Mensuration

Given that, the radius of cone(r) is 7 cm and height(h) is

24 cm.

And, Surface area of cone is - πrl

Also, slant height(l) = √ r² + h²

⇒ l = √ 7² + 24²

⇒ l = √ 49 + 576

⇒ l = √ 625

⇒ l = 25

⇒ Surface area of joker’s cap

= 550cm²

⇒ ∴ the area of the sheet required to make 10 such caps

= 550 × 10 cm²

= 5500 cm²

Given that, the radius of cylinder(r) required is 7 cm and height (h) is 35 cm.

And, Surface area of cylinder is - 2πrh

⇒ Surface area

= 1540cm²

⇒The required area of thick paper sheet needed to make 100 cylinders = 1540 × 100 cm²

= 154000 cm²

Given that, the radius of cone(r) is 6 cm and height(h) is 7 cm.

And, volume of the cone

⇒ Volume of the right circular cone

= 264 cm³

Given that, the lateral surface area of a cylinder is equal to the curved surface area of a cone and their bases are same.

⇒Let r = radius of cylinder = radius of cone, h = height of cylinder and l = slant height of the cone.

And, lateral surface area of cylinder = 2πrh

Also, curved surface area of cone = πrl

⇒2πrh = πrl

⇒ The ratio of the height of the cylinder to the slant height of the cone = 1:2

Given that, the radius(r) of cone is 3 cm and height (h) is

4 cm.

And, the available paper sheet is- 1000 cm²

And, Surface area of cone is - πrl

Also, slant height(l) = √ r² + h²

⇒ l = √ 3² + 4²

⇒ l = √ 9 + 16

⇒ l = √ 25

⇒ l = 5

⇒ Surface area of joker’s cap

And, the available paper sheet is- 1000 cm²

⇒ No. of caps that can be manufactured from that paper sheet

= 21.21 cm²

= 21 cm²

Given that, A cylinder and cone have base of equal radii and are of equal heights and their volumes are in the ratio of 3 : 1.

⇒ Let r = radius of cylinder = radius of cone,

and h = height of cylinder = height of cone.

⇒ And, volume of cylinder = πr²h

Volume of cone

⇒ Their volumes are in the ratio of 3 : 1.

Given that, height(h) of cylinder is 11 cm and diameter(d) is 7 cm.

⇒ radius(r) of cylinder

And, the volume of cylinder = πr²h

⇒ volume of one rod

⇒ volume of one rod

= 423.5

⇒ the total volume of 50 such rods = 423.5 × 50

= 21,175cm³

Given that, the diameter(d) of cone is 12 cm and heigh(h)t is 8 cm.

⇒ radius(r)

And, Surface area of cone is - πrl

Also, volume of cone

Also, slant height(l) = √ r² + h²

⇒ l = √ 6² + 8²

⇒ l = √ 36 + 64

⇒ l = √ 100

⇒ l = 10

⇒ volume of heap

= 301.44 cm³

⇒ Surface area of cone = 3.14 × 6 × 10

= 188.4 cm²

⇒ ∴ canvas cloth required to cover the heap = 188.4 cm²

Given that, curved surface area of a cone is 4070 cm² and diameter(d) is 70 cm.

⇒ radius(r)

Let the slant height be l.

And, curved surface area of a cone = πrl

⇒ πrl = 4070

⇒ l = 37 cm

Given that, diameter(d) is 6 cm and height(h) is 4 cm.

⇒ radius(r)

Also, slant height(l) = √ r² + h²

⇒ l = √ 3² + 4²

⇒ l = √ 9 + 16

⇒ l = √ 25

⇒ l = 5

And, surface area of the toy = surface area of the cone + surface area of hemisphere

⇒ Surface area of toy = πrl + 2πr²

= πr(l + 2r)

= 103.71 cm²

The figure is shown below:

Given that, radius(r) is 8 cm, height of cylinder(H) is 10 cm and height of cone(h) is 6cm.

Also, l = √ r² + h²

⇒ l = √ 8² + 6²

⇒ l = √ 64 + 36

⇒ l = √ 100

⇒ l = 10

Now, total surface area of solid = surface area of cone + surface area of cylinder + surface area of sphere

⇒ total surface area of solid = πrl + 2πrH + 2πr² = πr(l + 2H + 2r)

= 3.14 × 8(10 + 2 × 10 + 2 × 8)

= 3.14 × 8 × 46

= 1155.55 cm²

Given that, the length of the capsule is 14 mm and the width is 5 mm.

⇒ Radius(r)

⇒ height of cylinder(h) = total height-2 × radius of hemisphere

= 14-5 = 9mm

⇒ surface area of capsule = surface area of cylinder + 2 × surface

Area of one hemisphere

= 2πrh + 2 × 2πr²

= 2πr(h + 2r)

= 220 mm²

Given that, volume of cube is 64 cm³.

Also, volume of cube = a³ (where, a is side)

⇒ a³ = 64

⇒a = 4

⇒ Length(l) of cuboid = 2 × a = 8 cm,

Breadth(b) = height(h) = a = 4 cm

⇒ the total surface area of the resulting cuboid = 2(lb + bh + hl)

⇒ 2 × (8 × 4 + 4 × 4 + 4 × 8)

= 2 × 80

= 160 cm²

The figure is shown below:

Given that, diameter(d) of cylinder is 1.4 m and length(h) is 8 m.

⇒ the radius of sphere(r) = radius of cylinder

⇒ surface area of tank = surface area of cylinder + 2(surface

Area of hemisphere)

= 2πrh + 2(2πr²⁾

= 2πr(h + 2r)

= 41.36 m²

⇒ ∴ the cost of painting it on the outside at rate of D20 per m²

= 41.36 × 20

= D827.20

Given that, a sphere, a cylinder and a cone have the same radius(say r) and same height(say h).

⇒ ∵ they have same height,

⇒ diameter of sphere = height of cylinder

⇒ radius of sphere

⇒ h = 2r

Ratio of their volumes = vol. of sphere : vol. of cylinder : vol. of cone

Multiplying the whole by 3, we get-

⇒ Ratio of their volumes = 4:6:2

Give that, A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube.

Now, let the side of cube be ‘l’

⇒ radius of hemisphere(r)

⇒ surface area of remaining solid = surface area of cube- surface area of hemisphere

= 6l² - 2πr²

Given that, radius(r) of base is 3.5 cm and height(h) of cylinder is 10 cm.

⇒ the total surface area of the article = surface area of cylinder +

2(Surface area of hemisphere)

= 2πrh + 2(2πr²)

= 2πr(h + 2r)

= 374 cm²

Given that, height of cylinder(H) = 2.8m = 280 cm, diameter(d) is 20 cm and height of cone(h) is 42 cm.

⇒ radius(r)

⇒ vol. of pole = vol. of cylinder + vol. of cone

= 92400 cm³

If 1 cm³ of iron weighs 7.5 g.

⇒ the weight of the pillar = 92400 × 7.5

= 693,000 g

= 693 kg (∵ 1kg = 1000g)

Given that, The radius of the base(r) of the cone is 7 cm and its volume is of the hemisphere.

∵ circular base of cone is joined with the plane surface of the hemisphere,

⇒ radius of hemisphere = radius of base of cone = 7 cm

Also,vol. of cone = vol. of hemisphere

⇒ h = 3r

⇒h = 3 × 7

⇒h = 21 cm

Also, slant height(l) = √ r² + h²

⇒ l = √ 7² + 21²

⇒ l = √ 490

⇒ l = 22.13

⇒ Slant Height of cone (l) = 22.13 cm

Now, surface area of toy = surface area of cone + surface area of hemisphere

⇒ surface area of toy = πrl + 2πr²

= πr(l + 2r)

= 794.86 cm²

The figure is shown below:

Give that, edge of cube is 7 cm

⇒ largest possible height of cone(h) is 7 cm and diameter of base(d) is 7 cm

⇒ radius(r) =

⇒ the volume of the largest right circular cone that can be cut out of a cube whose edge is 7 cm

= 89.83 cm²

Given that, radius of cylinder(r) is 5 cm and height(h) is 9.8 cm. Also, radius of hemisphere(R) is 3.5 cm and height(H) of cone outside the hemisphere is 5 cm.

⇒ vol. of solid = vol. of hemisphere + vol. of cone

= 154 cm³

⇒ vol. of tub = πr²h

= 770 cm³

⇒ the volume of water left in the tub = vol.of tub-vol. of solid

= 770-154

= 616 cm³

Given that, height of cylinder is 10 cm and diameter is 7 cm.Also, radius of cone is 3 cm and height is 4 cm.

⇒ radius of cylinder

⇒ the volume of the remaining solid = vol. of cylinder-2(vol.

Of cone)

= 309.57 cm³

Given that, diameter(d) of sphere is 1.4 cm and diameter(D) of cylindrical beaker is 7 cm and height(h) required is 5.6 cm.

⇒ vol. of one spherical marble

(∵ )

= 1.437 cm³

⇒ vol.of water reqired(cylindrical) = πR²h

(∵ )

= 215.7 cm³

∴ the number of marbles that should be dropped in to the beaker, so that water level rises by 5.6 cm

= 150

Given that, the dimensions of the cuboid are l = 15 cm by b = 10 cm by h = 3.5 cm and the radius(r) of each of the depression is 0.5 cm and the depth(h) is 1.4 cm.

⇒volume of wood in entire stand = vol.of cuboid-3(vol. of cones)

= 525-1.1

= 523.9 cm³

Given that, radius of sphere(r) is 4.2 cm and radius of cylinder(R) is 6 cm.

Let the height of cylinder be h.

∵ the sphere is melted and recast into cylinder

⇒ vol. of sphere = vol. of cylinder

⇒ h = 2.74 cm

Given that, radii of spheres melted are 6 cm, 8 cm and 10 cm.

Let the radius of solid sphere be R

Also, volume of sphere

∵ these are melted to make a single sphere

⇒ vol. of solid sphere = sum of vol. of all melted sphere

⇒ R³ = 1728

⇒ R = 12 cm

Given that, height of cylindrical well(H) is 20 m and diameter(d) is 7 m. And length(l) and breadth(b) of rectangular platform are 22m and 14 m respectively.

Now, let the height of platform be h.

⇒ radius of cylinder(r)

⇒ vol. Of well = vol. of platform

⇒ πr²H = lbh

⇒ h = 2.5 cm

Given that, height of cylindrical well(h) is 15 m and diameter(d) is 14 m. And width of circular embankment is 7 m.

⇒ radius(r) of well

⇒ outer radius of Embankment(R1) = (7 + 7) = 14 m

⇒ inner radius(r1) = 7 m

Let the height of the embankment be H.

⇒ vol. of well = vol. of embankment

⇒ πr²h = π(R1²-r1²)H

⇒ H = 5 cm

Given that, diameter(d) of cylindrical container is 12 cm and height(h) is 15 cm. and, diameter(D) of cone is 6cm and height(H) is 12 cm

We know that,

⇒ radius of cylinder(r)

And, radius of cone(R)

⇒ volume of container = πr²h

= 540π

⇒ vol. of 1 such cone = vol.of cone + vol. of hemisphere

= 54π

⇒ the number of such cones which can be filled with ice cream

= 10

Given, that dimensions of cuboid formed is 5.5 × 10 × 3.5 and the diameter(d) of cylindrical coin is 1.75 cm and height(h) is 2mm = 0.2cm.

⇒ l = 5.5, b = 10 and h = 3.5

⇒ radius(r) of coin (∵ )

⇒ vol. of silver coin = πr²h

= 3.14 × 0.878 × 0.875 × 0.2

= 0.48

⇒ vol. of cuboid = lbh

= 5.5 × 10 × 3.5

= 192.5

⇒ ∴ no. of silver coins

= 400

Given that, radius(r) of cone is 5 cm and height(h) is 8 cm.And, radius(R) of spheres is 0.5 cm.

Also, When lead shots are dropped into the vessel, of the water flows out.

⇒ vol. of water in cone

⇒ vol. of 1 spherical lead shot

⇒ (vol. of water in cone) = n × vol. of 1 lead shot

(where, n is no. of lead shots)

⇒ n = 100

Given that, diameter(d) of sphere is 28 cm and diameter(D) of cone is and height(h) is 3 cm.

⇒ radius(r) of sphere

And, radius(R) of cone

⇒vol. of spheres

⇒ vol. of cone

⇒ vol. of sphere = n × vol. of 1 cone

(where, n is no. of cones)

⇒ n = 672