It can be observed that four walls and the ceiling of the room are to be white washed. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Volume of the cone so formed is 100π cm3. The height of a cone is 15cm. How many square meters of the sheet are required for the same? (ii) the cost of plastering this curved surface at the rate of Rs. (i) Rs 20 is the cost of painting 1 m2 area. A hemi spherical tank is made up of an iron sheet 1cm thick. 13.22). Find the outer curved surface of the bowl. 384.34. What fraction of the volume of the earth is the volume of the moon? The chapter includes topics on the Surface Area and Volume of Class 9 … Our subject experts and teaching faculty together have prepared this NCERT maths solution chapter wise, so that is beneficial for the students to solve the problems easily while using it as a reference. It is to be covered with a decorative cloth. Find the volume of the solid so obtained. Therefore, the cost of the canvas required to make such a tent is Rs 137280. (Assume π =22/7 ). Radius of the circular end of roller = r = (84/2) cm = 42 cm. A metal pipe is 77 cm long. If the cost of white-washing isRs20 per square meter, find the, (i) inside surface area of the dome (ii) volume of the air inside the dome, (i) Cost of white-washing the dome from inside = Rs 4989.60, CSA of the inner side of dome = 498.96/2 m2  = 249.48 m2. curved surface area of the cone is 2200 cm2. Therefore, 47 m2 tarpaulin will be required. The exercise questions further challenge this newly-developed expertise. 3. Tarpaulin will be required for the top and four wall sides of the shelter. 2. 9. NCERT Solutions Class 9 Maths Chapter 13 Surface area and volumes Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volumes, curated by Vedantu equip students with all the necessary skills. Radius of wooden sphere, r = diameter/ 2 = (21/2) cm = 10.5 cm, Formula: Surface area of wooden sphere = 4πr2, Radius of the circular end of cylindrical support = 1.5 cm, Area of the circular end of cylindrical support = πr2, Area to be painted silver = [8 ×(1386-7.07)] = 8×1378.93 = 11031.44, Cost for painting with silver colour = Rs(11031.44×0.25) =Rs 2757.86, Area to be painted black = (8×66) cm2 = 528 cm2, Cost for painting with black colour =Rs (528×0.05) = Rs26.40. NCERT Solutions for Class 9 Maths Exercise 13.1. (Assume π = 22/7), Base radius, r = diameter/2 = 14/2 m = 7m, Cost of white-washing 550 m2 area, which is Rs (210×550)/100. Volume of godown = (40×25×15) m3 = 15000 m3, Volume of a wooden crate = (1.5×1.25×0.5) m3 = 0.9375 m3, Let us consider that, n wooden crates can be stored in the godown, then, Volume of n wooden crates = Volume of godown. (4340+1935) = Rs. It is 30cm long, 25 cm wide and 25 cm high. 2. Hence, volume of the sphere is 1.05 m3 (approx). A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. (see fig. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π = 22/7), Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm, Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm, Height of pipe, h = Length of pipe = 35cm, Now, the Volume of pipe = π(r22-r12)h cm3. 6. = [4(30+25+25)] (after substituting the values). List of Exercises in class 9 Maths Chapter 13 :Exercise 13.1 solution (9 questions)Exercise 13.2 solution (8 questions)Exercise 13.3 solution (9 questions)Exercise 13.4 solution (5 questions)Exercise 13.5 solution (5 questions)Exercise 13.6 solution (8 questions)Exercise 13.7 solution (9 questions)Exercise 13.8 solution (10 questions)Exercise 13.9 solution (3 questions). The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. (ii)How much of tape is needed for all the 12 edges? Find its, Let r1 and r2 Inner and outer radii of cylindrical pipe, Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm, (i) curved surface area of outer surface of pipe = 2πr1h, (ii) curved surface area of outer surface of pipe = 2πr2h. Rate of water flow = 2km per hour = 2000m/60min = 100/3 m/min, Now, Volume of water flowed in 1 min = (100/3) × 40 × 3 = 4000m3. Answer! 210 per 100 m2. If the diameter of the base is 28cm, find, Volume of a right circular cone = 9856 cm3, (i) Radius of cone, r = (28/2) cm = 14 cm. A solid cube of side 12 cm is cut into eight cubes of equal volume. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22/7), Formula: Total Surface area of the cone = πr(l+r), Total Surface area of the cone = (22/7)×12×(21+12) m2, 3. 10. Your email address will not be published. Consumption of the water per head per day = 150 litres, Water consumed by the people in 1 day = (4000×150) litres = 600000 litres …(1), Formula to find the capacity of tank, C = l×b×h, Water consumed by all people in d days = Capacity of tank (using equation (1)), Therefore, the water of this tank will last for 3 days. The diameter of a roller is 84 cm and its length is 120 cm. It helps students to tackle the questions in the best possible way. A right circular cylinder just encloses a sphere of radius r (see fig. Find the capacity in litres of a conical vessel with, (i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 12 cm, Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm3), Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm3), 3. The inner radius of the bowl is 5cm. Therefore, hemispherical bowl can hold 0.303 litres of milk. Therefore, 95.04m2 steel was used in actual while making such a tank. Mathematics NCERT Grade 9, Chapter 13: Surface Areas And Volumes- The chapter begins with recalling plane and solid figures. …(2). Area to be polished = (19100+2600) cm2 = 21700 cm2. If the lateral surface of a cylinder is 94.2cm2 and its height is 5cm, then find, (i) radius of its base (ii) its volume. 3. Total area to be white washed = Area of walls + Area of ceiling of room, Cost of white wash per m2 area = Rs.7.50 (Given), Cost of white washing 74 m2 area = Rs. NCERT Solutions for Class 9 Chapter 13 explains the formation of various geometrical objects. measuring 1.5m×1.25 m×0.5 m that can be stored in the godown. (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70. Therefore, 7.48 square meters of the sheet are required. Consequently, students can avail NCERT Solutions for Class 9 Maths Chapter 13 for a profound approach to solving exercises. The difference between both the lateral surfaces is, 40 cm2. Inner radius of circular well, r = 3.5/2m = 1.75m. 4400. (16×173.25)/100 = Rs 27.72. Find the diameter of the base of the cylinder. The volume of a right circular cone is 9856cm3.