Let the base radius of the cylindrical vessel be r cm. Solution: Ex 13.6 Class 9 Maths Question 1. Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. Solution: NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5 have been provided here for students to prepare well for the board and competitive examinations. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. = 2 x \(\frac { 22 }{ 7 }\) x 10 x 35 cm2 Internal surface area = Area of five faces of 3 cuboids each of dimensions 75 cm x 30 cm x 20 cm ∴ πrl = 308 ⇒ \(\frac { 22 }{ 7 }\) x r x 14 = 308 (Use π = 3.14) [∵ a2 – b2 = (a + b)(a – b)] 2. ∴ Volume of the earth = \(\frac { 4 }{ 3 }\)πr3 and, Ex 13.8 Class 9 Maths Question 5. Solution: Solution: Question 2. ∵ Area of four walls = Lateral surface area = 2(1 + b) x h, where h is the height of the hall = 250 h m2 ⇒ h = \(\frac { 380 }{ 80 }\) = 4.75m Hence, the cost of paint required = Rs. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Ex 13.4 Class 9 Maths Question 7. Depth (h) = 12m It takes 500 complete revolutions to move once over to level a playground. = 6 x 5 x \(\frac { 45 }{ 10 }\)m = 3 x 45m3 = 135m3 These ncert book chapter wise questions and answers are very helpful for CBSE board exam. = 2[(15 x 12) + (12 x 5) + (5 x 15)] cm2 ∴ Outer radius (R) = 1 m + 0.01 m = 1.01 m Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (ii) Which box has the smaller total surface area and by how much? In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. NCERT Solutions as well as NCERT Solutions Offline Apps are updated for new academic session 2020-2021 based on latest CBSE Syllabus 2020-21. (i) 14 cm Solution: Thus, the required volume of the iron used But lateral surface of the cylinder 94.2 cm2 [given] (ii) radius of the base, एक छोटा पौधा घर (Green house) सम्पूर्ण रूप से शीशे की पट्टियों से (आधार भी सम्मिलित है) घर के अंदर ही बनाया गया है और शीशे की पट्टियों को टेप द्वारा चिपका कर रोका गया है। यह पौधा घर 30 cm लंबा, 25 cm चौड़ा और 25 cm ऊँचा है। सभी 12 किनारों के लिए कितने टेप की आवश्यकता है? ∴ Total surface area of n bricks = n x \(\frac { 937.5 }{ 10000 }\) m2 Ex 13.3 Class 9 Maths Question 8. (i) Here, radius of the cone (r) =6 cm The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m? NCERT Solutions for class 9 chapter 13 is applicable for CBSE Delhi Board, MP Board, UP Board – High School, UK Board, Gujrat Board and other board using NCERT Books 2020-21. Important Questions on 9th Maths Chapter 13, NCERT Solutions for Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Physical Education, NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Physical Education, CBSE Sample Papers for Class 10 Session 2020-2021, CBSE Sample Papers for Class 12 Session 2020-2021, 9th Maths Exercise 13.1 Solutions in English, 9th Maths प्रश्नावली 13.1 का हल हिंदी में, 9th Maths Exercise 13.2 Solutions in English, 9th Maths प्रश्नावली 13.2 का हल हिंदी में, 9th Maths Exercise 13.3 Solutions in English, 9th Maths प्रश्नावली 13.3 का हल हिंदी में, 9th Maths Exercise 13.4 Solutions in English, 9th Maths प्रश्नावली 13.4 का हल हिंदी में, 9th Maths Exercise 13.5 Solutions in English, 9th Maths प्रश्नावली 13.5 का हल हिंदी में, 9th Maths Exercise 13.6 Solutions in English, 9th Maths प्रश्नावली 13.6 का हल हिंदी में, 9th Maths Exercise 13.7 Solutions in English, 9th Maths प्रश्नावली 13.7 का हल हिंदी में, 9th Maths Exercise 13.8 Solutions in English, 9th Maths प्रश्नावली 13.8 का हल हिंदी में, 9th Maths Exercise 13.9 Solutions in English, 9th Maths प्रश्नावली 13.9 का हल हिंदी में, NCERT Solutions for Class 9 Maths Chapter 1, NCERT Solutions for Class 9 Maths Chapter 2, NCERT Solutions for Class 9 Maths Chapter 3, NCERT Solutions for Class 9 Maths Chapter 4, NCERT Solutions for Class 9 Maths Chapter 5, NCERT Solutions for Class 9 Maths Chapter 6, NCERT Solutions for Class 9 Maths Chapter 7, NCERT Solutions for Class 9 Maths Chapter 8, NCERT Solutions for Class 9 Maths Chapter 9, NCERT Solutions for Class 9 Maths Chapter 10, JEE Main Answer Key 2021: Dates, How to Download and Score vs Rank, How Quarantine is Disrupting the Teenage Experience, Useful Tips To Score Good Marks On Programming Assignments. Since, a matchbox is in the form of cuboid. Solution: = [2(l + b)h] + [lb] (ii) volume of the air inside the dome. एक घनाकार डिब्बे का एक किनारा 10 cm लंबाई का है तथा एक अन्य घनाभाकार डिब्बे की लंबाई, चौड़ाई और ऊँचाई क्रमशः 12.5 cm, 10 cm और 8 cm हैं। किस डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल अधिक है और कितना अधिक है? गढ्ढे के आधार की त्रिज्या r = 3.5/2 = 1.75 m और ऊँचाई h = 12 m है।. It is to be covered with a decorative cloth. ∴ Volume (capacity) of the tank = l x b x h If the perimeter of … Find = 2[(225) + (75) + (168.75)] cm2 Diagonal of a cuboid = √3a units 4. Since, height of the vessel (h) = 25 cm = 8 x 4 x \(\frac { 22 }{ 7 }\) x (\(\frac { 21 }{ 2 }\))2cm2 We hope the UP Board Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (पृष्ठीय क्षेत्रफल एवं आयतन) help you. Find the diameter of the base of the cylinder. Here, diameter of a sphere = 21 cm ∴ 25b = 50 ⇒ b = \(\frac { 50 }{ 25 }\) = 2 ∴ Amount of water required by 4000 people per day = 150 x 4000 litres If the rate of polishing-is 20 paise per cm2 and the rate of pointing is 10 paise per cm2, find the total expenses required for palishing and painting the surface of the bookshelf. = 2 [500 + 100 + 125] cm2 What will be the volume of a packet containing 12 such boxes? Let the length and breadth of the hall be l and b respectively. Here, Length of a brick (l) = 22.5 cm The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Thus, the required capacity of the conical vessel is \(\frac { 11 }{ 35 }\) litres. (i) Total cost of painting = Rs. 9 in Hindi Medium to use offline for session 2020-21. = 22.45833 mm3 ∴ Volume of the wood = [Volume of the pendil] – [Volume of the graphite] ∴ Area of the canvas required, Ex 13.3 Class 9 Maths Question 5. = \(\frac { 22 }{ 7 }\) x 35 x (14+12) x (14 – 12)cm3 210 For the cuboidal box with dimensions, (i) Diameter of the ball =28 cm Volume of the given cube = (side)3 = (12)3 cm3 = 520000 cm2 + 26000 cm2 = 546000 cm2 The length of the pipe is 35 cm. = 4 x \(\frac { 25 }{ 10 }\) x \(\frac { 15 }{ 10 }\)cm3 Volume: Space occupied by an object (solid body) is called the volume of the object. = 2 x \(\frac { 22 }{ 7 }\) x 42 x 120 cm2 Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g. (i) What is the area of the glass? = [2(1.50 + 1.25)0.65] m2 + [1.50 x 1.25] m2 How much water will fall into the sea in a minute? (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm. Radius (r) = \(\frac { 1.40 }{ 2 }\)m = 0.70 m Height (h) = 8 cm = 2200 cm2 384.34 (approx.). 5. Radius of the base (r) = 24 m. (i) The slant height, l = \(\sqrt { { r }^{ 2 }-{ h }^{ 2 } }\) Now, curved surface area = 2πrh Curved surface area of a right circular cylinder is 4.4 m^2. The inner ft diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Surface area of new cube = 6 x 62 cm2 (i) Which box has the greater lateral surface area and by how much? ⇒ Radius(r) = \(\frac { 0.21 }{ 2 }\)m = \(\frac { 21 }{ 200 }\)m = 4 x 22 x 7 cm2 = 616 cm2, Case II: When radius (r2) = 14 cm2 = [Lateral surface area] + [Base area] CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. = 110 m2, (ii) Let r be the base radius of the cylindrical vessel. Inner diameter of the cylindrical pipe = 24cm Students can also refer to NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes for better exam preparation and score more marks. Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 (Assume π = \(\frac{22}{7}\), unless stated otherwise) Question 1. Exercise 13.4 Chapter 13 Class 9 Maths : Download the NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 PDF to know how to solve the problems related to the topic to score more marks. (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹70. Total cost NCERT Solutions for class 9 maths all chapters or other subjects NCERT Solutions are also available to free download. ∴ Cost of polishing external faces = Rs. A metal pipe is 77 cm long. \(\frac { 210 }{ 100 }\) x 550 = Rs. = d – \(\frac { 25 }{ 100 }\) x d Surface Area: Surface area of a solid body is the area of all of its surface together and it is always measured in square units. ⇒ Volume of 12 such boxes = 12 x 15 cm3 Sides of the right triangle ABC are 5 cm, 12 cm and 13cm. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Hence, the height of the cylinder is 1 m. Hence, the total surface area of hemisphere is 942 cm^2. Lateral surface area of cuboid = 4a² square units 3. ), Ex 13.8 Class 9 Maths Question 4. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Radius of the moon = \(\frac { 1 }{ 4 }\) (r) = \(\frac { r }{ 4 }\) (i) Here, diameter = 14 cm, Ex 13.4 Class 9 Maths Question 3. The NCERT Chapter 13 Maths Class 9 gave an activity to the students to understand the difference between cylinder and cone. (ii) how much steel was actually used, if \(\frac { 1 }{ 12 }\) of the steel actually used was wasted in making the tank. Ex 13.6 Class 9 Maths Question 4. Mass of wood in the pipe = (Mass of wood in 1 cm3 of wood) x (Volume of wood in the pipe) Cost of 1m² of white-washing = Rs. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. [Given], Ex 13.2 Class 9 Maths Question 7. Find the area of the playground in m2. 2200 ∴ 1mm = \(\frac { 1 }{ 10 }\) cm = 2[(225 x 1(0) + (10 x 7.5) + (7.5 x 22.5)] cm2 (ii) slant height of the cone Hence, the inner curved surface area is 110 m^2. CBSE Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise Questions with Solutions to help … Cost of 5.45 m2 sheet = Rs. A dome of a building is in the form of a hemisphere. Ex 13.6 Class 9 Maths Question 8. Thus, S : S’ = 1 : 9, Ex 13.8 Class 9 Maths Question 10. Height of the cylinder = Diameter of the sphere = 180 cm3. Solution: MCQs from CBSE Class 9 Maths Chapter 13: Surface Areas and Volumes. = \(\frac { 314 }{ 100 }\) x 6 x 10 m2 = 1884 m2 Length (l) = 2 km = 2000 m (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Solution: Now, surface area of a penholder = [Lateral surface area] + [Base area] The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. Solution: A matchbox measures 4 cm x 2.5 cm x 1.5 cm. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Here, length (l) = 85 cm, A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. = 2 x 22 x 2 x 11 cm2 = 968 cm2, (iii)Total surface area = [Inner curved surface area] + [Outer curved surface area] + [Area of two circular ends] ∴ Capacity of the conical vessel Thus, the required mass of the pipe is 3.432 kg. Question 1. Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Surface area of a sphere = 4πr2 Ex 13.1 Class 9 Maths Question 1. Students can also refer to NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes for better exam preparation and score more marks. ), Ex 13.9 Class 9 Maths Question 1. The frame has a base diameter of 20 cm and height of 30 cm. Total surface area = 6a2 = 6 x 102 cm2 Here, curved surface area = πrl = 308 cm2 Thus, 7920 cm2 of cardboard was required to be bought. (i) radius r’ of the new sphere, Download NCERT Solutions for Class 9 Maths in Hindi Medium for CBSE Board, UP Board (High School), MP Board, Gujrat Board and other board’s students who are following NCERT Books. ⇒ x = \(\frac { 1800000 }{ 4000\times 150 }\) = 3 Surface area = 4πr12 = 4 x \(\frac { 22 }{ 7 }\) x (7) cm2 All answers are solved in … The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. ∴ l = 50 = Rs. Total cost 498.96 2 Ans. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. Solution: Are you looking for Surface Areas and Volumes formulas or important points that are required to understand Surface Areas and Volumes for class 9 maths Chapter 13? A soft drink is available in two packs Solution: Solution: Curved surface area of the roller = 2πrh Assume π = unless stated otherwise. ∴ The surface area of the cuboid, excluding the base Solution: 2200 Ex 13.1 Class 9 Maths Question 5. = [2(l + b)h] + [l x b] Let the radius of the sphere be r cm. 20 ∴ Diameter of the base of the cone = (2 x 4)cm = 8 cm. It costs Rs. [Hint: Area of the four walls = Lateral surface area] Since, 150 litres of water is required per head per day. Diameter of the base 28 cm USD Length of the cylindrical pipe (h) = 28 m Let height of the cylinder be h m Inner radius (r) = 5 cm ∴ Volume of the spherical capsule = 4πr3 Surface area (or total surface area) of cuboid = 2 (lb + bh + hl) square units 2. ⇒ n3 = 63 ∴ Capacity of the rectangular pack = 300 cm3, (ii) For cylindrical pack, Here, diameter = 24 m 24 Find its curved surface area. Hene, total expenses = Rs. Cost of painting the inner curved surface of cylindrical vessel = ₹ 2200. (ii) ratio of S and S’. Solution: Solution: Solution: Slant Height = √[r² + h²] Curved Surface Area (CSA) = πrl square units Total Surface Area (TSA) = πr(r + l) square units Volume = 1/3 πr²h cubic units, Sphere: A sphere is three dimensional figure which is made up of all points in the space, which lie at a constant distance, form a fixed point called the centre of the sphere and the constant distant is called its radius.
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