSA = 9 LSA – 4 LSA 4 = 5 4 LSA ∴ Percentage of increased area = 5 4 LSA LSA × 100% = 125% 174 Allied The Leading Mathematics-9 Prism and Cylinder 175 Mensuration Mensuration

PRACTICE 6.1 Keeping Skill Sharp 1. Define the following terms related to prism: (a) Prism (b) Right Prism (c) Cross-section (d) Base (e) LSA (f) TSA 2. (a) If the length of a cube is a cm, what is the area of cross-section of the cube? (b) What is the formula to compute the LSA of the prism having perimeter of the base P cm and height h cm ? (c) What is the formula to compute the TSA of the prism in which its LSA and base area A are given ? (d) Write the height of a prism with volume a cm3 and base area b cm2 . 3. (a) What is the area of the cross-section of a prism having length of the side 3.5 cm ? (b) What is the volume of a prism having the length of the cross-section 5 cm ? (c) What is the TSA of a prism with LSA 96 cm2 and base area 16 cm2 ? 4. What is the area of the cross-section of the given prism : (a) 10 cm 5 cm 8 cm (b) 8 cm Check Your Performance Answer the given questions for each situation. 5. Find the area of the cross-section of the following prisms: (a) 14 cm 5 cm 5 cm 5 cm 6 cm 10 cm (b) 13 cm 3cm 3cm 5cm 5 cm 3cm (c) 9 cm 6 cm 5 12 cm cm 3 cm 3 cm 3 cm (d) 4cm 4cm 4cm 4cm 6cm 6cm 6cm 6cm 6. Find the total surface area of the following prisms: (a) 10 cm 7 cm 6 cm (b) 8 cm 8 cm 8 cm (c) 6 cm 9 cm 12 cm 5 cm (d) 8 cm 10 cm 12 cm 176 Allied The Leading Mathematics-9 Prism and Cylinder 177 Mensuration Mensuration

7. Find the volume of the following prisms: (a) 6 cm 10 cm (b) 14 cm 8 cm 6 cm (c) 6 cm 9 cm (d) 8 cm 6 cm 12 cm 8. (i) The dimension of a Sancho box is 6 cm × 4 cm × 2 cm. (a) Define a prism. (b) Find the volume of the Sancho box. (c) Find the volume of a box which contains such 8 Sancho boxes. (ii) A box contains 20 cubical soaps of the length 15 cm each. (a) Is the soap a prism ? Give reason. (b) Find the volume of the soap. (c) Find the volume of the box. 9. (i) A cubical wooden block has a length 12 cm. (ii) A cuboid wooden block has the length 12 cm, breadth 8 cm and height 5 cm. (a) What are cube and cuboid ? Define them. (b) Find the cross-section area and lateral surface area of the block. (c) How much wood does each block contain ? Find it. 10. (a) What isthe volume of a cuboid having the area of the cross-section 25 cm2 and height 4 cm? (b) What is the volume of a squared base prism having the perimeter of the cross-section 144 cm ? (c) If the height of a squared based prism having side 6 cm is 12 cm, find itstotalsurface area. (d) The area and perimeter of the cross-section of a cuboid are 125 cm2 and 60 cm respectively. If the height of the cuboid is 7 cm, find its total surface area. 11. (i) The length, breadth and height of a box of the volume 82944 m3 are in the ratio 3:2:1. (ii) The ratio of the length, breadth and height of a cuboid is 6:4:3. If the volume of the cuboid is 1944 cm3 , find (a) Find the its length, breadth and height. (b) Find its LSA and TSA. 12. (i) A closed wooden box 15 ft long, 10 ft wide and 6 ft high, externally is made up of 1.5 inches thick wood. (a) Find the outer volume of the box. (b) Find the capacity of the box. (c) Find the weight of the box, if 1 ft3 of wood weighs 6 kg. (d) Calculate the cost of the wood at Rs. 1225 per cu. ft. 176 Allied The Leading Mathematics-9 Prism and Cylinder 177 Mensuration Mensuration

(ii) An open rectangular cistern when measured from outside is 1.35 m long, 1.08 m broad and 90 cm deep and is made of 2.5 cm thick iron. (a) Find the outer volume of the cistern. (b) Find the capacity of the cistern. (c) Find the volume of the iron used. (d) Calculate the cost of the iron at Rs. 105 per kg if the cost of 1 cc iron is 50 gm. 13. (i) A box is made of iron 0.05 cm thick, its internal dimensions are 50cm × 20cm × 20cm. (a) Find the weight of the box if 1 cm3 weighs 7 gm. (b) Find the cost of the box if the iron costs Rs. 5.50/cm3 . (ii) The inner dimensions of a closed wooden box are 20 cm, 13 cm and 8 cm. Thickness of the wood is 1 cm. (a) Find the cost of wood required to make the box if 1 cm3 of wood costs Rs. 1.50. (b) Find the weight of the box if 1 cm3 wood weighs 5 gm. 5. (a) 120 cm2 (b) 93 cm2 (c) 60 cm2 (d) 72 cm2 6. (a) 344 cm2 (b) 384 cm2 (c) 410.52 cm2 (d) 449.67 cm2 7. (a) 155.89 cm3 (b) 250.32 cm3 (c) 162 cm3 (d) 576 cm3 8. (a) 384 cm3 (b) 67500 cm3 9. (i) (b) 144 cm2 , 576 cm2 (c) 1728 cm3 (ii) (b) 96 cm2 , 200 cm2 (c) 480 cm3 10. (a) 100 cm3 (b) 46656 cm3 (c) 360 cm2 (d) 670 cm2 11. (i) (a) 72 m, 48 m, 24 m (b) 5760 m2 , 9216 m2 (ii) (a) 18 cm, 12 cm, 9 cm (b) 540 cm2 , 756 cm2 12. (i) (a) 900 ft3 (b) 826.9219 cu.ft (c) 438.469 kg (d) Rs. 89520.70 (ii) (a) 1.3122 m3 (b) 1138150 cm3 (c) 174050 cm3 (d) Rs. 913,762.50 13. (i) (a) 1 kg 686 gm, Rs. 1324.96 (ii) Rs. 1830, 6.1 kg Answers Project Work Collectthreerectangularandcubicalwoodenblockshavingthesamebreadthorotherrectangularlike solidsorboxes.Arrange theseblocksorsolidsorboxesindifferentshapedprismslike asL-shaped, T-shaped, I-shaped, H-shaped, etc. Measure the dimensions of cross-section and height of the prism so formed. Calculate the following and fill the results in the table. SN Cross-section shape Dimension of crosssection Area of cross-section Perimeter of crosssection Height of prism Lateral surface area Total surface area Volume of prism 1. 2. 3. 4. 178 Allied The Leading Mathematics-9 Prism and Cylinder 179 Mensuration Mensuration

6.2 Surface Area and Volume of Cylinder At the end of this topic, the students will be able to: ¾ find the CSA, TSA and Volume of cylinder and solve the related problems. Learning Objectives As we increase the number of sides of the polygonal base of the n–gonal prism, the length of the sides becomes very small and furthermore if the number of sides is infinitely larger and larger the n–gonal polygonal base becomes approximately equal to the circle. In this condition, the n–gonal prism becomes equivalent to a cylinder of the same height and circumference equivalent to the perimeter of polygonal base. A cylinder is a prism with circular base of infinite countable sides. The different parts of the cylinder are shown in the adjoining figure. Curved Surface Area of Cylinder We take a hollow cylinder of chart paper and cut down vertically as shown in the adjoining figure. What do you find? It is a rectangle. This is made from the curve surface of the cylinder and whose area is called the curved surface area (CSA). From the adjoining figure, Curved Surface Area (CSA) = ch = 2πrh = πdh, where h = height/length of cylinder, c = circumference r = radius and d = diameter of base. Total Surface Area of Cylinder We take a cylinder closed with both ends of chart paper and cut down vertically and around the circle as shown in the adjoining figure. From the adjoining figure, Total Surface Area (TSA) = 2A + CSA = 2πr2 + 2πrh = 2πr(r + h), where h = height/length of cylinder, A = Area of circular base. SOUP S O U P r 2πr h h h h Lateral surface area Curved surface c = 2πr CSA = Area of rectangle = l × b = ch = 2πrh O O A B D C radius radius h h h r r A=πr2 c=2πr c=2πr CSA=2πrh 178 Allied The Leading Mathematics-9 Prism and Cylinder 179 Mensuration Mensuration

Volume of Cylinder We take a cylindrical radish and cut down diagonally into 16 equal pieces and arrange them into opposite to each other as shown in the adjoining figure. This is like as a cuboid. What is the volume of this cuboid? From the figure, the volume of the cylinder, V = l × b × h = πr × r × h = πr2 h. = Area of circular base × height of prism = Ah. Remarks: (i) The curved surface area of hemi–cylinder is πrh. (ii) The total surface area of hemi–cylinder is semi–curved surface area + area of the rectangle. i.e., πr(r + h)+2rh. (iii) The volume of hemi-cylinder is 1 2 πr2 h. Volume of Hollow Cylinder Some of the most common examples of a right circular cylinder are the ordinary water pipe, the garden roller, a circular well. It may be solid or hollow. If it is hollow it has inside cylindrical surface and outside cylindrical surface. In case the cylinder is hollow with an external radius R, an internal radius r and height or length h, we have 1. Thickness of its wall/material = R – r. 2. External curved surface = 2πRh 3. Internal curved surface = 2πrh. 4. Volume of material = External volume – Internal volume = πR2 h – πr2 h = πh(R2 – r2 ) Example-1 The given cylindrical object has the radius 2.1 ft and height 7 ft. (a) What is cylinder ? Define it. (b) Find the curved surface area of the cylinder. (c) Find its total surface area. (d) Find its volume. Solution: Here, Radius of the base (r) = 2.1 ft Height of the cylinder (h) = 7 ft. (a) Now, we know that πr πr r h r πr h r R h 7 ft 2.1 ft 180 Allied The Leading Mathematics-9 Prism and Cylinder 181 Mensuration Mensuration

Curved surface area (s) = 2πrh = 2 × 22 7 × 2.1 × 7 = 92.40 ft2 (b) Total surface area = 2πr (r + h) = 2 × 22 7 × 2.1 (2.1 + 7) = 13.20 × 9.1= 120.12 ft2 (c) Volume = πr2 h = 22 7 × 2.1 × 2.1 × 7 = 97.02 ft3 Example-2 Observe the given solid. (a) Write the formula to find the total surface area of the hemi-cylinder. (b) Calculate the volume of the solid. (c) Find the total surface area of the solid. Solution: (a) Total surface area of the hemi-cylinder = πr(r + h) + 2rh (b) Here, diameter of the base (d) = 14 cm ∴ Radius of the base (r) = 14 2 = 7 cm Height of the cylinder (h) = 60 cm. Now, we know that Volume (V) = 1 2 πr2 h = 1 2 × 22 7 × 7 × 7 × 60 = 4620 cm3 . (c) Total surface area = 1 2 curved surface area + area of the rectangle = πr(r + h) + 2rh = 22 7 × 7 (7 + 60) + 2 × 7 × 60 = 1474 + 840 = 2314 cm2 . Example-3 Water supply department of metropolis office wants to build a 50,000 litres capacity cylindrical water tank at the top of the tower. The height of the tank cannot exceed 4 m. Find the radius of the tank. (a) Write the formula to find volume of cylindrical tank. (b) Find the volume of the tank. (b) Find the radius of the tank Solution : Here, Capacity of the tank = 50,000 litre = 50000 1000 m3 = 50m3 [ 1000 litre = 1 m3 ] Now, the volume of cylinder (V) = πr2 h or, 50 = 3.14 × r 2 × 4 or, r 2 = 50 3.14 × 4 or, r2 = 50 12.56 = 3.9809 ∴ r = 1.99 m 2m. 60 cm 14 cm 180 Allied The Leading Mathematics-9 Prism and Cylinder 181 Mensuration Mensuration

Example-4 A cylindrical card-board container of badminton shuttle cork with radius 3.5 cm and length 40 cm has a lid of 2 cm depth over one of its ends as shown in the diagram. (a) What area of cardboard is necessary to make it without lid? (b) What area of cardboard is necessary to its lid? (c) What area of cardboard is necessary to make it? Solution: This question is much easier to visualize if you draw its net. We can easily visualize from the below drawing that the lengths of the curved surface of the container and its lid are the same. Given, the radius of circular base/lid (r) = 3.5 cm The height of card-board container (h1) = 40 cm The height of lid (h2) = 2 cm Now, we have (a) Area of cardboard is necessary to make it without lid = Curved surface area of cylindrical container = 2πrh1 = 2 × (3.14) × (3.5) × 40 = 879.2 cm2 (b) Area of cardboard is necessary to its lid = Curved surface area of the lid = 2πrh2 = 2 × (3.14) × (3.5) × 2 = 43.96 cm2 Area of 2 circles = 2πr2 = 2. (3.14) × (3.5)2 = 76.93 cm2 (c) The area of required card-board = CSA of cylindrical container + CSA of lid + Area of 2 circles = 879.2 + 43.96 + 76.93 = 1000.06 cm2 . Example-5 (a) The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3. Calculate the ratio of their volumes. (b) 50 circular plates, each of radius 7cm and thickness 1 2 cm are placed one above the other to form a solid right cylinder. Find the total surface area. Solution : (a) Let r and R be the radii of the two cylinders whose heights are h and H respectively. Then, r R = 2 3 and h H = 5 3 Suppose v and V are their respective volumes. Then, we have v V = πr2 h πR2 H = 2 3 2 × 5 3 = 20 27 That is, the ratio of the volumes of the two cylinders is 20:27. (b) No. of circular plates (n) = 50; Radius of each circular plate (r) = 7 cm. 3.5cm 2cm 40 cm h Axis r H R Axis 182 Allied The Leading Mathematics-9 Prism and Cylinder 183 Mensuration Mensuration

Thickness of circular plate (t) = 1 2 cm. Then, the height of the cylinder formed by 50 plates (h) = nt = 50 × 1 2 = 25 cm. Total surface area of the cylinder = 2πr(r + h) = 2 × 22 7 × 7 × (7 + 25) = 1408 cm2 . Example-6 A 20 m deep well with diameter 6 m is dug up and the earth from digging is spread evenly to form a platform 18m × 10m. (a) Find the volume of the earth dug out. (b) Find the area of platform made by the earth. (c) Find the height of the platform. Solution : (a) Here, Diameter of the well (d) = 6 m. ∴ Radius (r) = 1 2 × 6 m = 3 m, Depth (l) = 20 m Volume of the earth dug out (V) = πr2 l = π × 3 × 3 × 20 = 180π m3 . (b) Area of the platform = 18 × 10 = 180 m2 . (c) Let the height of the platform be h m. Then, the volume of the platform = 180h m3 . Since, volume of the platform formed = Volume of earth dug out, i.e., 180h = 180π ∴ h = π m Hence, the height of the platform is π m. Example-7 A rectangular piece of paper is 71 cm long and 10 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder. (Take π = 355 113 ) Solution : When the paper is rolled along its length, the length (71 cm) becomes the circumference of the base of the cylinder and the breadth of the paper i.e. 10 cm becomes its height. If r denotes the radius of the base circle, then 2πr = 71 or, 2 × 355 113 × r = 71 ∴ r = 113 10 cm. Since its height (h) = 10 cm and its volume is πr 2 h = 355 113 × 113 10 × 113 10 × 10 = 4011.5 cm3 . Example-8 10 cylindrical pillars of a building have to be painted. The diameter of each pillar is 50 cm and the height 4 m. (a) Find the curved surface area of each pillar. (b) Find the curved surface area of 10 pillars. (c) What will be the cost of painting at the rate of 50 paisa per sq. m? Solution : Here, Diameter of each pillar (d) = 50 cm. ∴ Its radius (r) = 50 2 = 25 = 0.25 m 182 Allied The Leading Mathematics-9 Prism and Cylinder 183 Mensuration Mensuration

Height of each pillar (h) = 4 m Now, we have (a) Curved surface area of each pillar = 2πrh = 2 × 3.14 × 0.25 × 4 m2 = 6.28 m2 (b) Curved surface of 10 pillars = 10 × 2 × 3.14 × 0.25 × 4 m2 = 62.8 m2 (c) Cost of painting at the rate of 50 paisa per sq. m = Rs.0.5 × 62.8 = Rs. 31.4 Example-9 The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of 30 paisa per sq. m. (use π = 22 7 ) Solution : Given, Diameter of the roller (d) = 84 cm ∴ Radius of the roller (r) = 84 2 = 42 cm = 42 100 m = 0.42 m Length of the roller (l) = 120 cm = 1.2 m When the roller makes one complete revolution, the area of the playground leveled by it = its curved surface area = 2πrl = 2 × 22 7 × 0.42 × 1.2 m2 In 500 revolutions, the area of the playground leveled by the roller = 500 × 2 × 22 7 × 0.42 × 1.2 = 1584 m2 ∴ Cost of leveling at the rate of 30 paisa per sq. m = Rs. 1584 × 0.30 = Rs. 475.20. Example-10 A cylindrical underground tank of the base diameter 28 m has a capacity of 3080 cu. m. (a) Find its depth. (b) Calculate the cost of painting its inside curved surface at the rate of Rs. 4 per sq. m. Solution : Given, Capacity of the cylindrical tank (V) = 3080 m3 Diameter of the cylindrical tank (d) = 28m ∴ Its radius (r) = 28 2 = 14 m. (a) Let the depth of the tank be h m. Internal vol. of the cylinder = Capacity of the cylindrical tank or, πr2 h = 3080 or, 22 7 × 142 × h = 3080 ∴ h = 5 m. Hence, the depth of tank is 5 m. (b) Curved surface area of the cylinder = 2πrh = 2 × 22 7 × 14 × 5 = 440 m2 . ∴ Cost of painting its inside curved surface at the rate of Rs 4/m2 = Rs 440 × 4 = Rs. 1760. 184 Allied The Leading Mathematics-9 Prism and Cylinder 185 Mensuration Mensuration

Example-11 The difference between outside and inside surface areas of a cylindrical 14 cm long metallic pipe is 44 sq. cm. If the pipe is made of 99 cubic cm of metal, find the outer and inner radii of the pipe. Solution : Let R and r be the outer and inner radii of the pipe respectively. Length of the pipe (l) = 14 cm. Outer surface area = 2πRl; Inner surface area = 2πrl. Difference = 2πRl – 2πrl = 2πl(R – r) By the question, we have 44 = 2 × 22 7 × 14(R – r) ⇒ R – r = 1 2 ................. (i) Volume of the metal = Volume of external cylinder – Volume of internal cylinder = πR2 l – πr 2 l = πl(R2 –r2 ) = 22 7 × 14(R + r)(R – r) or, 99 = 22 7 × 14 × (R + r) × 1 2 or, R + r = 9 2 ...................... (ii) Adding (i) and (ii), we get 2R = 5. => R = 2.5 cm. From the eqn (ii), 2.5 + r = 4.5 ∴ r = 2 cm. Hence, the outer and inner radii of the pipe are 2.5 cm and 2 cm respectively. Example-12 A well with inner diameter 8 m and 26.25 m deep is dug. Earth taken out of it is spread evenly all around it to a width of 7 m to form an embankment. (a) Find the capacity of well below the ground level. (b) Find the circular face area of the embankment. (c) Find the height of the embankment. Solution : Here, Diameter of the well (d)=8 m Radius of the well (r)= 8 2= 4 m Depth of the well (h) = 26.25 m (a) Volume of the well below the ground level = πr2 h = 22 7 × 42 × 26.25 = 1320 m3 . So, volume of the earth dug out = 1320 m3 . Radius of the well with embankment (R) = 4 + 7 = 11 m. (b) Circular face area of the embankment = face area of well with embankment – face area of well without embankment. = πR2 – πr2 = π(R2 – r2 ) = π(R + r) (R – r) = 22 7 × (11 + 4) (11 – 4) = 330 m2 . (c) Let H be the required height of the embankment. So, volume of embankment = Circular area of embankment × height = 330 × H. Since, Volume of the embankment = Volume of the earth dug 330 × H = 1320 => H = 4 m Hence, the height of the embankment is 4 m. R r 8m 7m Ground Level 26.25 m Well 184 Allied The Leading Mathematics-9 Prism and Cylinder 185 Mensuration Mensuration

PRACTICE 6.2 Keeping Skill Sharp 1. (a) What is cylinder ? Define it. (b) Write the formula to calculate the curved surface area of the cylinder. (c) What is the total surface area of a cylinder having height h cm and radius of its base r cm? (d) What is the volume of a right cylinder having height h ft and radius r cm? 2. Find the curved surface area of the given cylinders : (a) (b) (c) (d) 3. Calculate the volume of the given cylinders : (a) (b) (c) (d) Check Your Performance 4. (i) The height of a cylindrical tin can of the base circumference 42 cm is 24 cm. (ii) The diameter of the base of a cylinder log of the height 20 cm is 28 cm. (a) Find the radius of the circular base of the can. (b) Find the curved surface area of the can. (c) Find the total surface area of the can. (d) Find the volume of the can. 5. (a) If the lateral surface of a right circular cylinder is 3960 cm2 and its base radius is 10 cm, find its height. (b) If the volume of a right circular cylinder of height 28 cm is 13200 cm3 , find the radius of the base. (c) The circumference of the base of a cylinder is 33 cm and the area of its curved surface is 528 cm2 . Find its height. (d) The volume of a cylinder is 3080 cu. cm and its diameter is 14 cm. Find its height. 6. Observe the following objects and answer the given questions. (i) (ii) (iii) (iv) 40 cm 7 cm 9 in 20 in 10 cm 12 cm 13 cm 21 cm CSA = 2112 cm2 24 cm Diameter = 21 cm 24 cm r Circumference = 44 in20 in Area of base = 12 cm2 30 cm 14 cm 30 cm 24 cm 10.5cm 21 cm 40 cm 20 cm 6 cm 90° 186 Allied The Leading Mathematics-9 Prism and Cylinder 187 Mensuration Mensuration

(a) Calculate the curved surface area of the given solids. (b) Calculate the total surface area of the given solids. (c) Calculate the volume of the given solids. 7. (i) Find the quantity of water in liter containing in the given cylindrical drum. (ii) The cross section of a 240 m long tunnel is a semicircle of radius 4 meters. (a) Calculate the area of the cross section of the tunnel, (b) How much air can contain in the tunnel ? Calculate it.. 8. (i) Find the requirement of raw materials to make the 14 brisket fire woods with radius 5 inches and height 3 inches as shown in the adjoining figure (neglect pores). (ii) The thickness of the CD is 0.15 cm and radius 6 cm. (a) Find the volume of one CD. (a) Find the volume of such 24 CDs. (iii) A tin of beans is cylindrical shape whose base has radius of 4 cm and height of 10 cm. (a) What is its capacity? (b) What area of lid is required to cover it ? 9. (i) The cylindrical tea urn is 90 cm high, and has a base of radius 30 cm. (a) Calculate capacity in litres. (b) Find the number of 200 ml cups needed to fill it. (ii) Observe the given cylindrical oil-chamber of the tanker. (a) Calculate the capacity of the tanker. (b) Find the number of 200 l drums needed to fill it. (iii) A manufacturer is making a batch of 10000 plastic cylindrical knitting needles. Each one is 6 mm in diameter and 7 cm long. (a) Find the volume of each needle. (b) What volume of plastic will be needed altogether? 10. (a) The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm3 . Find its radius. (b) The radii of two cylinders are in the ratio 4 : 3. Find the ratio of their volumes if the heights are taken to be the same. (c) The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 4. Calculate the ratio of their curved surface areas. (d) The radius of the base of a right circular cylinder is halved. What is the ratio of the volume of the reduced cylinder to that of the original one? 130 cm 60 cm 240 m 4 m 4.5 m 1.5 m 186 Allied The Leading Mathematics-9 Prism and Cylinder 187 Mensuration Mensuration

11. (i) A solid cylinder has the total surface area of 462 sq. cm. Its curved surface area is onethird of its total surface area. (ii) A solid cylinder has a total surface area of 231 cm2 . Its curved surface area is 2 3 rd of the total surface area. (a) Find the height of the solid. (b) Find the base radius of the solid. (c) Find the volume of the cylinder. 12. (i) A rectangular sheet of aluminium foil is 44 cm long and 20 cm wide. A cylinder is made out of it by rolling the foil along its wide. (a) Find the volume of the cylinder. (b) Calculate the cost of painting on its outer and inner surfaces at Rs. 1.25/cm2 . (ii) A rectangular sheet of a 44 cm long and 20 cm broad metal is rolled along its length into a cylinder so that the cylinder has 20 cm as its height. (a) Calculate the cost of iron to use in it at Rs. 18.50 per sq. cm. (b) If its outer surface is painted, how much cost for it at Rs. 1.25 per sq. cm will heed? 13. (i) A cylindrical hollow iron pipe has an external diameter 20 cm. It is 35 cm long and thickness of iron is 2 cm. (a) Find the volume of the metal used. (b) Calculate the cost of iron at Rs. 5.50 per cu. cm. (ii) A cylindrical road roller made of iron, is 1 m long. Its inner diameter is 54 cm and the thickness of the iron sheet rolled into the road roller is 9 cm. (a) Find the weight of the roller if 1 c.c. of iron weights 8 gm. (b) Calculate the cost of iron at Rs. 5.50 per cu. cm. 2. (a) 1760 cm2 (b) 1131.43 sq.in (c) 377.14 cm2 (d) 858 cm2 3. (a) 360 cm3 (b) 8316 cm3 (c) 3080 cu.in (d) 14784 cm3 4. (i)(a) 6.68 cm (b) 1008 cm2 (c) 1288.56 cm2 (d) 3365.80 cm3 (ii)(a) 14 cm (b) 1760 cm2 (c) 2992 cm2 (d) 12320 cm3 5. (a) 126 cm (b) 12.25 cm (c) 16 cm (d) 20 cm 6. (i)(a) 1584 cm2 (b) 2277 cm2 (c) 8316 cm3 (ii) (a) 1320 cm2 (b) 1628 cm2 (c) 4620 cm3 (iii)(a) 1320 cm2 (b) 2506.50 cm2 (c) 6930 cm3 (iv)(a) 377.14 cm2 (b) 617.14 cm2 (c) 1697.14 cm3 7.(i) 367.71 l (ii) (a) 25.14 m2 (b) 6034.28 m3 8. (i) 3299.94 cu.in (ii) (a) 16.97 cm3 (b) 407.28 cm3 (iii) (a) 502.86 cm3 (b) 251.43 cm2 9. (i)(a) 254.57 l (b) 1272 pcs (ii)(a) 7960 l (b) 39 pcs (iii)(a) 1.98 cm3 (b) 19800 cm3 10. (a) 5 m (b) 16:9 (c) 5:6 (d) 1:4 11. (i)(a) 3.5 cm (b) 7 cm (c) 539 cm3 (ii)(a) 7 cm (b) 3.5 cm (c) 269.5 cm3 12. (i)(a) 1398.40 cm3 (b) Rs. 1298.74 (ii)(a) Rs. 56980 (b) Rs. 1100 13. (i)(a) 3960 cm3 (b) Rs. 21780 (ii)(a) 1425.6 kg (b) Rs. 980100 Answers 188 Allied The Leading Mathematics-9 Prism and Cylinder PB Mensuration Mensuration

7.1 Surface Area and Volume of Sphere At the end of this topic, the students will be able to: ¾ find the surface area and volume of sphere and solve related problems. Learning Objectives Uses of Sphere in Our Daily Activities What is sphere ? It is perfect round solid. How many vertices, edges and plane surfaces are in the sphere ? There are no vertices, no edges and no plane surfaces. But, what does it have ? It has only curved rounded surface. It contains something. It is knowingly or unknowing used in our daily life activities as mentioned below: Surface Area of Sphere A solid as a locus of points in space equidistant from a fixed point, is called a sphere. The fixed point is called its center and the constant distant is called a radius. The examples of the sphere are football, volleyball, cricket ball, tennis ball, marble, glob, etc. In these solids, all the points on their surface are equal from the center. The locus obtained by the set of points is called its surface, which is always curved. A line segment passing through the center of the sphere is called diameter. If we cut down at any part of the sphere, we obtain a circle. If it is cut down along diameter, we find a circle greater than others. Such circle is called a great circle and its circumference is called a circumference of the sphere. In other hand, when we revolve a circle along its diameter, we obtain a sphere and each circle obtained by revolving the given circle is great circle. Thus, the generation of the CHAPTER 7 SPHERE PB Allied The Leading Mathematics-9 Sphere 189 Mensuration Mensuration

revolving circle along the diameter ‘d’times is also the surface of the sphere. Hence, the surface of the sphere is the product of the circumference and the diameter of the great circle of the sphere. i.e., The surface area (SA) = Circumference (c) × diameter (d). Also, we know that the ratio of the circumference and diameter of the same circle is called a Pi (π), where π = 3.14159265... ≈ 22 7 . i.e., π = c d = c d × d d = c × d d2 ∴ c × d = πd2 or, Surface Area of Sphere (SA) = 2πrd = πd2 = π(2r)2 = 4πr2 sq. units. Again, SA = c2 π sq. units [ c = 2πr] where d = diameter, r = radius of sphere, c = circumference of great circle. Alternatively, the occupied space by the outer cover of a sphere is the area of the surface of the sphere, which is curved. We can’t easily change into plane surface. But, the great mathematician discovered that a cylinder that circumscribes a sphere, as shown in the following diagram, has a curved surface area equal to the surface area, S, of the sphere. h = 2r r h = 2r 2πr ∴ Surface area of sphere (SA) = CSA of a cylinder = 2πrh = 2πr × 2r = 4πr2 . In next method, take a cylinder circumscribing a sphere and wrap a rope around the surface of the sphere and it can be wrapped with the same rope around the curved surface of the cylinder, OR making new circle by the same rope as shown in the following diagram. From this activity, we conclude that, Surface Area of Sphere (SA) = CSA of a cylinder = 2πrh = 2πr × 2r = 4πr2 . OR, Surface Area of Sphere (SA) = Area of new circle = π(2r)2 = 4πr2 . In other hand, take a sphere with radius ‘r’ units and attach the mustard seeds by using glue/mustard oil temporarily. Put out these seeds and make a circle by them in the plane surface. Now, measure the length of the new circle which is exactly double of the radius of the sphere. Hence, the surface area of the sphere with radius ‘r’ units is equal to the area of the circle with radius ‘2r’ units. i.e., Surface Area of Sphere (SA) = Area of a circle = π(2r)2 = 4πr2 . CircleDiameter . r R=2r 190 Allied The Leading Mathematics-9 Sphere 191 Mensuration Mensuration

Volume of Sphere Take a sphere and cylinder with the same diameter and the height of cylinder equal to its height. Mark the cylinder into three equal parts and fill the cylinder with water. Now merge the sphere into the cylinder and collect overflowed water in another same shaped and sized cylinder. After that put out the sphere from the cylinder. What you notice that? The sphere displaces the two–thirds of water containing in the cylinder and remains the one–third water in it. That means the volume of the sphere is equal to the displaced water by the sphere. Therefore, Volume of sphere (V) = 2 3 × Volume of cylinder = 2 3 × πr2 h, where height of cylinder, h = d = 2r = 2 3 × πr2 × 2r = 4 3 πr 3 3 h=d=2r d = 2r Sphere and cylinder with h = d Dropping sphere Overflowed water Remaining water 2 1 Alternatively, if four points on the surface of a sphere are joined to the centre of the sphere, then a pyramid of perpendicular height r is formed, as shown in the diagram. Consider the solid sphere to be built with a large number ofsuch solid pyramidsthat have a very small base which represents a small portion of the surface area of a sphere. We know that the volume of the pyramid is equal to the one– third of the product of base area and height. i.e., Volume of Pyramid = 1 3 Ah. But in the case of sphere, Volume of Pyramid = 1 3 Ar. If A1, A2, A3, ......... , An represent the bases areas of the pyramids on the surface of the sphere, the volume of sphere (V) = Sum of volumes of all pyramids on the surface of the sphere. = 1 3 A1r + A2r + A3r + ........ + 1 3 Anr = 1 3 (A1 + A2 + A3 + ........ + An) r = 1 3 (Surface area of sphere) r = 1 3 4πr2 . r = 4 3 πr3 . ∴ V = 4 3 πr3 cu. units. r 190 Allied The Leading Mathematics-9 Sphere 191 Mensuration Mensuration

Hemisphere When a sphere is cut down along diameter, it halves into two pieces. Each piece is called a hemisphere whose radius is the same as the radius of the sphere. In the hemisphere, we can easily see the two types of surfaces, one is curved and another is plane circle as shown in the figure below: Sphere r r Hemisphere r Now, for the hemisphere with radius r, we can identify the following: (i) Curved Surface Area of hemisphere (CSA) = 1 2 × Surface Area of Sphere = 1 2 × 4πr2 = 2πr2 (ii) Total Surface Area of hemisphere (TSA) = CSA of Hemisphere + Area of circle = 2πr2 + πr2 = 3πr2 (iii) Volume of hemisphere (V) = 1 2 × Volume of Sphere = 1 2 × 4 3 πr3 = 2 3 πr3 Example-1 A spherical iron ball has the radius 9.6 ft. (a) Write the formula to calculate curved surface area of a hemi-sphere. (b) Find the surface area of the sphere. (c) Find the volume of the sphere. (d) Calculate the cost of painting on its surface at the rate of Rs. 5.75 per sq. ft. Solution: (a) Curved surface area of a hemi-sphere = 2πr2 . (b) Here, the radius of the sphere (r) = 9.6 cft Now, we have, Surface area of the sphere (SA) = 4πr2 = 4 × 22 7 (9.6)2 = 1158.58 ft2 (c) Volume of the sphere (V) = 4 3 πr3 = 4 3 × 22 7 (9.6)3 = 3707.47 ft3 . (d) Cost of painting on its surface (C) = Rs. 5.75 × 1158.58 = Rs. 6661.84. 192 Allied The Leading Mathematics-9 Sphere 193 Mensuration Mensuration

Example-2 The curved area of the given hemisphere is 308cm2 . (a) Write the formula to calculate surface area of a sphere. (b) Find the radius of the hemisphere. (c) Find the volume of the hemisphere. Solution: (a) Surface area of a sphere = 4πr2 (b) Given, The curved area of the hemisphere (CSA) = 308cm2 . or, 2πr2 = 308 or, 2 × 22 7 × r 2 = 308 or, r2 = 49 => r = 7cm (c) Volume of the given hemisphere (V) = 2 3 πr3 = 2 3 × 22 7 × 73 = 718.67 cm3 . Example-3 Observe the adjoining spherical lead ball. (a) Find he volume of the given lead ball. (b) How many lead balls each of radius 1 cm can be made from the given spherical lead? (c) How many times the surface area of the new small spherical lead ball is less than the surface area of the given sphere? Solution: (a) Here, radius of the given spherical lead ball (R) = 8 cm. Then, the volume of the given lead ball (V) = 4 3 πR3 = 4 3 π × 8 × 8 × 8 cm3 (b) Radius of each small lead ball (r) = 1 cm. Then, the volume of each lead ball (v) = 4 3 πr3 = 4 3 π.13 = 4 3 π cm3 . ∴ Number of balls (N) = Volume of the sphere (V) Volume of each ball (v) = 4 3 π × 8 × 8 × 8 4 3 π = 512. (c) Surface area of the given lead ball (SA1) = 4πR2 = 4 × π × 82 = 256π cm2 . Surface area of the new small lead ball (SA2) = 4πr2 = 4 × π × 12 = 4π cm2 . ∴ Ratio of the new small lead ball and the given lead ball = 4π 256π = 1:64 Hence, the surface area of the new small spherical lead ball is 64 times less than the surface area of the given sphere. Example-4 A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. (a) What is the volume of the given spherical ball? (b) Find the volume of the 1st ball and 2nd ball. (c) What is the ratio of the radius of these three new balls. Solution: (a) Here, volume of the large ball of radius 3 cm = 4 3 π × 3 × 3 × 3 cm3 [. . . V = 4 3 πr3 ] rr 8 cm 192 Allied The Leading Mathematics-9 Sphere 193 Mensuration Mensuration

(b) Volume of the 1st new ball of radius 1.5 cm = 4 3 π × 1.5 × 1.5 × 1.5 cm3 . Volume of the 2nd new ball of radius 2 cm = 4 3 π × 2 × 2 × 2 cm3 . (c) Let r cm be the radius of the 3rd new ball. Then, the volume of the third new ball = 4 3 πr3 cm2 . But, volume of the large ball = Vol. of 1st ball + Vol. of 2nd ball + Vol. of 3rd ball or, 4 3 π × 33 = 4 3 π × 1.53 + 4 3 π × 23 + 4 3 π × r 3 . or, 4 3 π × 33 = 4 3 π × [1.53 + 23 + r 3 ] or, 27 = 1.53 + 8 + r 3 or, r3 = 15.625 = 125 8 => r = 5 2 = 2.5 cm. Hence, the ratio of three new balls = 3 2 : 2 : 5 2 = 3 : 4 : 5. "Alternative method for finding radius of the third ball" Let R = Radius of big sphere, r1 = Radius of 1st small new sphere, r2 = Radius of 2nd small new sphere, r3 = Radius of the 3rd small new sphere Then, R3 = (r1) 3 + (r2) 3 + (r3) 3 or, 33 = (1.5)3 + (2)3 + (r3) 3 or, 27 = 3.375 + 8 + (r3) 3 or, 27 – 11.375 = (r3) 3 or, 15.375 = (r3) 3 or, (2.5) = (r3) 3 => r3 = 2.5. Example-5 The diameter of a spherical aluminum ball is 42 cm. It is melted and drawn into a cylindrical rod of 28 cm diameter. (a) What is the volume of the spherical aluminum ball? (b) Find the length of the rod. (c) Write the relation between the length and diameter of the rod. Solution: (a) Here, diameter of the sphere (d) = 42 cm, Radius of the sphere (r) = 42 2 = 21 cm. Now, volume of the sphere (v) = 4 3 πr3 = 4 3 π × 213 = 38808 cm3 . (b) Diameter of cylindrical rod (D) = 28 cm Radius of cylindrical rod (R) = 28 2 = 14 cm. Let h be the length of the cylindrical rod. Then, the volume of the cylindrical rod (V) = πr2 h = π × 142 × l. r h 194 Allied The Leading Mathematics-9 Sphere 195 Mensuration Mensuration

But, Volume of the cylindrical rod (V) = Volume of the sphere (v). or, π × 14 × 14 × l = 4 3 π × 21 × 21 × 21 ∴ The length of the rod l = 63 cm. (c) Length of the rod (l) Diameter of rod (d) = 63 28 or, l d = 9 4 ∴ 4l = 9d. Example-6 A cylindrical tub of radius 16 cm contains water to a certain depth. When a spherical iron ball is dropped into the tub, the level of water is raised by 9 cm. (a) Find the volume of water displaced by the ball. (b) Find the radius of the ball. (c) How many parts of the ball will be wet ? Find it. Solution: (a) Here, radius of the tub (r) = 16 cm, Level of water raised (h) = 9 cm. Volume of water displaced by the iron ball = πr2 h = π × 162 × 9 = 7241.14 cm3 . (b) Let the radius of the iron ball be R cm. Then, Volume of the iron ball = 4 3 πR3 . But, volume of the iron ball = Volume of water displaced by iron ball. So, 4 3 πR3 = π × 16 × 16 × 9 or, R3 = 64 × 27 => R = 4 × 3 = 12 cm. Hence, the radius of the iron ball is 12 cm. (c) All outer surface part of the ball will be wet. PRACTICE 7.1 Keeping Skill Sharp 1. (a) Define sphere. (b) Define hemisphere. (c) What is the area of great circle of a sphere having radius r ? (d) Write down the formula to compute the surface area of a sphere with radius r. (e) Write down the formula to calculate the curved surface area of a hemisphere when its radius is given. (f) What isthe formula to calculate the totalsurface area of a hemisphere having radius‘r’units? (g) What is the volume of a sphere having radius 21 cm? (h) Write the curved surface area of a hemisphere having radius 14 cm. (i) What is the surface area of a hemisphere having radius 3.5 cm? (j) What is the radius of a hemisphere with curved surface area 8 square units? 9 cm 194 Allied The Leading Mathematics-9 Sphere 195 Mensuration Mensuration

Check Your Performance 2. Observe the following solids and answer the following questions. (i) (ii) (iii) (iv) (a) Write the definition of sphere. (b) Find the surface area of the above spherical solids. (c) Find the volume of the above spherical solids. (d) Find the relation between the total surface area and volume of each spherical solid. 3. Observe the following solids and answer the following questions. (i) (ii) (iii) (iv) (a) Write the formula to calculate the volume of a sphere. (b) Find the curved surface area and the total surface area of the above hemispheres: (c) Find the volume of the above hemispheres. (d) Write the ratio of the curved surface area and total surface area of the hemisphere. 4. (i) A sphere has a circumference of the great circle 66 cm. (a) Write the relation between the surface area of a sphere and the curved surface area of a hemi-sphere. (b) Find the radius of the sphere. (c) Find the surface area and volume of the sphere. (ii) A spherical iron ball has diameter 7 cm. (a) Write the relation between the volume of a sphere and its hemi-sphere. (b) Find the radius of the ball. (c) Find the surface area and volume of the sphere. (iii) A sphere has the surface area 154 cm2 . (a) Write the formula to calculate the volume of a sphere. (b) What is the radius of the given sphere. (c) Find the volume of the sphere. (iv) The volume of a sphere is 38808 cm3 . (a) Write the formula to calculate the total surface area of a sphere. (b) Find the diameter of its great circle. (c) How many outer layer of the sphere are there? Find it. Radius = 12.1 cm Radius = 12.5 cm Diameter = 7 cm Diameter = 10 cm r Radius = 14 cm Diameter = 3.5 ft Circumference = 44 cm Diameter = 12742 km 196 Allied The Leading Mathematics-9 Sphere 197 Mensuration Mensuration

5. (a) Find the radius of a sphere whose surface area is equal to the area of a circle with diameter 5.6 cm. (b) Find the volume of a sphere whose area of the great circle is equal to the curved surface area of a hemisphere with circumference of great circle 44 cm. (c) Calculate the volume of a sphere whose surface area is 36π m2 . 6. (i) A manufacturer company wants to make the small lead balls each of radius 1 cm from a sphere of the diameter 16 cm. (a) How much lead does it have ? Find it. (b) What is the volume of each small lead ball ? Find it. (c) How many small lead balls can he form from the lead of sphere ? Find it. (d) If it decreases 10% on the radius of the small lead ball, how many more small lead balls will form ? (ii) How many solid spheres each 6 cm in diameter can be moulded from a solid metal cylinder whose height is 45 cm and diameter 4 cm without any loss? (a) Calculate the volume of the sphere. (b) Calculate the volume of the cylinder. (c) Find the number of sphere formed from the cylinder. (iii) How many spherical bullets can be made from a cylinder 90 mm long and 4 mm in radius, each bullet being 12 mm in diameter? (iv) How many spherical lead shots each 4.2 cm in diameter can be obtained from a rectangular solid of lead with dimensions 66 cm, 42 cm and 21 cm? 8. (i) A solid right circular cylinder has a base radius of 12 cm and height of 16 cm. It is melted and made into eight spherical balls of equal size. (a) Find the volume of the cylinder and spherical ball. (b) Find the radius of the spherical ball. (c) Calculate the surface area of each spherical ball. (ii) The diameter of a copper sphere is 6 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 36 cm, find its curved surface area. (iii) The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire. 9. (i) Three solid spheres of lead are melted into a single solid sphere. If the radii of the three spheres is 1 cm, 6 cm and 8 cm respectively, find the radius of the new sphere. (ii) A sphere of lead of diameter 6 cm is melted and recast into three spherical balls. The diameters of the two of them are 3 cm and 4 cm respectively. Find the diameter of the third. 10. (i) A solid metallic sphere of radius 7 cm is cut into two halves. (a) Find the volume of a hemisphere. (b) Find the total surface area of the two hemispheres. 4 cm 45 cm 6 cm 196 Allied The Leading Mathematics-9 Sphere 197 Mensuration Mensuration

(c) What is the ration of the surface area of the sphere and the total surface area of the two hemispheres. (ii) There are two solid hemispherical objects made of earth having the same shape and size. If the total surface area of one hemispherical object is 462cm2 , find the total surface area of the sphere formed by attaching those two hemispherical objects. 11. (i) A spherical object of diameter 14 cm is dropped into water contained in a right circular cylindrical vessel of diameter 42 cm. (a) Are the volume of spherical object and displaced water by it equal ? Give reason. (b) Find the volume of a spherical object. (c) If the object is completely immersed, find how much level of the water is raised. (ii) A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and the level of water is raised by 6.75 cm. Find the radius of the ball. (iii) A cylindrical jar of radius 6 cm contains oil. Some iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by 2 cm? 2. (i) (b) 1840.58 cm2 (c) 7423.68 cm3 (d) 3V = 12.1 SA (ii) (b) 1964.29 cm2 (c) 8184.52 cm3 (d) 3V = 12.5 SA (iii) (b) 154 cm2 (c) 179.67 cm3 (d) 3V = 3.5 SA (iv) (b) 314.29 cm2 (c) 523.81 cm3 (d) 3V = 5 SA 3. (i) (b) 1232 cm2 , 1848 cm2 (c) 5749.33 cm3 (d) 2:3 (ii) (b) 19.25 ft2 , 115.5 ft2 (c) 89.83 ft3 (d) 1:6 (iii) (b) 308 cm2 , 462 cm3 (c) 718.67 cm3 (d) 2:3 (iv) (b) 255134886.29 km3 , 382702329.43 km2 (c) 5.418214535 × 1011 4. (i) (b) 10.5 cm (c) 1386 cm2 , 4851 cm3 (ii) (b) 3.5 cm (c) 154 cm2 , 179.67 cm3 (iii) (b) 3.5 cm (c) 179.67 cm3 (iv) (b) 42 cm (c) 5544 cm2 5. (a) 1.4 cm (b) 409.88 cm3 (c) 113.14 cm3 6. (i) (a) 2145.52 cm3 (b) 4.19 cm3 (c) 512 pcs (d) 702 pcs (ii) (a) 36π cm3 (b) 180π cm3 (c) 5 pcs (iii) 5 pcs (iv) 150 pcs 7. (i) (a) 7241.14 cm3 (b) 6 cm (c) 452.57 cm2 (ii) 226.29 cm2 (iii) 360 m 8. (i) 9 cm (ii) 5 cm 9.(i) (a) 7.18.67 cm3 (b) 924 cm2 (c) 2:3 (ii) 616 cm2 10. (i) (b) 1437.33 cm3 (c) 1.04 cm (ii) 9 cm (iii) 16 pcs Answers Project Work Find the surface area of a metallic spherical ball by emerging in the cylinder with specific measures filled with water. If the total surface area of a cone having specific slant height is equal to the surface area of the ball, find the volume of the cone. 20 cm 6.75 cm 198 Allied The Leading Mathematics-9 Sphere 199 Mensuration Mensuration

Additional Practice – III 1. The area of the four walls of a cuboidal room of a school is 81 square meters. The length and breadth of the room are 8 meters and 5 meters respectively. The room has one door with area 3 square meters and 2 windows with total area 2.25 square meters. (a) Write the formula to find the area of four walls. (b) Find the height of the room. (c) What is the total cost of painting the four walls of room excluding windows and door at the rate of Rs. 175 per square meter? Find it. (d) If the paper is pasted at the rate of Rs. 250 per square meter excluding the windows and door, by how much more rupees does it require than it is colored, compare it. 2. A house has a room of 8 m long, 6 m wide and 4m high, with one door of 2.5 m × 1 m and two windows of 1.5 m × 2 m. (a) Write the formula for finding the area of celing. (b) How much is the area of the four walls of the room. Calculate it. (c) Compare the area of the celing and the area of the windows and the door? (d) Find the total cost of painting in the four walls of room excluding windows and door at the rate of Rs. 45 per square meter? Find it. 3. The curved surface area of the cylinder is 748 sq. cm and the sum of radius and height is 24 cm. (a) Write the formula to find the curved surface area of cylinder. (b) Find the volume of cylinder. (c) How much total surface area is greater than curved surface area of the cylinder? 4. A prism is shown below in the figure. (a) Write the formula for finding the total surface area of the prism. (b) Find the volume of the prism. (c) Which is greater in the lateral surface area of the prism and the area of the cross section? Find it. 5. In the given two cylinder with their measurement, the sum of diameters and heights are equal. (a) Write the formula to find the base area of cylinder. (b) Are the total surface areas of both cylinders equal? Compare it. (c) Are the volumes of both cylinder equal? Compare it. 6. The length of the edge of the isosceles triangular surface is 10 feet and two equal side is 11 feet with the length of the tent is 12 feet. (a) Find the total surface area of triangular faces of the tent. (b) What should be the capacity of the tent? (c) The cost of cloths of the wide 3.75 ft of the tent is Rs. 15 per feet. Is Rs. 1500 sufficient for the cloths? Calculate and write with reason. 6 cm 15 cm 5 cm 4 cm 5 cm 14 cm 7 cm 14 cm 7 cm 12 ft 10 ft 11 ft 11 ft 198 Allied The Leading Mathematics-9 Sphere 199 Mensuration Mensuration

7. In the figure a cylindrical can is shown. The radius and height of that can are 14 cm and 21 cm respectively. (a) Write the formula to find the total surface area of a cylinder. (b) What is the cost of coloring on the surface area of that cylindrical can at the rate of 50 paisa per square cm? Find it. (c) Is there a difference in the full capacity of the cylinder if the top of the cylinder is covered or not? Write with reasons. 8. The volume of a triangular prism is 240 cm3 . The length of its triangular bases are 6 cm, 8 cm and 10 cm respectively. (a) Find the length of the triangular prism. (b) Find the ratio of the total surface area of the triangular prism to its rectangular surface area. Calculate. (c) Is the base of the triangle being right angled triangle? Justify with reason. 9. The diameter of the circular part of the hemispherical bowl is 10.5 cm. (a) Write the formula for finding total surface area of the hemisphere. (b) Find the differences between the total surface area and curved surface area of the hemispherical bowl. (c) Is the capacity of bowl 300 cm3 ? Calculate. 10. In the below two advertisement boards A and B, the length of sides of triangles and cost of painting in the board are given in the figure. (a) Write the formula for finding area of scalene triangle. (b) Find the area covered by advertisement boards A and B. (c) Compare the cost of painting in the advertisement boards A and B. 1. (b) 3.5 m (c) Rs. 14962.50 (d) Rs. 6412.50 2. (b) 112 m2 (c) 39.50 m2 more (d) Rs. 4657.50 3. (b) 2618 cm2 (c) 308 cm2 4. (b) 180 cm2 (c) 228 cm2 greater 5. (b) No (c) No 6. (a) 362 ft2 (b) 587.88 ft3 (c) Yes 7. (b) Rs. 1540 8.(a) 10 cm (b) 6:5 (c) Yes 9. (b) 86.625 cm2 (c) No 10. (b) 239.25 m2 , 216 m2 (c) Rs. 5811.68 more on A. Answers 7 cm 21 cm d = 10.5 cm 20 m 24 m 30 m A Advertisement 24 m 18 m B Advertisement Rs. 250/m2 200 Allied The Leading Mathematics-9 Sphere PB Mensuration Mensuration