Description : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th
Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165
Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th
Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.
Description : The total surface area of a cone whose radius is r/2 and slant height 2l is -Maths 9th
Last Answer : Total surface area of cone = πr(r+l) Given, radius = r/2 and slant height = 2l Therefore, new total surface area of cone = πr/2(r/2+2l) = π(r/4^2+rl) = πr(l+r/4)
Last Answer : Radius (r)=r/2 & slant height=2l TSA (S)=PIE R (l+r) =22/7×r/2(2l+r/2) =11/7×r(2l+r/2)
Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th
Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.
Description : The circumference of the base of 9 m high wooden solid cone is 44 m. Find the slant height of the cone. -Maths 9th
Last Answer : Circumference of the base of a cone = 2πr
Description : A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. -Maths 9th
Last Answer : : Ncert solutions class 9 chapter 13-5 Let ABC be a conical tent Height of conical tent, h = 10 m Radius of conical tent, r = 24m Let the slant height of the tent be l. (i) In right triangle ... (13728/7) 70 = Rs 137280 Therefore, the cost of the canvas required to make such a tent is Rs 137280.
Description : The slant height and base diameter.... -Maths 9th
Last Answer : Radius of the base of the conical tomb (r) = 14/2 m = 7 m Slant height of conical tomb (l) = 25 m Curved surface area of conical tomb = πrl = 22/7 x 7 x 25 = 550 m2 Cost of white-washing 1 m2 = ₹ 210/100 = ₹ 2.1 ∴ Cost of white-washing 550 m2 = ₹ 550 x 2.1 = ₹ 1155
Description : The radius and slant height of a cone... -Maths 9th
Last Answer : Let the radius of cone (r) = 4x cm and the slant height of the cone (l) = 7x cm Curved surface area of cone = πrl ∴ πrl = 792 cm2 ⇒ 22/7 x 4x x 7x = 792 ⇒ x2 = 792/22 x 4 = 9 ⇒ x = 3 cm ∴ Radius of the cone = 4 x 3 = 12 cm
Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th
Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.
Description : A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. -Maths 9th
Last Answer : Diameter of cone = 10.5 m Radius of cone (r) = 5.25 m Height of cone (h) = 3 m Volume of cone = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 5.25 × 5.25 × 3 = 86.625m3 Cost of 1m3 of wheat = 10 ∴ Cost of 86.625 m3 of wheat = 10 × 86.625 = 86.625
Description : A cone of height 24 cm has a curved surface -Maths 9th
Last Answer : Height of the cone (h) = 24 cm Let r сm be the radius of the base and l cm be the slant height of the cone. Then, l = root under (√r2+ h2 ) = root under (√r2 + 242) = root under (√r2 + 576) Now, Curved surface ... ⇒ r = 7 cm ∴ Volume of the cone = 1/3πr2h = 1/3 x 22/7 x 72 x 24 = 1232 cm3
Description : Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th
Last Answer : (c) 450 cm3. Required difference in capacities = 227227 x (5)2 x 21~ (10)2 x 21 = (1650 ~ 2100) cm3 = 450 cm3
Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th
Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.
Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th
Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.
Description : If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th
Last Answer : answer:
Description : A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th
Description : Find the capacity in litres of a conical vessel having height 8 cm and slant height 10 cm. -Maths 9th
Last Answer : Height of conical vessel (h) = 8 cm Slant height of conical vessel (l) = 10 cm ∴ r2 + h2 = l2 r2 + 82 = 102 r2 = 100 - 64 = 36 r = 6 cm Now, volume of conical vessel = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 6 × 6 × 8 = 301.71 cm3 = 0.30171 litre
Description : Diameter of the base of a cone is 10.5 cm -Maths 9th
Last Answer : Radius of cone (r) = 10.5/2 cm Slant height of cone (l) = 10 cm Curved surface area of cone = πrl = 22/7 x 10.5/2 x 10 = 165 cm2
Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th
Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3
Description : A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th
Last Answer : 92239223% Volume of cone = 1313πr2h = 13×227×1×7=22313×227×1×7=223 cu. cm Volume of cubical block = (10 × 5 × 2) cm3 = 100 cm3 ∴ Wastage of wood = (100−227)100×100(100−227)100×100 = 27832783% = 92239223%
Description : It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th
Last Answer : Radius of the closed cylindrical tank = 140/2 cm = 70 cm = 0.7 m Height of the closed cylindrical tank = 1 m Area of metal sheet required = 2πr(r + h) = 2 x 22/7 x 0.7 (1 + 0.7) = 7.48 m2
Description : if the slant height (l) of a cone is equal to the square root of the sum of the squares of radius (r) and height (h) then,
Last Answer : if the slant height (l) of a cone is equal to the square root of the sum of the squares of radius (r) and height (h) ... (2))` D. `r^(2)-h^(2)=l^(2)`.
Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th
Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required
Description : A heap of wheat is in the form of a cone whose diameter is 10.5 m -Maths 9th
Last Answer : Radius of the conical heap of wheat (r) = 10.5/2 m Height of the conical heap of wheat (h) = 3 m Volume of the conical heap of wheat = 1/3 πr2h = 1/3 x 22/7 x (10.5/2)2 x 3 = 173.25/2 = 86.625 ... = 6.05 m Area of canvas required = curved surface area of cone πrl = 22/7 x 10.5/2 x 6.05 = 99.825 m2
Description : A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th
Last Answer : Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.
Description : Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th
Last Answer : Required sheet = T.S.A. of cyclinder = 2πr (h+r) = 2 × 22 / 7 × 70 / 100(1 + 70 / 100) = 2 × 22 × 0.1 × 1.7 = 7.48 m2
Description : he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th
Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.
Description : It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th
Last Answer : Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface are of tank = 2πr( ... [2 (22/7) 0.7(0.7+1)] = 7.48 Therefore, 7.48 square meters of the sheet are required.
Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th
Last Answer : 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 ... = 2 (22/7) 0.025 28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2.
Description : For minimum curved surface area and given volume, the ration of the height and radius of base of a cone is :
Last Answer : For minimum curved surface area and given volume, the ration of the height and radius of base of a cone is : A. ` ... : 1` C. `1 : 2` D. None of these
Description : Curved surface area of a cone is -Maths 9th
Last Answer : Slant height of the cone (l) = 14 cm Curved surface area of the cone = 308 cm2 Let 'r' be the radius of the base of cone (i) Curved surface area of cone = πrl ∴ 22/7 x r x 14 = 308 ⇒ r = 308 x 7/22 x 14 ... ii) Total surface area of cone = πr(r + l) = 22/7 x 7(7 + 14) = 22/7 x 7 x 21 = 462 cm2
Description : A right circular cylinder and a right circular cone have equal bases and equal volumes. But the lateral surface area of the right circular cone is -Maths 9th
Description : The radius of a right circular cone is 3 cm and its height is 4 cm. The curved surface of the cone will be (1) 12 sq. cm (2) 15 sq. cm (3) 18 sq. cm (4) 21 sq. cm
Last Answer : (2) 15 sq. cm
Description : Find the volume of cone of radius r/2 and height ‘2h’. -Maths 9th
Last Answer : Volume of cone = 1 / 3π × (r / 2)2 × 2h = 1 / 3π × r2 / 4 × 2h = 1 / 6 πr2h cu. units.
Description : The radius and height of a cone are in the ratio 3 : 4 -Maths 9th
Last Answer : Let the radius ofthe cone (r) = 3x cm Height of the cone (h) = 4x cm Volume of the cone = 1/3 πr2h ⇒ 301.44 = 1/3 x 3.14 x (3x)2 .4x ⇒ x3 = 301.44/3.14 x 12 = 8 ⇒ x3 = 23 ⇒ x = 2 ... 4 x 2 = 8 cm Slant height of the cone (l) = root under (√r2 + h2 ) = root under (√62 + 82)= √100 = 10 cm
Description : The height of a cone is 15 cm. -Maths 9th
Last Answer : Let, the radius of the base of cone be r cm Height of the cone = 15 cm Volume of the cone = 1570 cm3 ⇒ 1/3πr2h = 1570 ⇒ 1/3 x 3.14 x r2 x 15 = 1570 ⇒ r2 = 1570 x 3/3.14 x 15 = 100 ⇒ r = √100 = 10 cm Thus, the diameter of the base of the cone = 2r = 2 x 10 cm = 20 cm
Description : In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th
Description : A sphere, a cylinder and a cone respectively are of the same radius and same height. Find the ratio of their curved surfaces. -Maths 9th
Description : A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th
Last Answer : Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r2 = 8 8 r = 8 cm l2 = h2 + r2 = (9)2 + (8)2 l = = 12 cm C.S.A (cone) = = 301.71 cm2
Description : Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th
Last Answer : Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of cylindrical tank is 2πrh = 2 (22/7) 2.1 4.5 m2 = (44 0.3 ... ) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was used in actual while making such a tank.
Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th
Last Answer : Volume of cone = Volume of sphere 1 / 3π(2.1)2 × 8.4 = 4 / 3 πr3 ⇒ r3 = (2.1)2 × 8.4 / 4 = (2.1)3 ⇒ r = 2.1 cm ∴ Radius of the sphere = 2.1 cm