Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th

Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th
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Area is two-dimensional while 4 πr represents a length.
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Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

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

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

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

From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

Description : From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

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For annual day, Sakshi and Nidhi were asked to make one rangoli -Maths 9th

Description : For annual day, Sakshi and Nidhi were asked to make one rangoli -Maths 9th

Last Answer : In △ABC and △PQR, BC = QR (Given) ⇒ 1/2BC = 1/2QR ⇒ BM = QN In triangles ABM and PQN, we have AB = PQ (Given) BM = QN (Proved above) AM = PN ... BC = QR (Given) ∴ ΔΑΒC ≅ ΔPOR (SAS congruence criterion) Participation, beauty, hardworking.

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

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.

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

The surface area of a sphere of radius 5 cm -Maths 9th

Description : The surface area of a sphere of radius 5 cm -Maths 9th

Last Answer : Radius of the sphere (r1) = 5 cm Radius of the base of cone (r2) = 4 cm Let r сm be the height of the cone. Surface area of sphere = 4 πr2 ⇒ 4 π(5)2 = 100 π cm2 Curved surface area of cone = πrl = 4 πl ... ∴ Volume of cone = 1/3 πr2h = 1/3 x 22/7 x 42 x 3 = 352/7 cm3 = 50.29 cm3 (Approximately)

A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Description : A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Last Answer : (c)\(\bigg(rac{147 imes\sqrt3}{4}\bigg)\) cm2Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF ... {1}{2}\)x \(rac{3\sqrt3}{2}\) x 49 = \(rac{147 imes\sqrt3}{4}\) cm2.

Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th

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

A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. -Maths 9th

Description : A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. -Maths 9th

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The diameter of a sphere is decreased by 25%. -Maths 9th

Description : The diameter of a sphere is decreased by 25%. -Maths 9th

Last Answer : Let the original diameter of the sphere be 2x. Then, original radius of the sphere = x Original curved surface area = 4πr2 Decreased diameter of the sphere = 2x - 25% of 2x = 2x - x/2 = 3/2x Decreased ... Hence, percentage decreases in area = 7/4πx2/4πx2 x 100% = 7/16 x 100% = 175/4% = 43.75%

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 : 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

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The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

Description : The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

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How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Description : The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Last Answer : This answer was deleted by our moderators...

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Description : The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Last Answer : This answer was deleted by our moderators...

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th

Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Last Answer : NEED ANSWER

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th

Last Answer : According to question find the radius of the sphere

A cube of side 5 cm contain a sphere -Maths 9th

Description : A cube of side 5 cm contain a sphere -Maths 9th

Last Answer : Each side of the cube (a) = 5 cm Diameter of the sphere (2r) = 5 cm . ∴ Radius of the sphere (r) = 5/2 cm Volume of the cube = a3 = 53 cm3 = 125 cm3 Volume of the sphere = 4/3 πr3 = 4/3 x ... /2 x 5/2 = 65.476 cm3 Volume of gap between cube and sphere = 125.000 cm3 - 65.476 cm3 = 59.524 cm3

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 : 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

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The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Last Answer : The roller is in the form of a cylinder of diameter = 84 cm ⇒ Radius of the roller(r) = 842 cm = 42 cm Length of the roller (h) = 120 cm Curved surface area of the ... roller = 31680 cm2 = 3168010000m2 ∴ Area of the playground levelled in 500 revolutions = 500 x 3168010000m2 = 1584m2

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

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

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Last Answer : A roller is shaped like a cylinder. Let h be the height of the roller and r be the radius. h = Length of roller = 120 cm Radius of the circular end of roller = r = (84/2) cm = 42 cm Now, CSA of roller = 2πrh = ... , we have 2 (22/7) 0.7 h = 4.4 Or h = 1 Therefore, the height of the cylinder is 1 m.

Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

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

Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

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

A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Description : A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Last Answer : External radius of hemispherical bowl = 5 cm Internal radius of the bowl = (5 – 1) cm = 4 cm Surface area of external portion = 2π(5)2 = 50 p sq. cm Surface area of internal portion = 2π(4)2 = ... = 91π sq. cm = (91×227)(91×227) sq. cm = 13 × 22 sq. cm = 286 cm2

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

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.

Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Last Answer : Step-by-step explanation: ◾As we have given the two sides of triangle, let the three sides of triangle are (a) , (b), (c) . ◾And perimeter of given triangle is 10.5 cm ◾were, let us assume the sides are, ... . ◾So, the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 [Area ]=

Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Last Answer : Using the formulas A=s(s﹣a)(s﹣b)(s﹣c) s=a+b+c 2Solving forA A=1 4﹣a4+2(ab)2+2(ac)2﹣b4+2(bc)2﹣c4=1 4·﹣124+2·(12·6)2+2·(12·15)2﹣64+2·(6·15)2﹣154≈34.19704cm²

Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Description : Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Last Answer : Area of a trapezium = 1/2(sum of parallel sides) x (perpendicular diatance between them) = 1/2(25 + 13) x 8 = 152cm2

Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Description : Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th

Description : What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th

Last Answer : Lateral surface area of a prism = Perimeter of base Height ⇒ 840 = (5 + 5 + 8) Height ⇒ Height = 8401884018 = 46 cm. = Semi perimeter of the triangular base = 182182 = 9 cm ∴ Area of triangle = 9(9- ... 4 1 = 12 cm2 ∴ Required volume of prism = Area of base Height = (12 46) cm3 = 552cm3

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

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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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

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.

Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2

Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Description : Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th

Last Answer : Required ratio = 4πr2 / 4/3.πr3 = 3 x 4 x π x (1)2 / 4 x π x (1)3 = 3/1 (Since, r = 1) i.e., 3 : 1

A sphere and a cube have the same surface area. What is the ratio of the square of volume of the sphere to the square of volume of the cube ? -Maths 9th

Description : A sphere and a cube have the same surface area. What is the ratio of the square of volume of the sphere to the square of volume of the cube ? -Maths 9th

Last Answer : answer:

A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

Description : A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

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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

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.

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. -Maths 9th

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

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. -Maths 9th

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

A heap of wheat is in the form of a cone whose diameter is 10.5 m -Maths 9th

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

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

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.

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

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.