ncert solutions for class 9 maths chapter 13

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. This chapter explains how the area is found by multiplying the length and breadth of various objects. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. The capacity of a cuboidal tank is 50000 litres of water. Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. (Assume π = 22/7), Radius of the circular end , r = (4.2/2)m = 2.1m, (i) the lateral or curved surface area of cylindrical tank is 2πrh, (ii) Total surface area of tank = 2πr(r+h). We know that, formula to find Curved surface area of cylinder is 2πrh. 7. (see fig. Details of the cuboid, cube, right circular cone, cylinder, hemisphere, and sphere are given. Ex 13.3 Class 9 Maths Question 1. 7. How many square meters of the sheet are required for the same? 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. Let r be the radius of the cylindrical vessel. 5. 13.12, you see the frame of a lampshade. This exercise of Chapter 13 of NCERT Solutions for Class 9 Maths helps students to learn how to find the surface area and volume of various geometrical objects in an easy and smart way. The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m. Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula), Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320. Find its curved surface area (Assume π=22/7), Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm. Answer! The diameter of the moon is approximately one-fourth of the diameter of the earth. Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. = [4(30+25+25)] (after substituting the values). [Use π=3.14], Formula: Total surface area of hemisphere = 3πr2. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? Let length, breadth, and height of the rectangular hall be l, b, and h respectively. (ii) the cost of plastering this curved surface at the rate of Rs. In this chapter, TopperLearning gives you access to some of the best textbook solutions to understand the surface area of objects and spaces using the given data. 1155 while white-washing tomb. Two sizes of boxes were required. 13.32. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Therefore, the slant height of the tent is 26 m. Cost of (13728/7)m2 canvas is equal to Rs (13728/7)×70 = Rs 137280. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. The external faces are to be polished and the inner faces are to be painted. [Hint: Area of the four walls = Lateral surface area.]. 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. 2. (Lateral surface area of cubical box – Lateral surface area of cuboidal box=400cm2–360cm2 = 40 cm2), (ii) Find the total surface area for both the figures, The total surface area of the cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2…(3), This implies, Total surface area of cuboidal box is 610 cm2..(4). (Assume π =22/7), Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm, Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of hemisphere. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Therefore, the total surface area of the cone is 462 cm2. External surface area of shelf while leaving out the front face of the shelf, = [85×110+2(85×25+25×110)] = (9350+9750) = 19100, External surface area of shelf is 19100 cm2, Area of front face = [85×110-75×100+2(75×5)] = 1850+750. Find its curved surface area. (ii) … Therefore, the cost of the canvas required to make such a tent is Rs 137280. Volume of the cone so formed is 100π cm3. 4. Answer: The cost for polishing and painting the surface of the book shelf is Rs. A matchbox measures 4 cm×2.5cm×1.5cm. If the inner radius is 1 m, then find the volume of the iron used to make the tank. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. curved surface area of the cone is 2200 cm2. NCERT Solutions for Class 9 Maths Chapter 13 includes the following topics: Surface area of a cuboid and a cube Surface area of a right circular cylinder Surface area of a right circular cone Surface area of a sphere Volume of a cuboid Volume of a cylinder Volume of a right circular cone Volume … The slant height and base diameter of conical tomb are 25m and 14 m respectively. (74×7.50). Find the outer curved surface of the bowl. 12.50 per m2 is Rs 68.75. Curved surface area of a right circular cylinder is 4.4 m2. The heap is to be covered by canvas to protect it from rain. Now, using relationship between, density, mass and volume. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? 7. What fraction of the volume of the earth is the volume of the moon? 11. The bigger of dimensions 25 cm×20cm×5cm and the smaller of dimension 15cm×12cm×5cm. Cost of plastering 110 m2 area = Rs (110×40). Each penholder was to be of radius 3 cm and height 10.5 cm. Therefore, the cost of plastering the curved surface of the well is Rs. It is a detailed and well-structured solution for a good grasp of concept-based knowledge. 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. Therefore, cost will be Rs. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. Dimensions of a matchbox (a cuboid) are l×b×h = 4 cm×2.5 cm×1.5 cm respectively, Formula to find the volume of matchbox = l×b×h = (4×2.5×1.5) = 15, Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3. A metal pipe is 77 cm long. Radius = half of diameter = (140/2) cm = 70cm = 0.7m, Area of sheet required = Total surface are of tank = 2πr(r+h) unit square. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3. 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. 6. Surface area of a right circular cylinder. The diameter of the moon is approximately one fourth of the diameter of the earth. (Use π = 3.14 and take √(1.04) =1.02), Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m, Slant height of cone is l, and l2 = (r2+h2), CSA of 50 such cones = (50×0.64056) = 32.028, Cost of painting 1 m2 area = Rs 12 (given), Cost of painting 32.028 m2 area = Rs (32.028×12). Difference in capacity = (385-300) cm3 = 85cm3, 4. Let h be the height and r be the radius of a cylindrical tank. (Assume π = 22/7), Volume of the iron used in the tank = (2/3) π(R3– r3), Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.013– 13) = 0.06348. Given: Curved surface area of a cone is 308 cm2, (ii) Total surface area of cone = CSA of cone + Area of base (πr2), Total surface area of cone = 308+(22/7)×72 = 308+154. (ii)How much of tape is needed for all the 12 edges? Therefore, 7920 cm2 cardboard sheet will be needed for the competition. Download Chapter-wise PDF for NCERT Solutions Class 9 Maths. 2. 1. 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. The frame has a base diameter of 20 cm and height of 30 cm. Cost of polishing 21700 cm2 area Rs. 8. For all the overlaps, 5% of the total surface area is required extra. A river 3m deep and 40m wide is flowing at the rate of 2km per hour. (Assume π = 22/7), Height of cylindrical pipe = Length of cylindrical pipe = 28m, Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m, Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder. If the cost of painting is at the rate of Rs 20 per m2, find, (i) inner curved surface area of the vessel. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. So, Rs 2200 is the cost of painting = (1/20×2200) m2. Cost of painting 19350 cm2 area = Rs (19350 x 0.1) = Rs 1935, Total expense required for polishing and painting = Rs. 8. Find the area of the canvas. The diameter of a roller is 84 cm and its length is 120 cm.

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