Vedanta Excel in Mathematics Book 9 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 101 Vedanta Excel in Mathematics - Book 9 Facts to remember Conversion of Terai customary units into other local and international units (i) 1 Bigha = 20 Kattha (ii) 1 Kattha = 20 Dhur (iii) 1 Bigha = 400 Dhur (iv) 1 Bigha = 6772.63 m² = 72900 sq.ft. (v) 1 Kattha = 338.63 m² = 3645 sq.ft. (vi) 1 Dhur = 16.93 m² = 182.25 sq.ft. (vii) 1 Bigha = 13.31 Ropani Conversion of Hilly customary units into other local and international units (i) 1 Ropani = 16 Aana (ii) 1 Aana = 4 Paisa (iii) 1 Paisa = 4 Daam (iv) 1 Ropani = 508.72 m² = 5476 sq.ft. (v) 1 Aana = 31.80 m² = 342.25 sq. ft. (vi) 1 Paisa = 7.95 m² = 85.56 sq. ft. (vii) 1 Daam = 1.99 m² = 21.39 sq. ft. Worked-out examples Example 1: Convert the following area of the lands in to the given units. a) 2 Ropani 5 Aana 3 Paisa into Paisa b) 3 Bigha 10 Kattha 7 Dhur into Dhur c) 10 Ropani 8 Aana 2 Paisa into Ropani d) 6 Bigha 5 Kattha 10 Dhur into Bigha Solution: a) 2 Ropani 5 Aana 3 Paisa = 2 × 16 Aana + 5 Aana + 3 Paisa [∵ 1 Ropani = 16 Aana] = 37 Aana + 3 Paisa = 37 × 4 Paisa + 3 Paisa [∵ 1 Aana = 4 Paisa] = 151 Paisa b) 3 Bigha 10 Kattha 7 Dhur = 3 × 20 Kattha + 10 Kattha + 7 Dhur [∵ 1 Bigha=20 Kattha] = 70 Kattha + 7 Dhur = 70 × 20 Dhur + 7 Dhur [∵ 1 Kattha = 20 Dhur] = 1407 Dhur c) 10 Ropani 8 Aana 2 Paisa = 10 Ropani + 8 Aana + 2 4 Aana [∵ 4 Paisa = 1 Aana] = 10 Ropani + 8.5 Aana = 10 Ropani + 8.5 16 Ropani [∵ 16 Aana=1 Ropani] = 10.53125 Ropani d) 6 Bigha 5 Kattha 10 Dhur = 6 Bigha + 5 Kattha + 10 20 Kattha [∵ 20 Dhur = 1 Kattha] = 6 Bigha + 5.5 Kattha = 6 Bigha + 5.5 20 Bigha [∵ 20 Kattha=1 Bigha] = 6.275 Bigha Example 2: Mr. Majhi has a plot of land with area 2 Bigha 15 Kattha 4 Dhur. Convert it into Ropani-Anan-Paisa-Daam format. (1 Bigha = 13.31 Ropani) Solution: Here, 2 Bigha 15 Kattha 4 Dhur = 2 Bigha + 15 Kattha + 4 20 Kattha [∵ 20 Dhur = 1 Kattha] Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 102 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur = 2 Bigha + 15.2 Kattha = 2 Bigha + 15.2 20 Bigha [∵ 20 Kattha = 1 Bigha] = 2.76 Bigha = 2.76 × 13.31 Ropani [∵1 Bigha = 13.31 Ropani] = 36.7356 Ropani = 36 Ropani + 0.7356 × 16 Aana [∵ 1 Ropani = 16 Aana] = 36 Ropani + 11.7696 Aana = 36 Ropani+11 Aana+0.7696×4 Paisa [∵ 1 Aana = 4 Paisa] = 36 Ropani + 11 Aana + 3.0784 Paisa = 36 Ropani+11 Aana+3 Paisa+0.0784×4 Daam [∵1Paisa=4 Paisa] = 36 Ropani + 11 Aana + 3 Paisa + 0.3136 Daam Hence, it is written in Ropani – Aana – Paisa – Daam format as 36 – 11 – 3 – 0.3136 (Ropani). Example 3: Mr. Bhurtel bought a 7 Kattha 8 Dhur plot of land in Tulsipur, Dang. The area of the plot is given in square meter in the blue print map. What is the area of his plot in square meter? (1 Kattha = 338.63 m2 and 1 Dhur = 16.93 m2 ) Solution: Here, 1 Kattha = 338.63 m2 and 1 Dhur = 16.93 m2 Now, area of plot of land = 7 Kattha 8 Dhur = 7 × 338.63 m2 + 8 × 16.93 m2 = 2,505.85 m2 Hence, the area of his plot in the blue print map is 2,505.85 m2 . Example 4: The map of a few plots of land is shown in the given figure. If the scale drawing the map is 1: 500, answer the following questions. (i) What is the area of plot no. 88? (ii) Convert the area of plot no. 88 in Aana (1 Aana = 31.80 m2 ) (iii) What is the cost of the plot at Rs 10,00,000 per Aana? Solution: Here, the given scale is 1: 500 The lengths of sides of plot no. 88, measured by ruler, are 6.4 cm, 7.8 cm and 2 cm. ∴Actual lengths of sides are; a = 6.4 × 500 cm = 3200 cm = 32 m b = 7.8 × 500 cm = 3900 cm = 39 m c = 2 × 500 cm = 1000 cm = 10 m (i) Now, semi-perimeter (s) = 32 m + 39 m + 10 m 2 = 40.5 ∴ Area of the plot no. 88 (A) = s (s – a) (s – b) (s – c) = 40.5(40.5 – 32) (40.5 – 39) (40.5 – 10) 88 Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 103 Vedanta Excel in Mathematics - Book 9 = 40.5 × 8.5 × 1.5 × 30.5 = 15749.4375 = 125.5 m2 (ii) We have, 31.80 m2 = 1 Aana ∴The area of the plot = 125.5 31.80 Aana = 3.94654 Aana (iii) Again, the cost of 1 Aana = Rs 10,00,000 Thus, the cost of the plot = Area × Rate = 3.94654 × Rs 10,00,000 = Rs 39,46,540 Example 5: An internal airport field in Terai region is in the shape of a rhombus. The diagonals of the field are 2500 ft and 2000 ft respectively. (i) What is the formula to find the area of rhombus? (ii) Find the area of the land covered by the airport? (iii) Convert the area of the land covered by the airport in Bigha-KatthaDhur format. Solution: (i) The formula of finding the area of rhombus = 1 2 d1 × d2 (ii) Here, in a rhombus ABCD, diagonal AC (d1 ) = 2500 ft and diagonal BD (d2 ) = 2000 ft Now, the area of the rhombus ABCD = 1 2 AC (d1 ) × BD (d2 ) = 1 2 × 2500 ft × 2000 ft. = 25,00,000 sq. ft Thus, the area of land covered by the airport is 25,00,000 sq. ft (iii) We have, 72900 sq.ft = 1 Bigha or, 1 sq. ft = 1 72900 Bigha ∴ 25,00,000 sq. ft = 25,00,000 72900 Bigha Thus, area of land covered by the airport is 34.293552812 Bigha Converting it into Bigha-Kattha-Dhur format = 34 Bigha + 0. 293552812 × 20 Kattha [∵ 1 Bigha = 20 Kattha] = 34 Bigha + 5. 87105624 Kattha = 34 Bigha + 5 Kattha + 0. 87105624 × 2 Dhur [∵ 1 Kattha = 20 Dhur] = 34 Bigha + 5 Kattha + 17.42 Dhur Hence, it is written in Bigha-Kattha-Dhur format as 34 – 5 – 17.42 (Bigha). EXERCISE 5.2 General section 1. a) How many Aana are there in 1 Ropani? b) How many Paisa are there in (i) 1 Aana? (ii) 1 Ropani? c) How many Daam are there in (i) 1 Paisa? (ii) 1 Aana? (iii) 1 Ropani? A B C D 2500 ft 2000 ft Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 104 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur d) How many Kattha are there in 1 Bigha? e) How many Dhur are there in (i) 1 Kattha? (ii) 1 Bigha? f) How many Ropani are there in 1 Bigha? 2. a) Convert the following area of the lands in to the units as indicated. (i) 6 Ropani into Aana (ii) 10 Ropani 7 Aana into Aana (iii) 12 Ropani 10 Aana 2 Paisa into Paisa (iv) 20 Ropani 15 Aana 3 Paisa into Paisa (v) 5 Bigha into Kattha (vi) 6 Bigha 9 Kattha into Kattha (vii) 7 Bigha 7 Kattha 7 Dhur into Dhur (viii) 8 Bigha 9 Kattha 10 Dhur into Dhur b) Convert the following area of the lands in to the units as indicated. (i) 32 Aana into Ropani (ii) 3 Ropani 8 Aana into Ropani (iii) 2 Ropani 8 Aana 4 Paisa into Ropani (iv) 20 Ropani 4 Aana 8 Paisa into Ropani (v) 100 Kattha into Bigha (vi) 2 Bigha 10 Kattha into Kattha (vii) 20 Bigha 10 Kattha 5 Dhur into Bigha (viii) 11 Bigha 4 Kattha 10 Dhur into Bigha Creative Section 3. a) Mr. Majhi has 5 Bigha 4 Kattha land in Jhapa district. Change it into Ropani-AanaPaisa-Daam format. (1 Bigha = 13.31 Ropani) b) Mr. Jha has 10 Bigha 10 Kattha 10 Dhur plot of land in Dhanusha district. Change it into Ropani-Aana-Paisa-Daam format. c) Mr. Deuba has 73 Ropani 8 Aana 2 Paisa 1.664 Daam land in Sudurpaschim province. Change it into Bihga-Kattha-Dhur system. d) Mrs. Gurung has 29 Ropani 13 Aana 0.4864 Daam land in Gandaki province. Change it into Bihga-Kattha-Dhur system. 4. Study the following relations and solve the questions given below. Unit In square meter In square feet 1 Ropani = 16 Aana = 64 Paisa 508.72 m2 5476 sq. ft 1 Bigha = 20 Kattha = 400 Dhur 6772.63 m2 72900 sq. ft a) A rectangular field is 50 m long and 40 m wide. Convert the area of the field in Ropani. Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 105 Vedanta Excel in Mathematics - Book 9 b) A rectangular field is 185 ft long and 74 ft wide. Convert the area of the field in Aana. c) The side of a squared ground is 180 m, find its area in Bigha. d) The edges of a triangular kitchen garden are 13 m, 14 m and 15 m respectively, find the area of the garden in Dhur. 5. The map of a few plots of land is shown in the given figure. Measure the dimensions of the plots with a ruler and answer the following questions. 167 168 169 170 171 172 Scale 1 : 500 a) (i) What are the actual lengths of sides of plot no. 170? (ii) What is the area of plot no. 170? (iii) Convert the area of plot no. 170 in Aana. (iv) Calculate the cost of the plot at Rs 15,00,000 per Aana. b) (i) What are the actual lengths of sides of plot no. 171? (ii) What is the area of plot no. 171? (iii) Convert the area of plot no. 171 in Aana. (iv) Find the cost of the plot at Rs 13,50,000 per Aana. c) (i) Write the actual lengths of sides of plot no. 167. (ii) Find the area of plot no. 167. (iii) Convert its area in Dhur. (iv) Find the cost of the plot at Rs 2,00,000 per Dhur. d) (i) Write the actual lengths of sides of plot no. 172. (ii) Find the area of plot no. 172. (iii) Convert its area in Dhur. (iv) Find the cost of the plot at Rs 2,50,000 per Dhur. 6. a) A municipality is building a hospital in the land of the shape of a rhomus having its diagonals 250 m and 480 m respectively in a mountainous district of Nepal. (i) What is the formula to find the area of rhombus? (ii) Find the area of the land covered by the hospital? (iii) Convert the area of the land covered by the hospital in Ropani-Aana-PaisaDaam format. b) Government is building an international airport in the land of the shape of a quadrilateral in Terai region. The longer diagonal of the land is 800 m. The perpendiculars drawn to the longer diagonal from the opposite vertices are 150 m and 200 m. Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 106 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (i) Write the formula to find the area of quadrilateral when the perpendiculars with lengths ‘a’ and ‘b’ are drawn from two opposite vertices on the diagonal ‘d’ of a quadrilateral. (ii) Find the area of the land covered by the airport? (iii) Convert the area of the land covered by the airport in Bigha-Kattha-Dhur format. c) Mrs. Gosain has a piece of land in the shape of rhombus in Bhaktapur. The diagonals of the field are 16 m and 15.9 m respectively. He sold the land to Mrs. Ojha at the rate of Rs 25,00,000 per Aana. How much did Mrs. Ojha pay to Mr. Gosain? Project work and activity section 7. a) Cut chart papers into a few number of triangles and quadrilateral shapes. Measure the dimensions of each triangular and quadrilateral piece. Then, find the area of each shape. b) Measure the length and breadth of your school ground, find its area in square meter and then convert it into (i) square feet (ii) Kattha - Dhur c) Prepare a report on the units of measuring the land in your locality. Then, present it in the class. 5.6 Area of 4 walls, floor and ceiling A rectangular room has four walls, a floor and a ceiling. The floor and ceiling are congruent. The walls along the length are congruent and the walls along the breadth are congruent. If l, b, and h be the length, breadth and the height of a room, (i) Area of floor = l × b (ii)Area of ceiling = Area of floor = l × b (iii) Area of opposite walls along the length = l × h + l × h = 2lh (iv) Area of opposite walls along the breadth = b × h + b × h = 2bh (v) Area of 4 walls of the room = 2lh + 2bh = 2h (l + b) Facts to remember 1) The perimeter of the floor of the room (P) = 2 (l + b) The area of four walls = 2h (l +b) = 2(l +b) × h = P × h 2) For a square room, length = breadth = l and height = h a) Area of floor = Area of ceiling = l × l = l2 b) Area of four walls = 2h (l + l) = 2h (2l) = 4lh c) Area of room including floor and ceiling = 2l 2 + 4lh 3. For a cubical room, length = breadth = height = l a) Area of floor = Area of ceiling = l × l = l2 b) Area of four walls = 2l (l + l) = 2l (2l) = 4l 2 c) Area of room including floor and ceiling = 2l 2 + 4l 2 = 6l 2 Ceiling Floor b h l Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 107 Vedanta Excel in Mathematics - Book 9 If l 1 and h1 be the length and height of a window, l 2 and h2 be the length and height of a door, (i) Area of 4 walls excluding a window = 2h (l + b) – l 1 h1 (ii) Area of 4 walls excluding a door = 2h (l + b) – l 2 h2 (iii) Area of 4 walls excluding a window and a door = 2h (l + b) – l 1 h1 – l2 h2 Facts to remember Let A = Area of floor, four walls, ceiling etc. a = Area of each piece of carpet, wallpaper etc. N = Number of pieces of carpet required for the floor, wallpaper for the walls etc. n = Number of designs made in 1 sq. units in the floor, walls etc. Then, (i) N = A a (ii) a = A N (iii) A = N × a (iv) N = n × A Worked-out examples Example 1: The length, breadth, and height of a rectangular hall are 20 m, 15 m and 6.5 m respectively. (i) Find the area of its floor. (ii) Find the area of its ceiling. (iii) Find the area of opposite walls along the length. (iv) Find the area of opposite walls along the breadth. (v) Find the area of its four walls. (vi) Find the area of the room including floor and ceiling. Solution: Here, the length of the hall (l) = 20 m The breadth of the hall (b) = 15 m The height of the hall (h) = 6.5 m (i) The area of the floor (A1 ) = l × b =20 m×15 m = 300 m2 (ii) The area of the ceiling (A2 ) = l × b =20 m×15 m = 300 m2 (iii) The area of the opposite walls along the length = 2 (l × h) =2 × 20 m × 6.5 m = 260 m2 (iv) The area of the opposite walls along the breadth = 2 (b × h)=2 × 15 m × 6.5 m = 195 m2 (v) The area of the four walls (A3 ) = 2h (l + b) = 2 × 6.5 m (20 m+15 m) = 13 m × 35 m = 455 m2 (vi)The area of the room including floor and ceiling (A) = A1 +A2 +A3 = 300 m2 + 300 m2 + 455 m2 = 1,055 m2 h1 h2 l 1 l 2 h b l 6.5 m 20 m 15 m Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 108 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: The guest room in Sandhya’s house is 10 m long, 8 m wide and 4.5 m high. It contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. If her parents wish to paste the walls with 3D paper, find the area of the walls covered with paper. If the size of each piece of paper is 8 sq. m, how many pieces of wallpaper are required for the room? Solution: Here, the length of the room (l) = 10 m The breadth of the room (b) = 8 m The height of the room (h) = 4.5 m Now, the area of its 4 walls = 2h (l + b) = 2×4.5 m (10 m + 8 m) = 162 m2 Also, the area of 2 windows =2 (l 1 ×h1 ) = 2 (2 m × 1.5 m) = 6 m2 The area of a door = l2 ×h2 = 1 m × 4 m = 4 m2 ∴The area of 4 walls excluding windows and door = (162 – 6 – 4) m2 = 152 m2 Hence, the area of the walls covered with paper is 152 m2 . Again, the size of each piece of wallpaper (a) = 8 m2 ∴ Number of paper required for the room (N) = Area of walls excluding windows and door (A) Area of each piece of wall paper (a) = 152 m2 8 m2 = 19 Hence, 19 pieces of 3D paper are required for pasting on the walls of the room. Example 3: The dining room of a restaurant is 25 m long, 18 m wide and 6 m high. It contains six windows of size 2.5 m × 2 m each and two doors of size 1.2 m × 5 m each. (i) How many dining table sets are required to arrange on its floor at 1 table set per 15 sq. meters? (ii) How many lamps are required to hang on its ceiling at 3 lamps per 30 sq. meters? (iii) How many paintings are required to hang on the walls at 5 painting per 79 sq. meters? Solution: Here, the length of the dining room of the restaurant (l) = 25 m The breadth of the dining room of the restaurant (b) = 18 m The height of the dining room of the restaurant (h) = 6 m (i) Area of floor = l×b = 25 m×18 m = 450 m2 Now, in 15 sq. meters, 1 table sets can be arranged. or, In 1 sq. meters, 1 15 table sets can be arranged. or, In 450 sq. meters, 1 15 × 450 = 30 table sets can be arranged. Hence, 30 table sets are required to arrange on its floor. (ii) Area of ceiling = l×b = 25 m×18 m = 450 m2 Also, in 30 sq. meters, 3 lamps can be hanged. Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 109 Vedanta Excel in Mathematics - Book 9 or, In 1 sq. meters, 3 30 lamps can be hanged. or, In 450 sq. meters, 1 10 × 450 = 45 lamps can be hanged. Hence, 45 lamps are required to hang on the ceiling. (iii) The area of its 4 walls = 2h (l + b) = 2×6 m (25 m + 18 m) = 516 m2 Also, the area of 6 windows = 6 (l 1 ×h1 ) = 6 (2.5 m × 2 m) = 30 m2 The area of 2 doors = 2(l 2 ×h2 ) =2 ×1.2 m × 5 m = 12 m2 ∴The area of 4 walls excluding windows and door = (516 – 30 – 12) m2 = 474 m2 Also, in 79 sq. meters, 5 paintings can be hanged. or, In 1 sq. meters, 5 79 paintings can be hanged. or, In 474 sq. meters, 5 79 × 474 =30 paintings can be hanged. Hence, 30 paintings can be hanged on the walls. Example 4: The rectangular room is 24 ft long. It contains three windows of size 4 ft. × 3 ft. each and a door of size 3 ft × 8 ft. If 4 pieces of carpets each of size 12 ft × 8 ft are required for carpeting its floor and 8 pieces of paper each of size 82.5 sq. ft. are required for pasting on its walls, determine the height of the room. Solution: Here, the length of the room (l) = 24 ft. Area of 4 pieces of carpet = 4 × (12 ft. × 8 ft.) = 384 ft2 . Now, area of floor = area of 4 pieces of carpet or, l × b = 384 ft2 . or, 24 × b = 384 ∴ b = 16 ft. Also, the area of its 4 walls = 2h (24 +16) = 80h ft2 . Again, the area of 3 windows =3 (l 1 ×h1 ) = 3 (4 ft. × 3 ft.) = 36 ft2 . The area of a door = l2 ×h2 = 3 ft. × 8 ft. = 24 ft2 . ∴The area of 4 walls excluding windows and door = (80h– 36–28) ft2 = (80h – 60) ft2 . But, area of walls excluding windows and door = area of 8 pieces of wallpapers or, 80h – 60 = 8 × 82.5 ft2 or, 80h – 60 = 660 ft2 or, 80h = 720 ∴ h = 9 ft. Hence, the required height of the room is 9 ft. h b 24 ft. https://www.geogebra.org/m/dbjdzpej Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 110 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 5.3 General section 1. a) What is the relation between the areas of floor and ceiling in a rectangular room? b) Write the relation between the areas of opposite walls of a rectangular room. 2. a) From the adjoining room, write down the area of its following parts. (i) Floor (ii) Ceiling (iii) Floor and ceiling (iv) Opposite walls along the length (v) Opposite walls along the breadth (vi) Four walls (vii) Walls and floor (viii) Walls and ceiling b) The perimeter of the floor of a rectangular room is P m and the height of the room is Q m, what is the area of its four walls? c) The length of a square room is x m and its height is y m, what is the area of its four walls? d) If the length of a cubical room is a ft., write down the area of its four walls and ceiling. e) In a rectangular room, if the area of the floor is A m2 and the area of its four walls is B m2 , what is the total surface area of the room? 3. a) The length of a square room is 10 m, find the area of its floor. b) A square room is 25 ft. long. Find the area of its ceiling. c) A cubical room is 12 m long. What is the area of its four walls? d) If the perimeter of a rectangular room having height 4 m is 36 m, calculate the area of its walls. e) If 5 pieces of carpets having dimensions 10 ft. × 6 ft. of each piece are required for carpeting the floor of a rectangular room, find the area of the floor. f) If 60 tiles each of having dimensions 2 ft. × 2 ft. are required for paving the floor of a rectangular room, find the area of the ceiling of the room. g) The area of 4 walls with two windows and a door is 220 m2 . If each window is 3 m2 and the door is 5 m2 , find the area of the walls excluding windows and door. Creative section-A 4. a) The dimensions of a room are given in the figure aside. (i) Find the area of its floor. (ii) Find the area of its ceiling. (iii) Find the area of opposite walls along the length. (iv) Find the area of opposite walls along the breadth. (v) Find the area of its 4 walls. (vi) Find the area of the rooms including floor and ceiling b) A square room is 12 m long and 5 m high. (i) Find the area of its floor and ceiling. (ii) Find the area of its 4 walls. (iii) Find the area of the room including floor and ceiling. y m x m z m 10m 15m 5m Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 111 Vedanta Excel in Mathematics - Book 9 c) The length of cubical room is 15 ft. (i) Find the area of its floor and ceiling. (ii) Find the area of its 4 walls. (iii) Find the area of the room including floor and ceiling. 5. a) The area of four walls of a room is 210 m2 . If the length and breadth of the room are 12 m and 9 m respectively, find the height. b) The area of four walls of a room is 639 ft2 . If the breadth and height of the room are 15.5 ft and 9 ft respectively, find the length. c) A room is 24 ft. long and 16 ft. wide. If the area of its floor and ceiling is equal to the area of four walls, find the height of the room. 6. a) A rectangular room is twice as long as it is broad and height is 4.5 m. If the area of its 4 walls is 216 m2 , find the area of the floor. b) The length of a rectangular hall is two times of its breadth and the breadth is two times of its height. If the area of its ceiling is 200 ft2 , find the area of the four walls. 7. a) A rectangular room is 8 m long, 6 m broad and 4 m high. It contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. Find the area of walls excluding windows and door. b) A square hall is 15 m long and 5 m high. It contains three square windows each of 2 m long and two doors of size 1.5 m × 4 m. If its walls are white washing, find the area of walls covered with white washing. c) Shambhu is painting the walls and ceiling of a recreation hall with length, breadth and height 40 ft., 25 ft. and 18 ft respectively. It contains three windows of size 6 ft. × 5 ft. m each and two doors of size 4 ft. × 10 ft. How much part of the walls and ceiling will be painted by him? Creative section-B 8. a) The length, breadth and height of Shashwat’s classroom are 9 m, 6 m and 4.5 m respectively. It contains two windows of size 1.7 m × 2 m each and a door of size 1.2 m × 3.5 m. (i) Find the area of four walls excluding windows and door. (ii) How many decorative chart papers are required to cover the walls at 2 chart paper per 8 sq. meters? b) A Taekwondo hall is 30 m long, 20 m wide and 5 m high.It contains four windows of size 2.5 m × 2 m each and two doors of size 1.5 m × 4 m each. (i) How many player's-posters are required to hang on the walls at 3 posters per 108 sq. meters? (ii) How many taekwondo mats are required to pave on the floor at 5 mats per 15 sq. meters? Mensuration (I): Area

Vedanta Excel in Mathematics - Book 9 112 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 9. a) The study room of Sunayana is 8 m long, 6 m wide and 4.5 m high. It contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. If she is papering the walls with 3D paper of size 4 m2 each, how many pieces of wallpaper are required for her room? b) There is an auditorium hall of length 50 m, breadth 30 m and height 10 m in Chhiring’s school. It contains six windows of size 3 m × 2 m each and three doors of size 2 m × 6.5 m each. If its walls and ceiling are covered with compressed fiberboard of size 5 m2 each, find the of fiberboard used in the hall. 10. a) The rectangular room is 20 ft long. It contains two windows of size 4 ft. × 3.5 ft. each and a door of size 3 ft × 8 ft. If 5 pieces of carpets each of size 10 ft × 6 ft are required for carpeting its floor and 8 pieces of paper each of size 81sq. ft. are required for pasting on its walls, find the height of the room. b) A rectangular room is twice its breadth. It contains four windows of size 2 m × 1.75 m each and two doors of size 1 m × 3.5 m. If 24 pieces of tiles of size 2 m × 1.5 m each are required for tiling its floor and 41 pieces of 3D paper of size 3 m2 each, are required for pasting on its 4 walls, find the height of the room. Project work and activity section 11. a) Measure the length, breadth, and height of your bedroom. Also measure the length and height of windows and doors using measuring tapes. Then, solve the following problems and preset in the classroom. (i) Find the area of the floor. (ii) Find the area of the ceiling. (iii) Find the area of the four walls including windows and doors. (iv) Find the area of the windows and doors. (v) Find the area of the four walls excluding windows and doors. b) Make a group of 10 friends and measure the length, breadth, and height of your classroom, its windows, doors and whiteboard and notice-board using measuring tapes. Then, prepare a report about the following problems and present in the classroom. (i) The area of the floor. (ii) The area of the ceiling. (iii) The area of the four walls including windows and doors. (iv) The area of windows, doors and white board. (v) The area of the walls and ceiling excluding windows and doors where the white washing is done. (vi) The estimated number of chart paper for papering on the walls excluding the windows, doors and whiteboard. Mensuration (I): Area

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 113 Vedanta Excel in Mathematics - Book 9 OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. The semi-perimeter of a triangle having sides 25 cm, 15 cm and 20 cm is (A) 60 cm (B) 30 cm (C) 45 cm (D) 15 cm 2. The amount of surface enclosed by the boundary line of a plane closed figure is its (A) Area (B) Perimeter (C) Volume (D) Semi-perimeter 3. What is the area of triangle whose sides are a, b and c cm and semi-perimeter is s cm? (A) s (s – a) (s – b) (s – c) cm2 (B) s(s + a)(s + b)(s + c) cm2 (C) (s – a) (s – b) (s – c) cm2 (D) s (s – a) (s – b) (s – c) cm2 4. The Heron’s formula to calculate the area of triangle is (A) 3 4 a2 (B) b 4 4a2 – b2 (C) s (s – a) (s – b) (s – c) (D) 1 2 × b × h 5. The area of triangular park is 900 m2 . What is the cost of covering the park with turfs at the rate of Rs 150 per square meter? (A) Rs 1,35,000 (B) Rs 5,13,000 (C) Rs 1,53,000 (D) Rs 3,15,000 6. If the area of an equilateral triangle is 9 3 cm2 , what is its perimeter? (A) 6 cm (B) 18 cm (C) 6 3 cm (D) 27 3 cm 7. The semi-perimeter of a triangle is 20 cm and two of its sides are 10 cm and 16 cm, what is the third side of the triangle? (A) 6 cm (B) 10 cm (C) 14 cm (D) 16 cm 8. In a rectangular room, which of the parts don’t have equal area? (A) The floor and ceiling (B) The opposite walls along length (C) The opposite walls along breadth (D) The four walls and the floor 9. A rectangular room is x m long, y m wide and z m high. What is the area of its four walls? (A) x (y + z) m2 (B) 2y (x + z) m2 (C) 2x (y + z) m2 (D) 2z (x + y) m2 10. The perimeter of the floor of a rectangular hall is 120 ft. and height is 12 ft. what is area of its four walls? (A) 132 ft2 . (B) 108 ft2 (C) 1,440 ft. (D) 1,440 ft2 .

Vedanta Excel in Mathematics - Book 9 114 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6.1 Prism – Looking back Classwork-Exercise Let's study the figures given below and answer the questions as quickly as possible. a) (i) What is the perimeter of the rectangle? (ii) What is the area of the rectangle? b) (i) What is the perimeter of the triangle? (ii) What is the area of the triangle? c) (i) What is the perimeter of base of the cuboid? (ii) What is the area of base of the cuboid? (ii) What is the total surface area of the cuboid? (iii) What is the volume of the cuboid? 6.2 Area and volume of solids Solids are three dimensional objects. Therefore, they occupy space. Cube, cuboid, cylinder, sphere, pyramid, cone, etc. are the examples of solid objects. Length, breadth and height (or thickness) are three dimensions of cube and cuboid. (i) Area and volume of a cuboid A cuboid has six rectangular faces. In the adjoining cuboid, ABCD, BGFC, EFGH, CDEF, ADEH and ABGH are the six rectangular faces of the cuboid. Here, the total surface area of the cuboid is the sum of the areas of its six faces. A = lb + lb + bh + bh + lh + lh = 2 (lb + bh + lh) So, the total surface area of cuboid (A) = 2 (lb + bh + lh) Similarly, the volume of cuboid (V) = Area of the base × height = l × b × h A B C 9 cm 20 cm D A B C 5 cm 3 cm 4 cm 10 cm 15 cm 5 cm D A l B b h C H G E F Unit 6 Mensuration (II): Prism

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 115 Vedanta Excel in Mathematics - Book 9 (ii) Area and volume of a cube In case of a cube, its length, breadth and height are equal. i.e., l = b = h So, the total surface area of a cube = 2 (l.l + l.l + l.l) = 2 × 3l 2 = 6l 2 Similarly, volume of a cube = l × l × l = l 3 6.3 Prisms In geometry, a prism is a flat solid which has two opposite congruent and parallel faces called bases. All prisms have any kind of polygonal bases and all the other faces are rectangular in shape. A few examples of prism are rectangular prism (cuboid), triangular prism, pentagonal prism etc. Triangular prism Square prism Pentagonal prism a) Cross-section of prism (i) If a piece of biscuit is taken out from a packet of biscuits, it is the shape of the piece of biscuit exactly the same to the shape of the biscuits remained in the same packet? (ii) If a 1000 rupee note is taken out from a bundle of 1000 rupees notes, is the note congruent to the notes remained in the bundle? A cross section of a prism is a plane surface which is congruent and parallel to the bases of the prism. A cross section is obtained by cutting the prism into plane surface (like a slice) perpendicular to the height or length of the prism. A prism has uniform and infinite of cross sections. The cross section of this prism is L-shaped. The cross section of this prism is -shaped. The cross section of this prism is H -shaped. The cross section of a cuboid is a rectangle. The cross section of a cube is a square. The cross section of a triangular prism is a triangle. A triangular prism with triangular bases A rectangular prism with rectangular bases A square prism with square bases Mensuration (II): Prism

Vedanta Excel in Mathematics - Book 9 116 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Facts to remember 1) The area of cross section or base of rectangular prism (cuboid) = l × b 2) The area of cross section or base of square prism = l 2 b) Volume of prism We have already discussed that a cross section of a prism is congruent to its base. Therefore, area of cross section (or cross sectional area) is, of course; equal to the area of the base of a prism. Then, the volume of a prism is the product of its cross sectional area and height. i.e., Volume of a prism = Area of cross section × height c) Lateral Surface Area (L.S.A.) of a prism The lateral surfaces of a prism are the rectangular faces excluding the bases of the prism. In the adjoining rectangular based prism (cuboid), Lateral surface area (L.S.A.) = Area of opposite walls along the length + Area of opposite walls along the breadth = 2lh + 2bh = 2h (l + b) = 2 (l + b) × h = Perimeter of base × height (h) L.S.A. of a prism = Perimeter of base × height, i.e. P × h d) Total Surface Area (T.S.A.) of a prism The total surface area of a prism is the sum of the lateral surface area and two times the area of its cross section (or two times the area of a base). i.e., Total surface area of a prism = lateral surface area + 2 × area of cross section Worked-out Examples Example 1: Calculate (i) the area of cross section (ii) lateral surface area (iii) total surface area (iv) volume of the adjoining prism. Solution: (i) Here, the area of cross section = Area of (ABCD – FGCE) = 10 × (2 + 2) cm2 – 5 × 2 cm2 = 30 cm2 Alternative process Area of cross section = Area of trapeziums (ABCD + ADEF) = 1 2 × 2 (10 + 5) cm2 + 1 2 × 5 (4 + 2) cm2 = 15 cm2 + 15 cm2 = 30 cm2 l b h 10 cm 2cm 5cm 5cm 2cm 8 cm 10 cm 2 cm 2 cm 5 cm A B E C F G D F 5cm E A B C D 2cm 2cm 10 cm 5cm Mensuration (II): Prism

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 117 Vedanta Excel in Mathematics - Book 9 (ii) Lateral surface area = Perimeter of cross section × length = (10 + 2 + 5 + 2 + 5 + 4) cm × 8 cm = 224 cm2 (iii) Total surface area = lateral surface area + 2 × cross sectional area = 224 cm2 + 2 × 30 cm2 = 284 cm2 (iv) Volume of the prism = Area of cross section × height = 30 cm2 × 10 cm = 300 cm3 Example 2: Calculate (i) the area of cross section (ii) lateral surface area (iii) total surface area (iv) volume of the adjoining prism. Solution: (i) Area of cross section = (9 × 4) cm2 + (3 × 4) cm2 = 48 cm2 (ii) Lateral surface area = Perimeter of cross section × length = (9 + 4 + 3 + 4 + 3 + 4 + 3 + 4) cm × 8 cm = 34 cm × 8 cm = 272 cm2 (iii) Total surface area = lateral surface area + 2 × cross sectional area = 272 cm2 + 2 × 48 cm2 = 368 cm2 (iv) Volume of the prism = Area of cross section × height = 48 cm2 × 8 cm = 384 cm3 Example 3: A rectangular metallic block is 30 cm long, 25 cm broad and 8 cm high. How many pieces of rectangular slices each of 5 mm thick can be cast lengthwise from the block? Solution: Here, length of the block = 30 cm Breadth of the block = 25 cm Height of the block = 8 cm ∴ Volume of block = 30 cm×25 cm×8 cm Also, volume of each slice = 0.5 cm×25 cm× 8cm ∴ Number of slices = volume of block volume of each slice = 30 × 25 × 8 0.5 × 25 × 8 = 60 Hence, the required number of rectangular slices is 60. 3 cm 9 cm 3 cm 3 cm 4 cm 4 cm 4 cm 8cm 30 cm 25cm Here, thickness of each slice = 5 mm = 0.5 cm Length of the block = 30 cm When it is cut lengthwise, number of slices = 30 cm÷0.5 cm = 60 Alternative process: Mensuration (II): Prism

Vedanta Excel in Mathematics - Book 9 118 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 4: A cubical vessel of length 10 cm is completely filled with water. If the water is poured into a rectangular vessel 25 cm × 20 cm × 10 cm, find the height of the water level in the rectangular vessel. How much more water is poured into the vessel to fill it completely? (1 l = 1000 cm3 ) Solution: Here,volume of water in cubical vessel = 10 cm × 10 cm × 10 cm Volume of water in rectangular vessel = 25 cm × 20 cm × h cm Now, 25 × 20 cm × h cm = 10 cm × 10 cm × 10 cm or, h = 10 cm × 10 cm × 10 cm 25 cm × 20 cm = 2 cm Again, remaining height of rectangular vessel to be filled = 10 cm - 2 cm = 8 cm Then, volume of empty space of the vessel = 25 cm × 20 cm × 8 cm = 4000 cm3 Now, 1000 cm3 = 1 l ∴ 4000 cm3 = 1 1000 × 4000 l = 4 l Hence, the required height of water level in the rectangular vessel is 2 cm and 4 l of water is needed to pour into it to fill it completely. EXERCISE 6.1 General section 1. a) What is a prism? Define with examples. b) Define a cross section of a prism. What is the shape of cross section of a cuboid? c) Define lateral surface of a prism. What are the lateral surfaces of a triangular prism? d) If the area of a base of a prism is 24 cm2 , what is its cross sectional area? 2. a) The area of cross section of a prism is 36 cm2 and its height is 15 cm. Find its volume. b) The perimeter of cross section of a prism of length 10 cm is 48 cm. Find its lateral surface area. c) The lateral surface area and the cross sectional area of a prism are 175 cm2 and 40 cm2 respectively. Find total surface area of the prism. d) The perimeter of cross section and the area of cross section of a prism are 36 cm and 54 cm2 respectively. If the length of the prism is 15 cm, find its total surface area. 3. a) The volume of a cube is 125 cm3 . Find its total surface area. b) The total surface area of a cube is 96 cm2 . Find its volume. c) The length and the breadth of a cuboid are 15 cm and 8 cm respectively. If the volume of the cuboid is 720 cm3 , find its height and calculate the total surface area. d) The area of the rectangular base of a metallic block is 192 cm2 and its volume is 768 cm3 . Find its thickness. 25 cm 10 cm 10 cm 20 cm 10 cm 10 cm Mensuration (II): Prism

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 119 Vedanta Excel in Mathematics - Book 9 e) A rectangular metallic block is 40 cm long, 24 cm broad and 10 cm high. How many pieces of rectangular slices each of 8 mm thick can be cast lengthwise from the block? Creative section-A 4. Calculate the cross sectional area, the lateral surface area, the total surface area and volume of each of the following prisms. 4 m 4 m 6 m 6 m 3 m 5 m 3 m 15 m 4cm 4cm 9 cm 6cm 3cm 8 cm 8cm 16 cm 6 cm4cm 4cm 2cm 6 cm 10 cm 5 cm 2cm 5 cm a) b) c) 2cm 2 cm 2 cm 3cm 3cm 2cm 3cm 5 cm 10cm 10cm 10cm 10cm 10cm 10cm 30 cm 30 cm d) e) f) 5. a) A cubical water tank is filled in 1296 seconds at the rate of 1 litre per 6 seconds. (i) Calculate the internal volume and the length of side of the tank. (ii) Calculate the total internal surface area of the tank. b) A lidless rectangular water tank made of zinc plates is 2 m long, 1.5 m broad and 1 m high. (i) How many square metres of zinc plates are used in the tank? (ii) How many litres of water does it hold when it is full? (iii) Find the cost of zinc plates at Rs 1200 per sq. m. c) A rectangular carton is 80 cm × 60 cm × 40 cm. (i) How many packets of soaps can each of 10 cm × 5 cm × 4 cm be kept inside the carton? (ii) By how many centimetres should the height of the carton be increased to keep 1200 packets of soaps? 6. a) A cubical wooden block of length 12 cm is cut into 8 equal cubical pieces. Find the length of edge of each piece. b) 8 metallic cubical blocks of equal size are melted and joined together to form a bigger cubical block. If each smaller block is 10 cm thick, find the thickness of the bigger block. c) A rectangular metallic block is 50 cm × 20 cm × 8 cm. If it is melted and re-formed into a cubical block, find the length of the edge of the cube. Mensuration (II): Prism

Vedanta Excel in Mathematics - Book 9 120 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur d) Vegetable ghee is stored in a rectangular vessel of internal dimensions 60 cm × 10 cm × 45 cm. It is transferred into the identical cubical vessels. If the internal length of each cubical vessel is 10 cm, how many vessels are required to empty the rectangular vessel? Creative section-B 7. a) The adjoining figure is a rectangular glass vessel of length 40 cm, breadth 30 cm, and height 20 cm. If it contains some water upto the height of 12 cm, how many litres of water is to be poured into it to fill the vessel completely? (1 l = 1000 cm3 ) b) In the given figure, a cubical vessel of length 20 cm is completely filled with water. If the water is poured into a rectangular vessel of length 32 cm, breadth 25 cm, and height 15 cm find the height of the water level in the rectangular vessel. How much more water is required to fill the rectangular vessel completely? (1l = 1000 cm3 ) 6.4 Surface area and volume of triangular prisms A triangular prism has two congruent and parallel triangular bases. It has 3 rectangular lateral faces. In the adjoining triangular prism, (i) Cross sectional area = Area of triangular base (ii) Lateral surface area = Area of 3 rectangular surface = l.a. + l.b + l.c = l (a + b + c) = perimeter of triangular base × length = p.l (iii) Total surface area = Lateral surface area + 2 × Area of triangular base = p.l + 2D (iv) Volume of prism = Area of triangular base × length (or height of prism) Facts to remember 1) The area of cross section or base of triangle prism = 1 2 × base (b) ×height (h) 2) The area of cross section or base of right angled triangular prism = 1 2 × b × p 3) The area of cross section or base of equilateral triangular prism = 3 4 a2 4) The area of cross section or base of isosceles triangular prism = b 4 4a2 – b2 5) The area of cross section or base of scalene triangular prism = s (s – a) (s – b) (s – c) where s is the semi-perimeter of the triangle 40 cm 20 cm 12 cm 30 cm 32 cm 15 cm 20 cm 25 cm 20 cm 20 cm Cross section Its base is a triangle. So, it is a triangular prism. a b l c Mensuration (II): Prism

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 121 Vedanta Excel in Mathematics - Book 9 Example 1: Find (i) the lateral surface area (ii) total surface area (iii) volume of the following triangular prism. a) b) c) Solution: a) The triangular base is a right angled triangle. Here, BC = 6 cm and AC = 10 cm By using Pythagoras theorem, AB = AC2 – BC2 = 102 – 62 = 64 = 8 cm Now, perimeter of the triangular base (P) = AB + BC + CA = 8 cm + 6 cm + 10 cm=24 cm Area of triangular base (A) = 1 2 × b × p = 1 2 × 6 cm × 8 cm = 24 cm2 (i) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l) = 24 cm × 15 cm = 360 cm2 (ii) Total surface area of the prism = Lateral surface area+2 × Area of triangular base (A) = 360 cm2 + 2 × 24 cm2 = 408 cm2 (iii) Volume of the prism (V) = Area of triangular base (A) × length (l) = 24 cm2 × 15 cm = 360 cm3 b) The triangular base is an equilateral triangle. Here, each side (a) = 2 3 cm Now, perimeter of the triangular base (P) = 3 × 2 3 cm = 6 3 cm Area of triangular base (A) = 3 4 × a2 = 3 4 × (2 3) 2 = 3 4 × 4 × 3 = 3 3 cm2 (i) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l) = 6 3 cm × 18 cm = 108 3 cm2 (ii) Total surface area of the prism = Lateral surface area+2 × Area of triangular base (A) = 108 3 cm2 + 2 × 3 3 cm2 = 114 3 cm2 (iii) Volume of the prism (V) = Area of triangular base (A) × length (l) = 3 3 cm2 × 18 cm = 54 3 cm3 c) The triangular base is a scalene triangle. Here, a = 13 cm, b = 14 cm and c = 15 cm Now, perimeter of the triangular base (P) = 13 cm + 14 cm + 15 cm = 42 cm Semi-perimeter (s) = 13 cm + 14 cm + 15 cm 2 = 21 cm Area of triangular base (A) = s (s – a) (s – b) (s – c) 15 cm 16 cm 14 cm 13 cm 10 cm 15 cm A B 6 cm C A' C' 18 cm 2 3 cm Mensuration (II): Prism

Vedanta Excel in Mathematics - Book 9 122 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur = 21(21 – 13)(21 – 14)(21 – 15) = 21 × 8 × 7 × 6 = 84 cm2 (i) Lateral surface area of the prism = Perimeter of the triangular base (P) × length (l) = 42 cm ×16 cm = 672 cm2 (ii) Total surface area of the prism = Lateral surface area+2 × Area of triangular base (A) = 672 cm2 + 2 × 84 cm2 = 840 cm2 (iii) Volume of the prism (V) = Area of triangular base (A) × length (l) = 84 cm2 × 16 cm = 1,344 cm3 Example 2: Mr. Gurung is making a wooden birdhouse having the dimensions as shown in the figure. If the volume of the birdhouse is 900 cubic inches, find its length. Solution: Here, the birdhouse has pentagonal cross-section. ∴Cross-sectional area = area of rectangle + area of triangle. Now, area of rectangular portion (A1 ) = l × b = 8 in. × 6 in. = 48 sq. inch Also, for isosceles triangular portion, base (b) = 8 inch and equal sides (a) = 5 inch ∴ Area of the triangle (A2 ) = b 4 4a2 – b2 = 8 4 4 × 52 – 82 = 2 100 – 64 = 2 36 = 2 × 6 = 12 sq. inch. ∴ The area of cross section of the birdhouse (A) = A1 +A2 = 48 + 12 = 60 sq. inch Again, volume of the birdhouse (V) = Area of cross section (A) × length (l) or, 900 = 60 × l or, l = 900 60 = 15 Hence, the length of the birdhouse is 15 inch. Example 3: Mrs. Rana manages the accommodation for 30 guests in her son’s birthday ceremony. For this purpose, she plans to make a tent in the shape of triangular prism of length 20 m in such a way that each person has the space of 4 square meters on the floor and 8 cubic meters of air to breathe. What is the height of the tent? 5 in. 5 in. 6 in. 8 in. A A' C' B' ? C B D 20 m Mensuration (II): Prism

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 123 Vedanta Excel in Mathematics - Book 9 Solution: Here, the volume of the tent (V) = 30 × 8 m3 = 240 m3 The area of rectangular floor = 30 × 4 m2 = 120 m2 But, area of the floor = l × b or, 120 m2 = 20 m × BC ∴ BC = 6 m Also, area of cross section, D ABC (A) = 1 2 × BC × AD = 1 2 × 6 × AD = 3 AD Again, the volume of the tent (V) = area of cross section (A) × length (l) or, 240 = 3AD × 20 or, 60 AD = 240 or, AD = 4 Hence, the height of the tent is 4 m. EXERCISE 6.2 General section 1. a) From the given triangular prism, write the formulae to find its (i) cross sectional area (ii) lateral surface area (iii) total surface area (iv) volume b) If p, q, r are the three sides of a triangle of a triangular base prism of length h, write the formulae to find its (i) cross sectional area (ii) lateral surface area (iii) total surface area (iv) volume c) The base of a prism of length h cm is an equilateral triangle with each side a cm, write the formulae to find its (i) cross sectional area (ii) lateral surface area (iii) total surface area (iv) volume 2. a) A prism has a right-angled triangular base with perpendicular 8 cm and base 6 cm. Find its cross sectional area. b) A right prism base is a triangle whose sides are 3 cm, 25 cm and 26 cm. Find the area of its cross section. c) If the perimeter of the triangular base of a prism is 18 cm and its length is 15 cm, find its lateral surface area. d) The perimeter of the triangular base of a prism is 14.5 cm and its lateral surface area is 290 cm2 . Find the length of the prism. e) The area of triangular base of a prism is 30.5 cm2 and its lateral surface area is 305 cm2 . Find the total surface area of the prism. f) The perimeter of the triangular base of a prism is 24 cm and the area of its cross section is 24 cm2 . If the prism is 10 cm long, find its total surface area. 3. a) If the area of the triangular base of a prism 25 cm long is 16.4 cm2 , find the volume of the prism. b) A triangular based prism is 30 cm long. If the length of the sides of its triangular base are 3 cm, 4 cm and 5 cm, find its volume. c) In a prism, if the volume is 400 cm3 and area of its triangular base is 50 cm2 , find the length of the prism. x z y l Mensuration (II): Prism

Vedanta Excel in Mathematics - Book 9 124 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative section-A 4. a) Find the volume of the following prisms. (i) (ii) (iii) (iv) (v) (vi) b) Calculate the lateral surface area and total surface area of the following prisms. (i) (ii) (iii) (iv) (v) (vi) 5. a) The given diagram is the solid prism of triangular base. If the volume of the prism is 48 cm3 , find its height. b) If the volume of the prism given aside is 432 cm3 , find its height. c) The volume of a prism having its base a right angled triangle is 6,300 cm3 . If the lengths of the sides of the right angled triangle containing the right angle are 20 cm and 21 cm, find the height of the prism. 6. a) The area of rectangular surfaces of a triangular prism having base sides 9 cm, 10 cm and 17 cm is 864 cm2 . Calculate the height of the prism. b) A prism with an equilateral triangular base is 18 cm long. If the area of its rectangular surfaces is 567 cm2 , find the length of each side of the triangular base. 10 3 cm N' L' L M N M' 30 cm 8 cm R' P' P Q Q' M R 25 cm A' B' C' A B C 12 cm 5 cm 6 cm 10 cm 18 cm 13 cm 10 cm 8 cm 17 cm 15 cm 20 cm 40 cm 12 cm 16 cm 8 cm 4 cm 5 cm A B H E C G 7 cm C 3.4 cm 3.5 cm C' B' A' A B 6 cm 8 cm 8 cm 6 cm 15 cm P' R' Q' P R Q A A' B' C 30cm B 12cm 5cm A B C E F D 15cm 10cm A 6 cm P R B Q C 7cm 20cm 3cm 4cm A B C A' B' C' 8cm 10cm A B C A' B' C' Mensuration (II): Prism

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 125 Vedanta Excel in Mathematics - Book 9 Creative section-B 7. a) The perimeter and area of the cross section of a triangular prism are 44 cm and 66 cm2 respectively. If the area of lateral surface area of the prism is 1,100 cm2 , (i) find its volume (ii) find its total surface area b) The area of cross section of a triangular prism is 126 cm2 and its volume is 5,040 cm3 . If the perimeter of its base is 54 cm, (i) find its lateral surface area (ii) find its total surface area. c) The perimeter and area of the cross section of a triangular prism are 24 cm and 24 cm2 . If its total surface area is 480 cm2 , find (i) the lateral surface area (ii) the volume 8. a) A trap for insects is in the shape of triangular prism which is shown alongside. (i) Find the total surface area of trap. (ii) Find the volume of the trap. b) Two campers made a tent shown alongside in the shape of triangular prism. (i) Find the amount of fabric required for making it including the floor. (ii) How much air is occupied in the tent? 9. a) Mrs. Ghale manages the accommodation for 16 guests in her daughter’s birthday ceremony. For this purpose, she plans to make a tent in the shape of triangular prism of length 12 m. If each person has the space of 6 square meters on the floor and 15 cubic meters of air to breathe in average inside the tent, what is the height of the tent? b) In the marriage ceremony of Suresh’s son, there was accommodation for 150 people. For this purpose, he made a tent in the shape of triangular prism in such a way that each person had the space of 4 square meters on the floor and 20 cubic meters of air to breathe in average inside the tent. If the tent was 30 m long, what was the height of the tent? 10. a) Find the volume of dog house given along side. b) Mr. Chemjung is making a wooden birdhouse having the dimensions shown in the figure. If the volume of the birdhouse is 432 cubic inches, find its length. 10 cm 6 3 cm 5 ft. 6 ft. 8 ft. 5 ft. 12 m ? ? 30m 3 ft. 3 ft. 4 ft. 6 ft. 5 ft. 5 ft. 5 in. 5 in. 7 in. 6 in. Mensuration (II): Prism

Vedanta Excel in Mathematics - Book 9 126 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Project work and activity section 11. a) Make the group of your 5 friends. Collect some objects in the shape of triangular prism which are available in your surrounding. Calculate their L.S.A., T.S.A. and volume. b) Prepare a model of prism with the base (triangle, L-shape, T-shape or U-shape) by using woods or cardboard paper. Then, find the cross-sectional area, L.S.A., T.S.A. and volume. OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. If A, h and V denote the area of cross section, length and volume of a prism, what is the relation among A, h and V? (A) V = A×h (B) A =V×h (C) h = V×A (D) V = A/h 2. The cross sectional area of a prism is 30 cm2 and height is 15 cm, what is its volume? (A) 45 cm3 (B) 450 cm3 (C) 900 cm3 (D) 225 cm3 3. What is the volume of a prism whose area of base is 600 cm2 and length is 30 cm? (A) 20 cm3 (B) 630 cm3 (C) 570 cm3 (D) 18,000 cm3 4. L, P and h represent the lateral surface area, perimeter of base and height of a prism respectively. The relation among L, P and h is (A) L = P × h (B) P =L × h (C) h = L × P (D) L = P + h 5. A prism of length 20 cm has the sides of base 8 cm, 6 cm and 10 cm. The area of its rectangular surfaces is (A) 160 cm2 (B) 120 cm2 (C) 200 cm2 (D) 480 cm2 6. If A and B represent the area of base and the area of cross section of a prism respectively, what is its total surface area? (A) 2(A+B) (B) A+2B (C) 2A + B (D) 2AB 7. The area of cross section and the lateral surface area of a prism are 30 cm2 and 300 cm2 respectively. What is the total surface area of the prism? (A) 330 cm2 (B) 360 cm2 (C) 9000 cm2 (D) 630 cm2 8. The capacity of a water tank of dimension 4 m × 3 m × 2 m is (A) 12,000 liters (B) 8,000 liters (C) 6,000 liters (D) 24,000 liters 9. A rectangular aquarium is 60 cm long, 40 cm wide and 50 cm high. How much water can it hold? (A) 24 liters (B) 150 liters (C) 120 liters (D) 100 liters 10. How many brick each of 15cm × 10cm × 5 cm are required to build a wall of 20 m long, 3 m high, and 20 cm wide? (A) 16000 (B) 18000 (C) 20000 (D) 24000 https://www.geogebra.org/m/g5mxqvn4 Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Mensuration (II): Prism

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 127 Vedanta Excel in Mathematics - Book 9 7.1 Cylinder –Looking back Classwork-Exercise Let's fill in the blanks with correct answers as quickly as possible. a) The circumference of a circle with radius r units is ….. b) The area of a circle with radius x units is ….. c) The circumference of a circle with radius 7 cm is ….. d) The area of a circle with diameter 28 cm is ….. 7.2 Cylinder Let’s observe a few real-life examples of cylindrical objects which are available in our surrounding. madal water jar rods battery wooden log A cylinder is a three dimensional solid that has two parallel circular bases. The two circular bases are joined by a curved surface. The distance between the two circular bases is called the height or length of the cylinder. Bottles, battery, LP-glass cylinder, rod, pipe, bucket, dustbin, water-jar, pipe, roller, candles, pole etc are a few real life examples of cylinder. Facts to remember 1. The line segment joining the centers of circular bases of a cylinder is called its axis. 2. A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides as axis. 3. The opposite circular bases are congruent and parallel. Unit 7 Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 128 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 7.3 Area of cylinder The surface area of a solid object is a measure of the total area that the surface of the object occupies. Let’s take a rectangular sheet of paper and revolve it about one of the sides to complete a full rotation (without overlapping) as shown in the figure. From the above figures, it is clear that, (i) The perimeter of the circular base(C) = 2πr (ii) The area of the circular base(A) = πr2 (iii) The curved surface area (C.S.A) of a cylinder is equal to area of rectangle with length 2πr and width h. Thus, C.S.A.= 2πr × h = 2πrh (iv) The total surface area of a cylinder = C.S.A. + Area of two opposite circular bases = 2πrh+ πr2 + πr2 = 2πrh+2πr2 = 2πr (r + h) where, r is the radius of a circular face and h is the height of the cylinder. 7.4 Volume of cylinder Let’s take a cylindrical radish or cucumber. Suppose the radius of base is r units and height is h units. Let’s cut it into 16 equal sections through centers of circular bases along its height. Such material may be made with wood or bought from the market and keep in the mathematics laboratory. Then, let’s place the segments alternatively as shown in the figure to form in to cuboid. Here, in the newly formed cuboid, length (l) = πr, Breadth (b) = r and height (h) = h ∴Volume of cuboid (V) = l × b × h = πr × r × h = πr2 h Now, the volume of cylinder (V) = Volume of cuboid = πr2 h Moreover, The area of circular base of a cylinder = πr2 Then, volume of the cylinder = Area of circular base × height = πr2 × h = πr2 h https://www.geogebra.org/m/aurfzgtg Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 129 Vedanta Excel in Mathematics - Book 9 7.5 Curved surface area, total surface area, and volume of a hollow cylinder The figure alongside is a hollow cylinder (or a pipe). Let R be the external radius, r the internal radius and h be the height of the hollow cylinder. Here, the external curved surface area = 2πRh The internal curved surface area = 2πrh ∴ The total curved surface area = 2πRh + 2πrh = 2πh(R + r) Again, the area of a ring-shaped circular base = πR2 – πr2 = π(R + r) (R – r) ∴ The area of two ring-shaped circular bases = 2π(R + r) (R – r) Now, the total surface area of a hollow cylinder = 2πh(R + r) + 2π(R + r) (R – r) = 2π(R + r) (h + R – r) Volume of material contained by a hollow cylinder: From the above diagram, The external volume of the cylinder = πR2 h The internal volume of the cylinder = πr2 h ∴ Volume of the material contained by the cylinder = πR2 h – πr2 h = πh(R2 – r2 ) = πh (R + r) (R – r) 7.6 Half cylinder or semicylinder When a cylinder is cut into half along it's axis it is called a half cylinder or a semi-cylinder. It is a horizontal cylindrical segment. A half cylinder has a curved surface, a rectangular flat surface and two semi-circular bases. Length of the rectangular flat surface = height = h Breadth of the rectangular flat surface = diameter of circular base = 2r ∴ Area of the rectangular flat surface = length × breadth = h × 2r = 2rh Also, the area of curved surface of a half cylinder = 1 2 × 2prh = prh And, the area of two semi-circles = 2 × 1 2 × pr2 = pr2 Now, lateral surface area of a half cylinder = Area of (curved surface + flat surface) = prh + 2rh = rh(p + 2) Total surface area of a half cylinder = lateral surface area + area of 2 semi-circles = rh(p + 2) + pr2 Also, volume of a half cylinder = 1 2 pr2 h R r h h d=2r h Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 130 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: The diameter of a cylindrical drum is 35 cm and its height is 30 cm. Find (i) its curved surface area (ii) total surface area (iii) volume. Solution: Here, diameter of the cylindrical drum (d) = 35 cm. ∴The radius of the drum (r) = 35 2 = 17.5 cm Height of the drum (h) = 30 cm (i) Now, the curved surface area (C.S.A.) = 2πrh = 2 × 22 7 × 17.5 × 30 cm = 3,300 cm2 (ii) Also, the total surface area (T.S.A.) = 2πr (r + h) = 2 × 22 7 × 17.5 cm (17.5 cm + 30 cm) = 2 × 22 7 × 17.5 cm × 47.5 cm = 5,225 cm2 (iii) And, volume (V) = πr2 h =22 7 × 17.5 cm × 17.5 cm × 30 cm = 28,875 cm3 Example 2: The perimeter of the circular base of a cylinder having height 20 cm is 44 cm. Calculate its: (i) curved surface area (ii) total surface area (iii) volume Solution: Here,the perimeter of circular base (C) = 44 cm. or, 2pr = 44 cm or, 2 × 22 7 × r = 44 cm or, r = 7 cm Height of the cylinder (h) = 20 cm (i) Now, the curved surface area (C.S.A.) = 2prh = 2 × 22 7 × 7 cm × 20 cm= 880 cm2 (ii) Also, the total surface area (T.S.A.) = 2pr (r + h) = 2 × 22 7 × 7 cm(7 cm + 20 cm) = 2 × 22 7 × 7 cm × 27 cm = 1,188 cm2 (iii) And, volume (V) = pr2 h = 22 7 × 7 cm × 7 cm × 20 cm = 3,080 cm3 Example 3: The radius and height of a cylindrical log are in the ratio 7:9. If its curved surface area is 6,336 sq. inch, find its volume. Solution: Let, the radius of circular base (r) = 7x inch and height (h) = 9x inch. Now, the curved surface area of the log = 6336 sq. inch or, 2πrh = 6336 or, 2 × 22 7 × 7x × 9x = 6336 or, 396x2 = 6336 or, x2 = 16 or, x = 4 ∴ The radius of the log (r) = 7 × 4 inch = 28 inch and height (h) =9 × 4 inch = 36 inch 35 cm 30 cm Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 131 Vedanta Excel in Mathematics - Book 9 Mensuration (III): Cylinder and Sphere Again, volume (V) = πr2 h = 22 7 × 28 inch × 28 inch × 36 inch = 88,704 cu. inch. Example 4: If a circular water tank having radius of base 3.5 m can hold 3.08 × 105 liters of water. Find the surface area of the tank. Solution: Let, the radius of circular base (r) = 3.5 m Capacity of the tank = 3.08 × 105 liters = 308000 liters ∴Volume of the tank (V) = 308000 1000 m3 = 308 m3 [ 1000 l = 1 m3 ] Now, volume of the tank (V) = 308 m3 or, πr2 h = 308 or, 22 7 × 3.5 × 3.5 × h = 308 or, 38.5 h = 308 or, h = 8 m Again, the total surface area (T.S.A.) = 2πr (r + h) = 2 × 22 7 × 3.5 m (3.5 + 8) m = 253 m2 Example 5: Suppose that you are going to buy a can of ghee in a dairy. The shopkeeper showed you the ghee of same quality in two cans of different size manufactured by a milk factory. The price of the ghee is not labeled in the can, which can is more expensive and why? Give reason with calculation. Solution: Here, for the can-A, base radius (r) = 10 2 cm = 5 cm and height (h) = 14 cm Now, volume of can-A (V1 ) = πr2 h = 22 7 × 5 cm × 5 cm × 14 cm= 1100 cm3 Also, for the can-B, base radius (r) = 14 2 cm = 7 cm and height (h) = 10 cm Now, volume of can-B (V2 ) = πr2 h = 22 7 × 7 cm × 7 cm × 10 cm = 1540 cm3 Since, the volume of the ghee in the can-B is more than the volume of ghee in the can-A, the can-B would be more expensive than can-A. Example 6: The external and the internal radii of a hollow cylindrical pipe 70 cm long are 5.4 cm and 5.1 cm respectively. Find (i) the volume of the material contained by the vessel and (ii) the total surface area of the pipe. Solution: Here, the external radius of the pipe (R) = 5.4 cm The internal radius of the pipe (r) = 5.1 cm The height of the pipe (h) = 70 cm (i) Now, the volume of the material = π (R2 – r2 ) h = 22 7 × (5.42 – 5.12 ) × 70 cm3 = 693 cm3 Hence, the required volume of the material contained by the vessel is 693 cm3 . (ii) Again, the T.S.A. of the hollow pipe = 2π (R + r) (h + R – r) 10 cm 14 cm 14 cm 10 cm Can - A Can - B

Vedanta Excel in Mathematics - Book 9 132 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur = 2 × 22 7 (5.4 + 5.1) (70 + 5.4 – 5.1) = 4,639.8 cm2 Hence, the surface area of the pipe is 4,639.8 cm2 . Example 7: A metal pipe is 1.5 m long. Its internal radius is 48 mm and thickness is 2 mm. If 1 mm3 of the metal weighs 0.02 gm, find the weight of the pipe. Solution: Here, the internal radius of the pipe (r) = 48 mm, thickness (t) = 2 mm ∴ The external radius of the pipe (R) = r + t = 48 mm + 2mm = 50 mm The length of the pipe (h) = 1.5 m = 1.5 × 100 × 10 mm = 1500 mm Now, the volume of the material used in the pipe = π (R2 – r2 )h = 22 7 × (502 – 482 ) × 1500 cm3 = 9,24,000 mm3 Also, 1 mm3 of metal weighs 0.02 gm ∴ 9,24,000 mm3 of metal pipe weighs 9,24,000 × 0.02 gm = 18480 gm = 18480 1000 kg = 18.48 kg Hence, the weight of the pipe is 18.48 kg. Example 8: The internal diameter and height of a cylindrical bucket are 14 cm and 35 cm respectively and it is filled with water completely. If the water is poured into a rectangular glass tray with internal length 28 cm and breadth 17.5 cm and it is completely filled with water, find the height of the tray. Solution: Here, the internal diameter of the cylindrical bucket = 14 cm. So, radius (r) = 7 cm Height of the bucket (h) = 35 cm Now, the internal volume of the bucket = πr2 h = 22 7 × 7 × 7 × 35 = 5,390 cm3 Thus, the volume of water = internal volume of bucket = 5,390 cm3 Again, volume of the rectangular glass tray = volume of water = l × b × h = 5,390 or, 28 × 17.5 × h = 5,390 or, h = 5,390 28 × 17.5 = 11 cm Hence, the height of the rectangular glass tray is 11 cm. Example 9: A roller of diameter 100 cm and length 140 cm takes 450 complete revolutions to level a playground. Calculate the area of the garden. Solution: Here, the diameter of roller (d) = 100 cm ∴ The radius of roller (r) = 100 cm 2 = 50 cm = 50 100 m = 0.5 m Length of roller (h) = 140 cm = 140 100 m = 1.4 m Now, the area covered by the roller in 1 revolution = curved surface area of the roller = 2πrh = 2×22 7 × 0.5 × 1.4 m2 = 4.4m2 ∴ Area of the ground = Area covered by the roller in 450 revolutions = 450 ×4.4m2 = 1,980 m2 Hence, the area of the ground is 1,980 m2 . Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 133 Vedanta Excel in Mathematics - Book 9 EXERCISE 7.1 General section 1. a) If x be the radius and y be the height of the cylinder given alongside, write the formulae to find it’s (i) curved surface area (C.S.A.) (ii) Total surface area (T.S.A.) (iii) volume b) A and a are the external and internal radii and b be the height of a hollow cylinder respectively. Write the formula to find it’s (i) curved surface area (C.S.A.) (ii) total surface area (T.S.A.) (iii) volume of material contained by the cylinder. c) From the given half-cylinder, write the formulae to find (i) curved surface area (ii) lateral surface area (ii) total surface area (iii) volume 2. a) The perimeter of the base of a cylinder is 44 cm and height is 10 cm, what is its curved surface area? b) What is the total surface area of a cylinder in which the circumference of the base is 22 inch and the sum of radius and height is 15 inch? c) What is the volume of a cylinder having area of base 38.5 cm2 and height 20 cm? 3. Find the (i) curved surface area, (ii) total surface area and (iii) volume of the given cylinders. a) b) c) d) Creative section-A 4. a) There is a cylindrical wooden log in a meat-shop. If the radius of its circular base is 21 cm and height is 50 cm, (i) find the curved surface area of the log. (ii) find the total surface area of the log. (iii) find the volume of the log. b) From the can of vegetables given alongside. (i) Find the area of the label on the can. (ii) Find the area of the metal sheet used to make the can. (iii) Find the volume of the can. 5. a) A cylindrical tank has diameter 1.4 m and height 2m. How many litres of water is required to fill the tank completely? b) The diameter and height of a cylindrical vessel are equal. If its radius is 35 cm and it is completely filled with diesel. How many litres of diesel is filled in the vessel? c) How many cubic meter of earth must be dug out to make a well 20 m deep and 1.4 m diameter? y cm x cm r=pcm q cm 25 cm 7cm 20cm 28 cm 14 cm 6 cm 50 in. 35 in. 5 in. 1.75 in. Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 134 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6. a) The circumference of circular base of a closed cylindrical juice can of height 10 cm is 22 cm. How much metal sheet is required to make it? b) The height of a cylindrical drum is 20 cm and the area of its base is 346.5 sq. cm. If it is to be covered with leather, find the minimum quantity of leather sheet needed for it in sq. cm. 7. a) The radius and height of a cylindrical container are in the ratio 5:7 and its curved surface area is 5500 sq.cm. Find its volume. b) The radius and height of a cylindrical jar are in the ratio 7:17 and its total surface area is 4224 sq.cm. Find its volume. c) A cylinder container water tank contains 4,62,000 litres of water and its radius is 3.5 m. Find its curved surface area. d) The volume of a cylindrical can is 3.08 litre. If the diameter of its base is 14 cm, find the total surface area of the can. 8. a) The area of curved surface of a solid cylinder is equal to 2 3 of the total surface area. If the total surface area is 924 cm2 , find the volume of the cylinder. b) The number of volume of a cylinder is half of its number of the total surface area. If the radius of its base is 7 cm, find the volume. 9. a) The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the total surface area of the cylinder is 1,628 cm2 , calculate its volume. b) The circumference of the base of a cylindrical drum is 44 cm and the sum of its radius and height is 27 cm, find the volume of the cylinder. c) The curved surface area and volume of a cylinder are 880 cm2 and 3,080 cm3 respectively, find its total surface area. 10. a) Calculate the lateral surface area, total surface area and volume of the given half-cylinder. b) The radius of the semi-circular base of a half-cylinder is 21 cm and its height is 50 cm. Find its lateral surface area, total surface area and volume. c) Mr. Binod made a vegetable tunnel as shown in the figure given aside. Find the volume of the tunnel. 11. a) A hollow cylindrical metallic pipe is 28 cm long. If the external and internal diameters of the pipe are 12 cm and 8 cm respectively, find the volume of metal used in making the pipe. b) The external and the internal radii of a cylindrical pipe of length 50 cm are 3 cm and 2.6 cm respectively. Find the volume of material contained by the pipe. c) A metal pipe is 1.4 m long. Its internal diameter is 25 mm and thickness 1 mm. If 1 mm3 of the metal weighs 0.02 gm, find the weight of the pipe. d) A metal pipe has inner diameter of 5 cm. The pipe is 5 mm thick all round and 2 m long. What is the weight of the pipe if 1 cm3 of the metal weighs 7.7 g? 12. a) The internal radius of a cylindrical bucket of height 50 cm is 21 cm. It is filled with water completely. If the water is poured into a rectangular vessel with internal length 63 cm and breadth 44 cm and it is completely filled with water, find the height of the vessel. b) Vegetable ghee is stored in a cylindrical vessel of internal radius 1.4 m and height 1.5 m. If it is transferred into the rectangular tin cans 33 cm × 10 cm × 5 cm, how many cans are required to empty the vessel? 20cm 14cm 6 m 21 m Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 135 Vedanta Excel in Mathematics - Book 9 Creative section - B 13. a) Mrs. Thapa has two cylindrical buckets. The bucket-A has radius of base 14 cm and height 30 cm. Likewise, bucket-B has the radius of base 16 cm and height 28 cm. Which of these buckets can hold more water and by how many litres? b) Mr. Rai has fixed two overhead cylindrical water tanks on the roof of his house. The diameter of the base of tank-X is 2.1 m and height is 3 m. Similarly, the diameter of the base of tank-Y is 3 m and height is 2.1 m. Which of these tanks holds more water and by how many litres? 14. a) Mr. Chaudhary drives a tanker in a milk supply company. He supplies 5 tankers of milk every day. If the milk tank is cylindrical in shape having radius 0.8 m and length 3.5 m. How many liters of milk does he supply in each day? b) Chari Maya is a juice seller. She serves the juice completely filled with the cylindrical glasses; each has diameter 7 cm and height 10 cm. If she sells 50 glasses of juice in a day, how many litres of juice does she sell in a day? 15. a) A roller of diameter 112 cm and length 150 cm takes 550 complete revolutions to level a playground. Calculate the area of the ground. b) A temple has 14 cylindrical pillars made of concrete. If the radius of base of each pillar is 20 cm and height is 5 m, how much concrete is required to make the pillars? Project work and activity section 16. a) Measure the different dimensions (such as diameter, circumference, height) of water tank, drum, cylindrical bucket, etc. in your house or in school and calculate their curved surface area, total surface area and capacity. b) Measure the circumference of the circular base of a mineral water bottle, then find its radius. Calculate the approximate capacity of the bottle. c) Measure the length and breadth of a rectangular sheet of chart paper and find its area. Now, roll the paper to form a cylinder and find its curved surface area. Then compare the two areas. 7.7 Area and Volume of a Sphere A sphere is a perfectly round geometrical object in three-dimensional space. It is generated by revolving a semi-circle about its diameter. A football, a cricket ball, etc. are said to have the shape of a sphere. Total surface area of a sphere Archimedes (c. 287 – 212 BCE) was an ancient Greek scientist and engineer, and one of the greatest mathematicians of all time. He discovered many concepts of calculus and worked in geometry, analysis and mechanics. While taking a bath, Archimedes discovered a way to determine the volume of irregular objects using the amount of water they displaced when submerged. He was so excited by this discovery that he ran out on the street, still undressed, yelling “Eureka!” (Greek for “I have found it!”). He discovered that the surface area of a sphere is the same as the lateral surface area of a cylinder having the same radius as the sphere and the height equal to the diameter of the sphere. Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 136 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Let, the radius of the given sphere is equal to the radius of the cylinder and the diameter of the sphere is equal to the height of the cylinder. Then, the sphere could easily be fitted in the cylinder. Here, Surface area of the sphere = curved surface area of the cylinder = 2πrh Now, let’s replace h by 2r, we get Surface area of the sphere = 2πr × 2r = 4πr2 . Volume of a sphere Step 1: Let’s take a hollow sphere and two identical hollow cylinders in which the diameter of base and height are equal to the diameter of the sphere. Step 2: Fill the hollow sphere with sand once and empty it into one of the cylinders. Also, fill the hollow sphere second time with sand and empty it into the second cylinder. Step 3: Again, fill the hollow sphere third time and empty it into the remaining spaces of the cylinders. r h = 2r r h = 2r From the above experiment, the volume of 3 spheres = volume of 2 cylinders = 2 × πr2 h = 2π × r2 × 2r [ h = 2r] = 4πr3 Thus, the volume of 1 sphere = 4 3 πr3 7.8 Hemisphere and great circle When a sphere is cut into two halves, one of the half portions is called hemisphere. A hemisphere has a plane circular face and a curved surface. The radius of the circular face is equal to the radius of the sphere. The circular face of a hemisphere is called a great circle. The area of a great circle of a sphere = πr2 or πd2 4 The circumference of the great circle = 2πr or πd Also, the surface area of a sphere = 4πr2 ∴ The curved surface area of a hemisphere = 1 2 × 4πr2 = 2πr2 or π 2 d2 https://www.geogebra.org/m/nvznmnek Vedanta ICT Corner Please! Scan this QR code or browse the link given below: r Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 137 Vedanta Excel in Mathematics - Book 9 And, the total surface area of a hemisphere = Area of curved surface + Area of great circle = 2πr2 + πr2 = 3πr2 or 3πd2 4 Similarly, the volume of a hemisphere = 1 2 of the volume of sphere = 1 2 × 4 3 πr3 = 2 3 πr3 or πd3 12 Worked-out Examples Example 1: The diameter of volleyball given alongside is 21 cm. Find (i) its surface area (ii) volume. Solution: Here, the volleyball is in the shape of a sphere. The diameter of the volleyball (d) = 21 cm. ∴ The radius of the volleyball (r) = 21 2 cm = 10.5 cm (i) Now, the surface area = 4πr2 = 4 × 22 7 × 10.5 cm × 10.5 cm = 1,386 cm2 (ii) And, volume (V) = 4 3πr3 = 4 3 × 22 7 × 10.5 cm × 10.5 cm × 10.5 cm = 4,851 cm3 Hence, the required surface area of the volleyball is 1,386 cm2 and its volume is 4,851 cm3 . Example 2: The surface area of a concrete sphere is 5544 cm2 , calculate its diameter. Solution: Here, the surface area of the sphere = 5544 cm2 , diameter (d) =? Now, the surface area of the sphere = 5544 or, 4πr2 = 5544 or, 4 × 22 7 × r2 = 5544 or, r2 = 441 or, r = 21 ∴ The diameter (d) = 2r = 2 × 21 cm = 42 cm Hence, the required diameter is 42 cm. Example 3: Given that 1 cm3 of iron weighs 7.8 g. What is the weight of shot-put if the circumference of its great circle is 33 cm? Solution: Here,circumference of the great circle = 33 cm or, 2πr = 33 or, 2 × 22 7 × r = 33 or, r = 5.25 cm Again, volume (V) = 4 3πr3 = 4 3 × 22 7 (5.25 cm)3 = 606.375cm3 https://www.geogebra.org/m/zacgudyp Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Mensuration (III): Cylinder and Sphere 21 cm

Vedanta Excel in Mathematics - Book 9 138 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur We have, 1 cm3 of iron weighs 7.8 g ∴ 606.375 cm3 of iron weighs 7.8 × 606.375 g = 4,729.725 g = 4,729.725 1000 kg = 4.729 kg Hence, the weight of the shot-put is 4.729 kg. Example 4: Find the curved surface area, total surface area and volume of the hemisphere given alongside. Solution: Here, the radius of the hemisphere (r) = 8.4 cm Now, the curved surface area = 2πr2 = 2 × 22 7 × 8.4 × 8.4 cm = 443.52 cm2 Also, the total surface area = 3πr2 = 3 × 22 7 × 8.4 × 8.4 cm = 665.28 cm2 And, volume (V) = 2 3πr3 = 2 3 × 22 7 × 8.4 cm × 8.4 cm × 8.4 cm = 1241.856 cm3 Hence, the curved surface area of the hemisphere is 443.52 cm2 , its total surface area is 665.28 cm2 and volume is 1241.856 cm3 . Example 5: Three metallic spheres of radii 3 cm, 4 cm, and 5 cm are melted and reformed into a single sphere. Find the radius of the new sphere. Solution: Solution: Here, the radius of the first sphere (r1 ) = 3 cm The radius of the second sphere (r2 ) = 4 cm The radius of the third sphere (r3 ) = 5 cm Now, the volume of the first sphere (V1 ) = 4 3πr1 3 = 4 3 π × 33 The volume of the second sphere (V2 ) = 4 3πr2 3 = 4 3 π × 43 The volume of the third sphere (V3 ) = 4 3πr3 3 = 4 3 π × 53 Again,the volume of new single sphere (V) = V1 + V2 +V3 or, 4 3πr3 = 4 3 π × 33 + 4 3 π × 43 + 4 3 π × 53 or, 4 3πr3 = 4 3 π (33 + 43 + 53 ) or, r3 = 216 ∴ r = 6 cm Hence, the radius of new sphere is 6 cm. Example 6: A solid metallic sphere of radius 6 cm is melted and drawn into a cylindrical wire of diameter 4 mm. Find the length of the wire. Solution: Here, the radius of the cylindrical wire (r) = 4 2 mm = 2 mm = 0.2 cm Now, the volume of the cylindrical wire = πr2 h = π(0.2 cm)2 × h = 0.008πh cm2 Also, the volume of the metallic sphere = 4 3πR3 = 4 3π(6 cm)3 = 288π cm3 Again, the volume of the cylindrical wire = the volume of the sphere 8.4 cm 3 cm ? 4 cm 5 cm Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 139 Vedanta Excel in Mathematics - Book 9 or, 0.008πh cm2 = 288 π cm3 ∴ h = 36000 cm = 36000 100 m = 360 m Hence, the required length of the wire is 360 m. Example 7: A shopkeeper has one spherical laddu of radius 5 cm. How many laddus of radius 2.5 cm can be made with the same amount of material? Solution: Here, the radius of bigger laddu (R) = 5 cm ∴Volume of bigger laddu (V) = 4 3πR3 = 4 3π(5 cm)3 = 500π 3 cm3 Also, the radius of each smaller laddu (r) = 2.5 cm ∴ Volume of each smaller laddu (v) = 4 3πr3 = 4 3π(2.5 cm)3 = 62.5π 3 cm3 Again, the number of smaller laddus = V v = 500π 3 62.5π 3 = 8 Hence, required number of smaller laddus is 8. EXERCISE 7.2 General section 1. a) Write the formula to find (i) surface area and (ii) volume of the given sphere. b) The radius of a hemisphere is p cm. Write the formulae to find its (i) curved surface area (ii) total surface area (iii) volume. c) If the radius of a sphere is 1 cm, what are its surface area and the volume? 2. a) Find the surface area and volume of the spheres given below. (i) (ii) (iii) (iv) b) Calculate the curved surface area, total surface area and volume of the following hemi-spheres. (i) (ii) (iii) (iv) 3. a) If the radius of a spherical orange is 3.5 cm, find its surface area and volume. b) The diameter of a basketball is 21 cm, find its surface area and the volume. c) The diameter of the moon is 3476 km. Find the volume of the moon. Creative section - A 4. a) The diameter of spherical iron ball is 14 cm. If 1 cm3 of iron weighs 7.8 gm, what is the weight of iron ball? x cm 7 cm 6.3cm 42 cm 8.4 cm 21 cm 14 cm 16.8 cm 5.6 cm Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 140 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) A spherical watermelon has diameter 21 cm. If it contains 90% of water, find the volume of water in it. c) The diameter of a cricket ball is 7 cm. How much sticker is required to paste on the surface of the ball? d) How much leather sheet is required to make a football with diameter 21 cm? 5. a) If the radius of a hemispherical cake is 10.5 cm, find its total surface area and volume. b) The diameter of the hemispherical top of a mushroom is 4.2 cm, find its curved surface area and volume. 6. a) A hemispherical bowl has radius 7 cm. If the bowl is completely filled with vegetable soup, find the volume of the soup. b) A hemispherical ‘Damaha’ has its diameter 35 cm. If it is tightly covered with a plastic, how much plastic is used to cover it? 7. a) If the perimeter of the great circle is p cm, find the volume of the hemisphere. b) If the circumference of the great circle is 44 cm, find the total surface area of the hemisphere. c) The total surface area of a hemisphere is 243p cm2 , find its volume. 8. a) A solid metallic sphere of radius 7 cm is cut into two halves. Find the total surface area of the two hemispheres so formed. b) A solid metallic sphere of diameter 42 cm is cut into two halves. Find the total surface area of the two hemispheres so formed. 9. a) If the volume of a spherical object is 9π 2 cm3 , find its diameter. b) If the total surface area of a solid sphere is 616 cm2 , what is its radius? c) If the volume of a hemisphere is 19,404 cm3 , find its radius. 10. a) If the surface area of a globe is 5,544 cm2 , find its volume. b) If the volume of a spherical ball is 38,808 cm3 , find its surface area. 11. a) If the circumference of the great circle of a solid hemisphere is 132 cm, find (i) its curved surface area (ii) total surface area (iii) volume b) If the total surface area of a hemisphere is 1,848 cm2 , find: (i) its curved surface area (ii) volume c) If the volume of a hemisphere is 2425.5 cm3 , find its curved surface area and total surface area. 12. a) If the radius of a sphere is doubled, by how much does its surface area increase? b) The radius of the earth is four times the radius of the moon, find the ratio of (i) their surface areas (ii) their volumes. c) If the volumes of two spherical objects are in the ratio of 8:27, find the ratio of (i) their radii (iii) their surface areas. 13. a) Three metallic spheres of radii 1 cm, 6 cm and 8 cm respectively are melted and re-formed to a single sphere. Find the radius of the new sphere. b) Three silver hemispheres of radii 6 cm, 8 cm and 10 cm respectively are melted and re-formed to a single hemisphere. Find the radius of the new hemisphere. c) 8 metallic spheres each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new sphere. 7 cm Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 141 Vedanta Excel in Mathematics - Book 9 Creative section - B 14. a) A hemi-spherical bowl of internal radius 9 cm is completely filled with water. If the water is poured into a cylindrical vessel of radius 6 cm and height 20 cm, what is the height of empty space in the vessel? b) A solid iron sphere of diameter 42 cm is dropped into a cylindrical drum partly filled with water. If the radius of the drum is 1.4 m, by how much is the surface of the water raised? c) A cylindrical jar of radius 6 cm contains water. How many iron solid spheres each of radius 1.5 cm are required to immerse into the jar to raise the level of water by 2 cm 15. a) A solid metallic sphere of radius 3 cm is melted and drawn into a cylindrical wire of radius 5 mm. Find the length of the wire. b) The diameter of a solid spherical metallic ball is 12 cm. If it is melted and drawn into a cylindrical wire of length 288 cm, find the thickness of the wire. c) How many solid spheres each 6 cm diameter can be made from a solid metallic cylinder of diameter 4 cm and height 45 cm? 16. a) Mrs. Thapa wishes to grow flowers on the hanging bowls at her home. She brings 16 hemispherical bowls from the market having external diameter 21 cm and uniform thickness 0.5 cm each and fills the bowls with compost completely. (i) If 1 cm3 of compost weighs 2.31 g, find the total weight of required compost. (ii) If she covers outer curved surface of each bowl with polythene sheet, find the total amount of polythene. b) Assume that the earth as a sphere of radius 6400 km. (i) If the average density of the earth is 5500 kg per cubic meter, estimate the mass of the earth. (ii) If 70% of the earth is covered by water, find the earth’s surface area covered by the land. Project work and activity section 17. a) Measure the radius or diameter of few spherical objects which are available in your house or school. Then, calculate their surface area and volume. b) Collect a few hemispherical objects then, calculate their curved surface area, total surface area and volume. 7.11 Cost estimation Let A = Area of floor, carpet, four walls, ceiling etc. R = Rate of cost of carpeting, plastering, coloring, papering etc. C = Total cost of carpeting, plastering, painting etc. N = Required number of pieces of carpets, paper etc. a = Area of each piece of carpet, paper, bricks. L = Total length of carpet/paper Now, (i) Total cost (C) = Area × Rate C = A × R . Then, A = C R and R = C A Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 142 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (ii) Total cost (C) = Length × Rate C = l × R (iii) Total cost (C) = Number of pieces of carpets/ paper × Rate C = N × R Worked-out Examples Example 1: A rectangular room is 9 m long and 6 m broad. The floor of the room is covered with the carpet each of 4 m long and 1.5 m wide. (i) Find the cost of carpeting the floor at the rate of Rs 125 per sq. meter. (ii) Find the cost of carpeting the floor at the rate of Rs 750 per piece. (iii) Find the cost of carpeting the floor at the rate of Rs 185 per meter. Solution: Here, Length of the room (l) = 9 m Breadth of the room (b) = 6 m Width of the carpet (b1 ) = 1.5 m Length of the carpet (l 1 ) = 4 m (i) Now, the area of the floor of the room (A) = l × b = 9 m × 6 m = 54 m2 Hence, the cost of carpeting the floor (C) = Area × Rate = 54 ×Rs 125=Rs 6,750 (ii) Also, the area of each piece of carpet (a) = l 1 × b1 = 4 m × 1.5 m = 6 m2 Now, the number of pieces of carpet = A a = 54 m2 6 m2 = 9 Hence, the cost of carpeting the floor (C) = of pieces of carpet × Rate = 9 × Rs 750=Rs 6,750 (iii) Again, let x m be the total length of carpet required to cover the floor. Here, area of carpet = Area of the floor or, x × 1.5 m = 54 m2 or, x = 54 m2 1.5 m = 36 m Thus, the total cost of carpeting the floor = length × rate = 36 × Rs 185 = Rs 6,660. Example 2: The length, breadth, and height of a rectangular room are 12 m, 10 m and 6 m respectively. a) Find the cost of painting its wall and ceiling at Rs 75 per sq. m. b) Find the cost of carpeting its floor at Rs 120 per sq. m. Solution: Here, the length of the room (l) = 12 m The breadth of the room (b) = 10 m The height of the room (h) = 6 m The rate of cost of painting (R) = Rs 75 per sq. m. The rate of cost of carpeting the floor (R') = Rs 120 per sq. m. a) Now, the area of 4 walls and ceiling (A) = 2h (l + b) + lb = 2 × 6 (12 + 10) + 12 × 10 = 384 m2 Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 143 Vedanta Excel in Mathematics - Book 9 Again, the cost of painting its wall and ceiling = A × R = 384 × Rs 75 = Rs 28,800 Also, the cost of carpeting the floor = Area floor × Rate = (12 × 10) × Rs 120 = Rs 14,400 Hence, the required cost of painting the walls and the ceiling of the room is Rs 28,800 and the cost of carpeting the floor is Rs 14,400. Example 3: A rectangular room is 15 m long, 10 m broad and 5 m high. If it contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m, find the cost of plastering its walls at Rs 50 per sq. metre. Solution: Here, the length of the room (l) = 15 m The breadth of the room (b)= 10 m The height of the room (h) = 5 m The rate of plastering walls (R) = Rs 50 per sq. m. Now, the area of its 4 walls = 2h (l + b) = 2 × 5 m (15 m + 10 m) = 250 m2 Also, the area of 2 windows = 2 (2 m × 1.5 m) = 6 m2 The area of a door = 1 m × 4 m = 4 m2 \ The area of 4 walls excluding windows and door = (250 – 6 – 4) m2 = 240 m2 Again, the cost of plastering its walls = A × R = 240 × Rs 50 = Rs 12,000 Hence, the required cost of plastering its walls is Rs 12,000. Example 4: A room is 12 m long and 5.5 m high. If the cost of carpeting its floor at Rs 80 per sq. metre is Rs 7,680, find the cost of colouring its walls at Rs 35.50 per sq, m. Solution: Here, the length of the room (l) = 12 m the height of the room (h) = 5.5 m Now, area of the floor = cost of carpeting the floor Rate of cost or, l × b = 7680 80 m2 or, 12 × b = 96 m2 or, b = 96 12 m = 8 m Again, area of 4 walls of the room (A) = 2h (l + b) = 2 × 5.5 m (12 m + 8 m) = 220 m2 \ The cost of colouring the walls = Area × Rate = 220 × Rs 35.50 = Rs 7810 Hence, the required cost of colouring the walls is Rs 7810. Example 5: The cost of carpeting a square room at the rate of Rs 75 per sq. metre is Rs 10,800. If the cost of plastering its walls at Rs 25 per sq. metre is Rs 6000, find the height of the room. Solution: Here, the area of the floor of the square room = cost of carpeting Rate of cost or, l 2 = 10800 75 m2 = 144 m2 or, l = 12 m and, also b = 12 m Again, area of 4 walls = cost of plastering the walls Rate of cost or, 2h (l + b) = 6000 25 m2 Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 144 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur or, 2h (12 + 12) = 240 m2 or, h = 5 m So, the required height of the room is 5 m. Example 6: The cost of carpeting the floor of a room, whose breadth is twice the height and the length is twice its breadth, at the rate of Rs 80 per sq. metre is Rs 10,240. What is the cost of plastering its walls at Rs 30 per sq. metre? Solution: Let, the height of the room (h) = x m Then, the breadth of the room (b) = 2x m And, the length of the room (l) = 2b = 2 × 2x m = 4x m Now, the area of the floor = cost of carpeting the floor Rate of cost or, l × b = 10240 80 m2 or, 4x × 2x = 128 m2 or, x2 = 16 m2 or, x = 4 m \ The height of the room (h)= x = 4 m, the breadth of the room (b) = 2x = 2 × 4 m = 8 m The length of the room (l) = 4x = 4 × 4 m = 16 m Again, area of 4 walls = 2h (l + b) = 2 × 4 m (16 m + 8 m) = 192 m2 \ The cost of plastering 4 walls = Area × Rate = 192 × Rs 30 = Rs 5760 So, the required cost of plastering its 4 walls is Rs 5760. Example 7: A square room contains 180 m3 of air. The cost of plastering its four walls at Rs 60 per sq. metre is Rs 7,200. Find the height of the room. Solution: Let, the length (l) = breadth (b) = x m Here, volume of the room = Volume of the air or, l × b × h = 180 m3 or, x × x × h = 180 m3 or, h = 180 x2 m ... equation (i) Again, area of 4 walls = Cost of plastering of 4 walls Rate of cost or, 2h (l + b) = 7200 60 m2 or, 2h (x + x) = 120 m2 or, h = 30 x m ... equation (ii) From equation (i) and (ii) we get, 180 x2 = 30 x or, x = 6 m Now, putting the value of x in equation (ii), we get, h = 30 6 = 5 m Hence, the required height of the room is 5 m. Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 145 Vedanta Excel in Mathematics - Book 9 EXERCISE 7.3 General section 1. a) If the area of floor of a room is x m2 , what is the cost of carpeting the room at Rs y per sq. m? b) A room requires P pieces of carpet. What is the cost of carpeting the room at Rs Q per piece of carpet? c) The total length of carpet required for a room is A ft. What is the cost of carpeting the room at Rs B per feet? d) What is the cost of papering the four walls of a room having area 180 m2 at Rs 120 per square meter? e) The area of four walls and ceiling of a rectangular room is 260 m2 . What is the cost of white painting the walls and the ceiling at Rs 150 per square meter? 2. a) The length of a hall is 30 ft. and the width is 18 ft. Find the cost of carpeting the room at Rs 65 per square feet. b) A rectangular hall is 12 m long and 10 m broad. Find the length of carpet 2 m wide required for covering its floor. If the rate of cost of carpet is Rs 110 per metre, find the cost of carpeting the floor. c) 6 pieces of carpet are required for a room. What is the cost of carpeting the floor at Rs 750 per piece? Creative section - A 3. a) A rectangular room is 10 m long and 9 m broad. The floor of the room is covered with the carpet each of 4 m long and 1.5 m wide. (i) Find the cost of carpeting the floor at the rate of Rs 150 per sq. meter. (ii) Find the cost of carpeting the floor at the rate of Rs 850 per piece. (iii) Find the cost of carpeting the floor at the rate of Rs 200 per meter. b) The guest room in Sunayana’s house is 24 ft. long and 15 ft. wide. The floor of the room is completely covered with a few numbers of pieces of carpet each of 9 ft. long and 5 ft. wide (i) Find the cost of carpeting the floor if the rate of cost of carpet is Rs 30 per sq. ft. (ii) Calculate the cost of carpeting the floor at Rs 1250 per piece. (iii) If the rate of cost of carpet is Rs 160 per ft, find the cost of carpeting the floor. 4. a) A rectangular room is 10 m long, 8 m wide, and 5 m high. Find the cost of colouring its walls and ceiling at Rs 65 per sq. metre. b) A square room is 15 feet long and 10 feet high. Find the cost of plastering its walls, ceiling and floor at Rs 18 per sq. ft. c) The cost of plastering the walls and the ceiling of a room at Rs 20 per sq. feet is Rs 14,400. Find the cost of colouring the walls and ceiling at Rs 16 per sq. feet. 5. a) A rectangular room is 8 m long, 6 m broad and 4 m high. It contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. Find the cost of painting its walls and ceiling at Rs 54 per sq. metre. b) A square hall is 15 m long and 5 m high. It contains three square windows each of 2 m long and two doors of size 1.5 m × 4 m. Find the total cost of plastering and colouring its walls and ceiling at Rs 50 per sq. metre and Rs 45 per sq. metre respectively. Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 146 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative section - B 6. a) A rectangular room is 10 m long and 8 m wide. It has two windows each of 2 m × 2 m and a door of size 1.5 m × 4 m. If the cost of plastering its walls at Rs 55 per sq. metre is Rs 9,130, find the height of the room. b) A square room is 4 m high. The cost of carpeting its floor at Rs 95 per sq. metre is Rs 6,080. Find the cost of painting its walls and ceiling at Rs 60 per sq. metre. c) A rectangular room is 10 m long and 5 m high. If the cost of paving marbles at Rs 180 per sq. metre is Rs 10800, find the cost of papering its walls at Rs 45 per sq. metre. d) The cost of carpeting a square room at Rs 110 per sq. metre is Rs 5,390. If the cost of plastering its walls at Rs 56 per sq. metre is Rs 7,840, find the height of the room. e) The cost of plastering the floor of a room, which is 10 m long, at Rs 54 pr sq. metre is Rs 4,320. If the cost of painting its walls at Rs 48 per sq. metre is Rs 10,368, find the height of the room. 7. a) A rectangular room is twice as long as it is broad and its height is 4.5 m. If the cost of papering its walls at Rs 40 per sq. metre is Rs 6,480, find the cost of paving marbles on its floor at Rs 150 per sq. metre. b) A rectangular room is three times as long as it is broad and its height is 4.6 m. If the cost of carpeting its floor at Rs 85 per sq. metre is Rs 6,375, find the cost of colouring the walls at Rs 50 per sq. metre. 8. a) A room is 12 m long and 8 m broad and it contains 480 cu. metre of air. Find the cost of colouring its walls at Rs 54 per sq metre. b) A room contains 600 cu. m. of air. If the cost of carpeting its floor at Rs 90 per sq. metre is Rs 13,500, find the height of the room. c) A square room contains 288 m3 of air. The cost of carpeting the room at Rs 105 per sq. metre is Rs 6,720. Find the cost of painting its walls at Rs 45 per sq. metre. d) A square room contains 256 cu. m. of air. If the cost of painting its 4 walls at Rs 50 per sq. metre is Rs 6,400, find the cost of carpeting its floor at Rs 99 per sq. metre. e) A square room contains 220.5 cu. m. of air. If the cost of colouring its 4 walls at Rs 55 per sq. metre is Rs 6,930, find the height of the room. 9. a) A room is two times longer than its breadth and it contains 360 m3 of air. If the cost of plastering its floor at Rs 40 per sq. metre is Rs 2,880, find the cost of plastering its walls at Rs 45 per sq. metre. b) The length of a room is two times its breadth and it contains 210 m3 of air. If the cost of painting its walls at Rs 54 per sq. m. is Rs 6,804, find the height of the room. Project work and activity section 10. a) Measure the length, breadth, height, and thickness of walls of your classroom. Also measure the length and height of windows and doors of the classroom. (i) Find the volume of wall excluding windows and doors. (ii) Find the average volume of a brick available in your surroundings. (iii) Find the number of bricks required to construct the wall excluding windows and doors. (iv) Estimate and cost of bricks required to construct the walls as per the local rate of cost. Mensuration (III): Cylinder and Sphere

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 147 Vedanta Excel in Mathematics - Book 9 b) Measure all the dimensions of a part of the compound wall of your school or your house. Calculate the volume of the part of the wall. Find the volume of a brick and calculate the number of such bricks required to construct the wall. Also, calculate the cost of bricks as per the local rate of cost. OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. What is the curved surface area of a cylindrical object whose radius of base is x cm and height is y cm? (A)πxy cm2 (B) 2πxy cm2 (C) πx (x + y) cm2 (D) πx2 cm2 2. If the area of base of a cylindrical object having curved surface area p cm2 is q cm2 , what is its total surface area? (A)(p + q) cm2 (B) (2p + q) cm2 (C) (p + 2q) cm2 (D) 2(p + q) cm2 3. What is the volume of a cylinder whose radius of base is 7 cm and height is 13 cm? (A) 2002 cm3 (B) 572 cm3 (C) 880 cm3 (D) 154 cm3 4. The internal and external radii of a cylindrical pipe of length ‘h’ cm are ‘r’ cm and ‘R’ cm respectively, what is the volume of material used to make the pipe? (A)π (R2 + r2 ) h cm3 (B) π (R2 – r2 ) h cm3 (C) 2π (R2 – r2 ) h cm3 (D) 2π(R+r)(R-r+h) cm3 5. The lateral surface area of a half-cylinder having the base radius r cm and height h cm is (A)πr2 cm2 (B) πrh cm2 (C) rh (π + 2) cm2 (D) πr (r + h) cm2 6. The surface area of a sphere having radius x cm is (A)πx2 cm2 (B) 2πx2 cm2 (C) 3πx2 cm2 (D) 4πx2 cm2 7. The diameter and height of a cylinder are equal to the diameter of a sphere. If the volume of cylinder is V1 and the volume of sphere is V2 then what is the relation between V1 and V2 is (A) V1 = V2 (B) V1 = 2V2 (C) 2V1 = 3V2 (D) 3V1 = 2V2 8. If the area of great circle of a sphere is A cm2 , what is the surface area of the sphere? (A)A cm2 (B) 2A cm2 (C) 3A cm2 (D) 4A cm2 9. The surface area of a hemisphere having radius x cm is (A)πx2 cm2 (B) 2πx2 cm2 (C) 3πx2 cm2 (D) 4πx2 cm2 10. The radius of a single sphere reformed by melting three metallic spheres of radii 1 cm, 6 cm and 8 cm is (A)9 cm (B) 15 cm (C) 18 cm (D) 30 cm Mensuration (III): Cylinder and Sphere

Vedanta Excel in Mathematics - Book 9 148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Assessment -III 1. An umbrella is made by stitching 8 triangular pieces of cloth, each piece measuring 60 cm, 63 cm and 39 cm. (a) Write down the formula for calculating the area of triangle having the sides a cm, b cm, c cm and semi-perimeter s cm? (b) What is the area of a triangular piece? (c) How much cloth is required for the umbrella? (d) If the cost of 1 cm2 of cloth is Rs 0.50, find the total cost of cloth. 2. In a school, the room of class 9 is 30 ft. long, 20 ft. wide and 12 ft. high. It contains a door of size 4 ft. × 8 ft. and two windows each of 6 ft. × 4.5 ft. (a) What is the area of its four walls? (b) What is the cost of plastering the walls at Rs. 150 per sq. feet? (c) If 80% of the floor is covered with seats, each seat covers 9 sq. ft. and no seat is found vacant, how many students are there in the class? 3. Mr. Rai built a house having two rectangular rooms of same width and height after the destruction of his old house by the earthquake. The first room having two windows each of size 4 ft × 4.5 ft and two doors each of size 3 ft × 6.5 ft is 18 ft long, 12 ft wide and 9 ft high, the second room having a window of size 4.5 ft × 4.5 ft and a door common to the first room is 16 ft long. (a) Find the cost of carpeting both the rooms at Rs 150 per sq. ft. (b) Find the cost of plastering the walls inside the rooms at the rate of Rs 80 per sq. ft. 4. A hemi-spherical bowl of internal radius 9 cm is completely filled with water. If the milk is poured from bowl into a cylindrical vessel of radius 6 cm and height 24 cm, answer the following questions. (a) What is the formula to find the volume of a hemispherical object having radius r units? (b) How many cubic centimeter of water is filled in the bowl? (c) Find the height of empty space of the vessel. (d) How many litre of water is required to fill the empty space of the vessel? 5. Mr. Yadav manages the accommodation for the guests in his daughter’s birthday ceremony. For this purpose, he plans to make an equilateral triangular tent. The edge of triangular surface of the tent is 20 feet and the length of the tent is 33 feet. (a) Find the total area of triangular surfaces of the tent. (b) How many guests can be accommodated at a time, if each guest requires 22 square feet of space on the ground? (c) If the triangular surfaces of the tent were unchanged and the canvas with area 3500 square feet was used to make the tent, by how much more or less would be the length of the new tent than the original tent? h

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149 Vedanta Excel in Mathematics - Book 9 8.1 Sequence - Introduction Let’s observe the patterns in the following figures. (i) (ii) (iii) In figure (i), the numbers of squares in each design are 1, 5, 9, … In figure (ii), the numbers of dots in each design are 3, 6, 9, 12, … In figure (iii), the total heights of the steps from the base are 10 cm, 20 cm, 30 cm, …. Let’s study the following lists of numbers and say the next two numbers (if possible) of each list. (i) 2, 5, 8, 11, … (ii) 3, 6, 12, 24, … (iv) 1, 4, 9, 16, .., (iv) 5, 6, 10, 19, … In (i), each number is formed by adding 3 to the just preceding number. In (ii), each number is formed by multiplying the just preceding number by 2. In (iii), each number is square of consecutive natural numbers. In (iv), the set of numbers does not follow any rule. So, we cannot guess the next to 19. Here, the patterns of numbers in (i), (ii) and (iii) are in a certain rule. So, these lists of numbers are sequences of numbers. But, the list of numbers in (iv) does not form any sequence. Facts to remember 1. An ordered list of numbers or objects following a definite rule is known as a sequence. 2. Each element of the sequence is called a term. The terms of the sequence are denoted by t1 , t2 , t3 , … 3. A sequence having a finite of terms is called a finite sequence. For example, 100, 90, 80, … 10 is a finite sequence. 4. If the of terms of a sequence is infinite then it is called an infinite sequence. For example, 2, 4, 6, 8, … is an infinite sequence. 10cm 10cm 10cm 10cm , , , , , , Unit 8 Sequence and Series

Vedanta Excel in Mathematics - Book 9 150 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: Find the next two terms of the following sequences. (i) 3, 8, 13, 18, 23, … (ii) 5, 2, –1, –4, –7, …(iii) 2, 4, 8, 16, … Solution: (i) 3, 8, 13, 18, 23 +5 +5 +5 +5 Here, each term is greater than its preceding term by 5. The required next two terms are, t6 = 23 + 5 = 28 and t7 = 28 + 5 = 33 (ii) 5, 2, –1, –4, –7 –3 –3 –3 –3 Here, each term is decreased from preceding term by 3. The required next two terms are, t6 = –7 – 3 = –10 and t7 = –10 – 3 = –13 (iii) 4, 8, 16, 32 ×2 ×2 ×2 Here, each term is 2 times its preceding term. The required next two terms are, t5 = 32 × 2 = 64 and t6 = 64 × 2 = 128 8.2 General term of sequence In a sequence 7, 11, 15, 19, … ; the numbers 7, 11, 15 are the first, second and third terms respectively. The nth term of the sequence is called its general term. It is denoted by tn or an. Let’s observe the following examples to investigate the ideas of finding the general term of the sequences. Finding the nth term of the sequences In 2, 4, 6, 8, … t 1 = 2 = 2×1 t 2 = 4= 2×2 t 3 = 6= 2×3 t 4 = 8= 2×4 …………… t n = 2n In 5, 9, 13, 17, … t 1 = 5 = 4×1 + 1 t 2 = 9 = 4×2 + 1 t 3 = 13 = 4×3 + 1 t 4 = 17 = 4×4 + 1 …………………… ∴t n = 4n + 1 In 1, 4, 9, 16, … t 1 = 1 = 12 t 2 = 4 = 22 t 3 = 9 = 32 t 4 = 16 = 42 …………… ∴t n = n2 In 2, 6, 12, 20, … t 1 = 2 = 12 + 1 t 2 = 6 = 22 + 2 t 3 = 12 = 32 + 3 t 4 = 20 = 42 + 4 …………… ∴t n = n2 + n In 6, 12, 24, 48, ... t 1 = 6 = 3×21 t 2 = 12 = 3×22 t 3 = 24 = 3×23 t 4 = 48 = 3×24 …………… ∴t n = 3×2n Example 2: Find the first four terms of a sequences whose nth terms are given by (i) tn = 2n + 3 (ii) tn = (–1) n+1(n + 7) Solution: (i) The given general term (tn) = 2n + 3. Putting n = 1, 2, 3, and 4 to get t1 , t2 , t3 and t4 . Sequence and Series