The curved surface area of a cylinder is 154 cm2. -Maths 9th

1 Answer

Answer :

Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 × curved surface area  2πrh + 2πr2 =  3 × 154  154 + 2πr2  = 462  2πr2  = 462 - 154 = 308 r2 = 308 × 7 / 2 × 22 = 49 r = 7 cm  Now, 2πrh = 154 2 × 22 / 7 × 7 × h = 154  h =  154 / 44 = 3.5 cm  ∴ Volume of cyclinder = πr2h   = 22 / 7 × 7 × 7 × 3.5 = 539 cm3

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Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th

Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3

Description : the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Last Answer : T.S.A = 3*154 = 462 cm² C.S.A = 154 cm² C.S.A = 2πrh T.S.A = 2πr(r+h) Now, In T.S.A = 2πrr + 2πrh 462 = 2πrr + 2πrh 462 = 2*22/7*r*r + 154 462 - 154 = 2*22/7*r*r 308*7/2*22 = r*r 49 = r*r R = 7 cm ... 7*h 154/44 = h 7/2 =h H = 3.5 cm or 7/2 cm Now volume = πrrh = 22/7 * 7* 7 *7/2 = 11*49 = 539 cm³

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 : 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 : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

Description : If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th

Last Answer : Let r and h be radius and height of the cyclinder, then C.S.A. = 2πrh Now, radius is doubled and height is halved. ∴ New radius = 2r and new height = h / 2 New C.S.A. = 2π × 2r × h / 2 = 2πrh .

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th

Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π cm3

Description : If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th

Last Answer : Let r and h be radius and height of the cyclinder, then C.S.A. = 2πrh Now, radius is doubled and height is halved. ∴ New radius = 2r and new height = h / 2 New C.S.A. = 2π × 2r × h / 2 = 2πrh .

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

Description : In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th

Last Answer : The curved surface area will remain same. So, there is no change in the curved surface area of cylinder . Hence the curved surface area will remain same.

Description : In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th

Last Answer : The curved surface area will remain same. So, there is no change in the curved surface area of cylinder . Hence the curved surface area will remain same.

Description : The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th

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Description : The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th

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Description : A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

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Description : If three cylinders of radius r and height h are placed vertically such that the curved surface of each cylinder touches the curved surfaces -Maths 9th

Last Answer : hr2 (3-√−π2)(3−π2) The bases of the three cylinders when placed as given are as shown in the figure : Let the radius of the base of each cylinder = r cm. We are required to find the volume of air. ... ∠C = 60º) = 3 x 60o360o πr2=πr2260o360o πr2=πr22 ∴ Required volume = (3-√r2−π2r2)h=(3-√−π2)r2h.

Description : Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Last Answer : Let 'r' be the radius of sphere Surface area of sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 154 = 4 x 22/7 x r2 ⇒ r 2 = 154 x 7/4 x 22 = 49/4 ⇒ r = 7/2 cm = 3.5 cm

Description : Find the volume of a sphere whose surface area is 154 cm sq. -Maths 9th

Last Answer : Let r cm be the radius of sphere. Surface area of the sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 4 x 22/7 x r2 = 154 r 2 = 154 x 7/4 x 22 = 72/22 ⇒ r = 7/2 Volume of sphere = 4/3 πr3 = 4/3 x 22/7 x 7/2 x 7/2 x 7/2 cm3 = 539/3 cm3 = 179.2/3 cm3

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

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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 : 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 : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

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Description : Curved surface area of a cone is -Maths 9th

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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

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Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

Description : The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

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Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

Description : The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

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Description : The total surface area of a cube is 96 cm2 . The volume of the cube is -Maths 9th

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Description : The total surface area of a cube is 96 cm2 . The volume of the cube is -Maths 9th

Last Answer : (c) Surface area of a cube = 96 cm2 Surface area of a cube = 6 (Side)2 = 96 ⇒ (Side)2 = 16 ⇒ (Side) = 4 cm [taking positive square root because side is always a positive quantity] Volume of cube = (Side)3 = (4)3 = 64 cm3 Hence, the volume of the cube is 64 cm3.

Description : A solid cylinder has total surface area of 462 cm square. -Maths 9th

Last Answer : Let r cm be the radius of the base and h cm be the height of the cylinder, Then, total surface area of cylinder = 2 πr (r + h) Curved surface area of cylinder = 2 πrh We have, Curved surface area = 1/3(Total surface ... x 22 = 7/2 cm Volume of the cylinder = πr2h = 22/7 x 7 x 7 x 7/2 = 539 cm3

Description : Define :Volume and surface area of a hollow cylinder. -Maths 9th

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Description : The magnitude of the volume of a closed right circular cylinder of unit height divided by the magnitude of the total surface area of the -Maths 9th

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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 : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Last Answer : Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm 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 ... 22/7) (5.25)2 = 173.25 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

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 outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

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Description : The outer curved surface areas of the hemisphere and sphere are in ratio 2:9. find their ratio of their raddii -Maths 9th

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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 : A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be : -Maths 9th

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Description : If the curved surface area of a closed cylinder is `2pirh` ad base area is `pir^(2)`, then total surface area of that cylinder A is ______

Last Answer : If the curved surface area of a closed cylinder is `2pirh` ad base area is `pir^(2)`, then total surface area of that cylinder A is ______

Description : if seve times the curved surface area (A) of a cylinder is equal to 44 times the proudct of base radius (r) and height (h) then what si the formula wi

Last Answer : if seve times the curved surface area (A) of a cylinder is equal to 44 times the proudct of base ... (h) then what si the formula with subject A?

Description : What is answer Curved surface area of a cylinder is 176cm square area of base is 38.5cm square fund volume?

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Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

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Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.