Surface Area And Volume

Given: A cuboid shape box of medicine having,

Length (l) = 20 cm

Breadth (b) = 12 cm

Height (h) = 10 cm

To find: Surface Area of vertical faces (SA) = ?

Total Surface Area of the box (cuboid) (TSA) = ?

To find Surface Area of vertical faces of cuboid:

It will be found by adding vertical sides of cuboid, i.e., front side, back side, left side and right side of the cuboid.

So, Surface Area of vertical faces = (Area of front face) + (Area of back face) + (Area of left face) + (Area of right face)

⇒ SA = lh + lh + bh + bh

⇒ SA = 2 lh + 2 bh

⇒ SA = 2(l + b)×h

⇒ SA = 2(20 + 12)×10

⇒ SA = 2 × 32 × 10

⇒ SA = 640

Thus, Surface Area of vertical faces of the box is 640 cm².

To find Total Surface Area of the box:

We have a formula for Total Surface Area of cuboid given by,

TSA = 2(lb + bh + hl)

⇒ TSA = 2(20×12 + 12×10 + 10×20)

⇒ TSA = 2(240 + 120 + 200)

⇒ TSA = 2(560)

⇒ TSA = 1120

Thus, Total Surface Area of the box is 1120 cm².

Hence, Surface Area of vertical faces of the box is 640 cm² and Total Surface Area of the box is 1120 cm².

Given: Total surface area of cuboidal box (TSA) = 500 sq. unit

Breadth of cuboidal box (b) = 6 unit

Height of cuboidal box (h) = 5 unit

To find length (l) of the cuboidal box, recall total surface area of cuboid given by

TSA = 2 (lb + bh + hl)

Substituting given values in the formula, we get

500 = 2 (l×6 + 6×5 + 5×l)

⇒ 500 = 2 (6l + 30 + 5l)

⇒ 6l + 5l + 30 = 500/2

⇒ 11l + 30 = 250

⇒ 11l = 250 – 30 = 220

⇒ l = 220/11 = 20

Thus, length of the cuboidal box is 20 unit.

Given: side of the cube (s) = 4.5 cm

We know that, cube has equal sides. This means, length = breadth = height.

So, to find surface area of all vertical faces of the cube, we need to add front side area, back side area, left side area and right-side area of the cube.

∴, Surface area (SA) = (Area of front face) + (Area of back face) + (Area of left face) + (Area of right face)

⇒ SA = s² + s² + s² + s² = 4s²

⇒ SA = 4(4.5)²

⇒ SA = 81

Also, Total surface area of cube is given by

TSA = 6 (side)²

⇒ TSA = 6 (s)²

⇒ TSA = 6 (4.5)²

⇒ TSA = 121.50

Thus, surface area of all vertical faces of the cube is 81 cm² and total surface area of the cube is 121.50 cm².

Given: Total surface area of cube = 5400 cm²

And we know total surface area of cube is given by,

TSA = 6 (side)²

⇒ 5400 = 6 (side)² [∵, it given that TSA = 5400 cm²]

⇒ (side)² = 5400/6

⇒ (side)² = 900

⇒ side = √900

⇒ side = 30

Now we know side = 30 cm.

Let side = s.

So, to find surface area of all vertical faces of the cube, we need to add front side area, back side area, left side area and right side area of the cube.

∴, Surface area (SA) = (Area of front face) + (Area of back face) + (Area of left face) + (Area of right face)

⇒ SA = s² + s² + s² + s² = 4s²

⇒ SA = 4 (30)²

⇒ SA = 3600

Thus, surface area of all vertical faces of the cube is 3600 cm².

Given: Volume of cuboid (V) = 34.50 m³

Breadth of cuboid (b) = 1.5 m

Height of cuboid (h) = 1.15 m

Volume of a cuboid is given by,

Volume = length × breadth × height

⇒ V = lbh

⇒ l = V/bh

⇒ l =

⇒ l = 20

Thus, length of cuboid is 20 m.

Given: length of edge (s) = 7.5 cm

To find volume of cube, we need to write the formula of volume of cube:

Volume of cube (V) = (length of edge)³

⇒ V = (7.5)³ [from given]

⇒ V = 421.88

Thus, volume of the required cube is 421.88 cm³.

Given: radius of base of cylinder (r) = 20 cm

Height of cylinder (h) = 13 cm

Since it is a cylinder, curved surface area is given by

CSA = 2πrh

Where CSA = Curved Surface Area

r = radius

h = height

⇒ CSA = 2 × 3.14 × 20 × 13

⇒ CSA = 1632.80 …(i)

Total surface area (TSA) of cylinder is given by

TSA = (Curved surface area) + 2 (Circular area of the base of the cylinder)

⇒ TSA = 2πrh + 2πr²

⇒ TSA = 1632.8 + (2 × 3.14 × (20)²) [from (i)]

⇒ TSA = 1632.8 + 2512 = 4144.80

Thus, curved surface area of the cylinder is 1632.80 cm² and total surface area of the cylinder is 4144.80 cm².

Given: Curved surface area of the cylinder (CSA) = 1980 cm²

Radius of the base of cylinder (r) = 15 cm

We need to write curved surface area of cylinder, that is

CSA = 2πrh

Where r = radius

h = height of the cylinder

So by substituting values in the formula, we get

1980 = 2 × 22/7 × 15 × h

⇒ h =

⇒ h = 21

Thus, height of the cylinder is 21 cm.

We have

Given that, height of cone = 12 cm and slant height of cone = 13 cm.

We need to find radius of the base of the cone, r.

In right-angled ∆AOB, using Pythagoras theorem,

AB² = AO² + OB²

⇒ OB² = AB² – AO²

⇒ r² = 13² – 12²

⇒ r² = 169 – 144 = 25

⇒ r = √25

⇒ r = 5

Thus, radius of the base of the cone is 5 cm.

Given: Total surface area of cone (TSA) = 7128 cm²

Radius of base of cone (r) = 28 cm

We know total surface area of cone is given by,

TSA = πrl + πr²

⇒ πrl = TSA – πr²

⇒

⇒

⇒

⇒

⇒ l = 53

So, slant height of cone = 53 cm

We have,

Using Pythagoras theorem in ∆AOB, we can find height of the cone.

AB² = AO² + OB²

⇒ AO² = AB² – OB²

⇒ h² = 53² – 28²

⇒ h² = 2809 – 784

⇒ h = √2025 = 45

We have got height (h) = 45 cm.

Volume of cone is given by,

V = πr²h/3

⇒

⇒ V = 36960

Thus, volume of cone is 36960 cm³.

We have

Given: Curved surface area (CSA) = 251.2 cm²

Radius of the base of cone (r) = 8 cm

We know the formula of curved surface area of cone,

CSA = πrl

Where l = slant height of cone

⇒ 251.2 = 3.14 × 8 × l

⇒

⇒ l = 10

So, we have found slant height of cone (l) = 10 cm

Now, using Pythagoras theorem in ∆AOB,

AB² = AO² + OB²

⇒ AO² = AB² – OB²

⇒ h² = 10² – 8²

⇒ h² = 100 – 64

⇒ h = √36 = 6

Thus, slant height of cone is 10 cm and height of cone is 6 cm.

Given: radius of base of cone (r) = 6 m

Slant height of cone (l) = 8 m

Since, this is a closed cone made of tin sheet. We need to find total surface area of the cone, it given by

TSA = πrl + πr²

⇒ TSA = πr (l + r)

Substituting values in the formula, we get

⇒

According to the question,

Cost of 1 m² = Rs. 10

Cost of 264 m² = 10 × 264

= Rs. 2640

Thus, cost of making the required cone is Rs. 2640.

Given: Volume of cone = 6280 cm²

Radius of the base of the cone = 30 cm

The figure is given as:

We know volume of cone is given by,

Volume = πr²h/3

⇒ 3×Volume = πr²h

⇒

⇒

⇒

⇒ h = 6.67

Thus, perpendicular height of cone is 6.67 cm.

Given: Surface area of cone (SA) = 188.4 cm²

Slant height of cone (l) = 10 cm

So, we need to write the formula of curved surface area of cone with slant height l,

Surface Area = πrl

Where, r = radius of base of cone

Substituting values in the formula, we get

188.4 = 3.14×r×10

⇒ 188.4 = 31.4r

⇒ r = 188.4/31.4

⇒ r = 6

So we have,

We need to find the perpendicular height, h.

Using Pythagoras theorem in ∆AOB, we get

AB² = AO² + OB²

⇒ AO² = AB² – OB²

⇒ h² = 10² – 6²

⇒ h² = 100 – 36 = 64

⇒ h = √64 = 8

Thus, perpendicular height of the cone is 8 cm.

Given: Volume of cone (V) = 1212 cm³

Height of cone (h) = 24 cm

We know the formula of volume of cone,

Volume = πr²h /3

⇒ 3V = πr²h

⇒ r² =

⇒

⇒

⇒

⇒ r² = 48.20 …(i)

⇒ r = √48.20 = 6.94

⇒ r ≈ 7 …(ii)

We know,

l = √(r² + h²)

⇒ l = √(48.20 + 24²) [from (i)]

⇒ l = √(48.20 + 576)

⇒ l = 24.98

⇒ l ≈ 25 …(iii)

⇒ Slant height of cone is 25 cm.

Curved surface area of cone is given by,

CSA = πrl

⇒ [from (ii) and (iii)]

⇒ CSA = 550 …(iv)

Thus, curved surface area of cone is 550 cm².

Given: Curved surface area of cone (CSA) = 2200 cm²

Slant height of cone (l) = 50 cm

For total surface area, we have the formula

TSA = πrl + πr²

⇒ TSA = CSA + πr² [∵, Curved surface area (CSA) = πrl] …(i)

Where, r = radius of the base of the cone

We have CSA, we just need to find radius of the base of the cone (r).

So, we know

CSA = πrl

Substituting values in above equation,

⇒

⇒

⇒ r = 14

⇒ Radius of the base of the cone = 14 cm

Thus, substituting value of CSA and r in equation (i),

⇒ TSA = 2200 + 616

⇒ TSA = 2816

Thus, total surface area of cone is 2816 cm².

Given that, the tent is conical in shape.

If 1 person occupies area inside the tent = 4 m²

Then 25 people will occupy area inside the tent = 25 × 4 = 100 m²

⇒ Base area of the cone = 100 m² …(i)

Also, base area of the cone is given by,

Base area of the cone = πr²

⇒ 100 = πr² [from (i)]

⇒

⇒

⇒ r² = 31.82 …(ii)

⇒ r = 5.64

We have, radius of the base of the cone (r) = 5.64 m

Height of the conical tent (h) = 18 m

Now we need to find volume of this conical tent. The formula is:

Volume = πr²h/3

Substituting values, we get

Volume = (3.14 × 31.82 × 18)/3 [from (ii)]

⇒ Volume = 599.5

⇒ Volume ≈ 600 m³

Thus, we can say volume of the conical tent is approximately 600 m³.

We have

Given: height of the cone (h) = 2.1 m

Diameter of the base of cone (d) = 7.2 m

⇒ Radius of the base of cone (r) = 7.2/2 = 3.6 m

Volume (V) of the conical fodder is given by,

Volume = πr²h/3

⇒

⇒ V = 28.512

Thus, volume of cone is 28.512 m³.

If it is covered by polythene in rainy season, then it will be covered on the curved surface area, not on the base.

Curved surface area (CSA) of cone is given by,

CSA = πrl …(i)

We need to find l.

In ∆AOB, using Pythagoras theorem,

AB² = AO² + OB²

⇒ l² = 2.1² + 3.6²

⇒ l² = 4.41 + 12.96

⇒ l² = 17.37

⇒ l = √17.37 = 4.17 [given that √17.37 = 4.17]

Substituting l = 4.17 in equation (i),

⇒ CSA = 47.18

Thus, curved surface area of cone is 47.18 m².

Hence, volume of cone is 28.51 m³ and curved surface area of cone is 47.18 m².

(i). When radius (r) = 4 cm

Surface area of the sphere is given by,

Surface area = 4πr²

⇒ Surface area = 4 × 3.14 × 4²

⇒ Surface area = 200.96

Volume of the sphere is given by,

Volume =

⇒ Volume =

⇒ Volume = 267.95

Thus, surface area of sphere is 200.96 cm² and volume of sphere is 267.95 cm³.

(ii). When radius (r) = 9 cm

Surface area of the sphere is given by,

Surface area = 4πr²

⇒ Surface area = 4 × 3.14 × 9²

⇒ Surface area = 1017.36

Volume of the sphere is given by,

Volume =

⇒ Volume =

⇒ Volume = 3052.08

Thus, surface area of sphere is 1018.29 cm² and volume of sphere is 3052.08 cm³.

(iii). When radius (r) = 3.5 cm

Surface area of the sphere is given by,

Surface area = 4πr²

⇒ Surface area = 4 × 3.14 × 3.5²

⇒ Surface area = 153.86

Volume of the sphere is given by,

Volume =

⇒ Volume =

⇒ Volume = 179.50

Thus, surface area of sphere is 153.86 cm² and volume of sphere is 179.50 cm³.

Given: Radius of hemisphere = 5 cm

Curved surface area is given by,

Curved surface area = 2πr²

= 2 × 3.14 × 5²

= 157

Total surface area is given by,

Total surface area = 3πr²

= 3 × 3.14 × 5²

= 235.5

Thus, curved surface area of hemisphere is 157 cm² and total surface area of hemisphere is 235.5 cm².

Surface area of sphere = 2826 cm²

Surface area of sphere is given by,

Surface area = 4πr²

Substituting the values in above equation,

⇒ 2826 = 4 × 3.14 × r²

⇒

⇒ r² = 225

⇒ r = √225

⇒ r = 15

Volume of sphere is given by,

Volume =

⇒ Volume =

⇒ Volume = 14130

Thus, volume of sphere is 14130 cm².

Given: Volume of sphere = 38808 cm³

Volume of sphere is given by,

Volume =

Substituting value of volume of sphere in above equation, we get

⇒

⇒

⇒

⇒ r³ = 9261

⇒

⇒ r = 21

Surface area of sphere is given by,

Surface area = 4πr²

⇒ Surface area =

= 5544

Thus, surface area of the sphere is 5544 cm².

Given: Volume of hemisphere = 18000π cm³

And we know volume of hemisphere is given by,

Volume =

Substituting value of volume of hemisphere in above equation, we get

18000π = 2/3 πr³

⇒ 18000 × 3 = 2r³

⇒ r³ = (18000 × 3)/2

⇒ r³ = 9000 × 3

⇒ r³ = 3 × 3 × 3 × 1000

⇒

⇒ r = 3 × 10

⇒ r = 30

If radius of hemisphere is 30 cm, then diameter of the hemisphere will be twice of the radius.

⇒ diameter = 2(radius)

⇒ diameter = 2(30)

⇒ diameter = 60

Thus, diameter of hemisphere is 60 cm.

We have the road roller which is cylindrical,

Now, given that

Length of the road roller, i.e., height of the cylinder (h) = 1.4 m

Diameter of the road roller = 0.9 m

⇒ Radius of base of the cylindrical road roller (r) = Diameter/2

⇒ Radius (r) = 0.9/2 m

Now, since the road roller is rotating, we can deduce that only the curved surface of the cylindrical road roller will be pressed against the field.

⇒ Area of the field pressed in 1 rotation = Curved surface area of the cylindrical road roller

Then, Area of the field pressed in 500 rotations = 500 (Curved surface area of cylindrical road roller) …(i)

So, curved surface area of cylinder is given by

CSA = 2πrh

Substituting values, we get

CSA =

⇒ CSA = 3.96

Now, using equation (i) we get

Area of the field pressed in 500 rotations = 500 × 3.96 = 1980 m²

Thus, area of field pressed by the road roller in 500 rotations is 1980 m².

The fish tank is cuboidal.

Although it is open from the top, it won’t affect our calculations for finding volume of the fish tank.

Outer dimensions:

Length (l’) = 60.4 cm

Breadth (b’) = 40.4 cm

Height (h’) = 40.2 cm

Width of the gauge is width of the tank (w) = 2 mm = 0.2 cm [∵, 1 mm = 10^-1 cm]

⇒ 2w = 2 × 0.2 = 0.4 cm

Inner dimensions:

Length (l) = l’ – 2w = 60.4 – 0.4 = 60 cm

Breadth (b) = b’ – 2w = 40.4 – 0.4 = 40 cm

Height (h) = h’ – 2w = 40.2 – 0.4 = 39.8 cm

Volume of the cuboidal fish tank is given by,

Volume = lbh

= 60 × 40 × 39.8

= 95520 cm³

Thus, volume of water that the fish tank will contain is 95520 cm³.

Given: Volume of cone = 314 m³

Radius of base of cone: height of cone = r : h

= 5 : 12

Let radius of base of cone (r) = 5x

And let height of cone (h) = 12x

Volume of cone is given by,

Volume = πr²h/3

Substituting values, we get

314 = (3.14 × (5x)² × 12x)/3

⇒

⇒ x³ = 1

⇒ x = 1

Thus, radius of the base of cone = 5×1 = 5 m

And height of cone = 12×1 = 12 m

For slant height, we have the relation

l = √(r² + h²)

⇒ l = √(5² + 12²)

⇒ l = √(25 + 144)

⇒ l = √169

⇒ l = 13

Hence, we have got perpendicular height of cone to be 12 m and slant height of the cone to be 13 m.

Given: Volume of the sphere = 904.32 cm³

And volume of the sphere is given by,

Volume = 4πr³/3

We have to find radius (r), so by rearranging the above equation, we get

Substituting values in the above equation, we get

⇒

⇒ r³ = 216

⇒ r = (216)^1/3

⇒ r = 6

Thus, radius of sphere is 6 cm.

Given: Total surface area of cube = 864 cm²

And total surface area of cube is given by,

Total surface area = 6(side)²

Substituting values in the above equation,

864 = 6(side)²

⇒ (side)² = 864/6 = 144

⇒ side = √144

⇒ side = 12

If the length of side of cube is 12 cm, then volume is given by

Volume = (side)³

⇒ Volume = (12)³

⇒ Volume = 1728

Thus, volume of cube is 1728 cm³.

Given: surface area of sphere = 154 cm²

And we know surface area of sphere is given by,

Surface area = 4πr²

Substituting values in the above equation, we get

⇒

⇒

⇒ r² = 12.25

⇒ r = √12.25

⇒ r = 3.5

If radius of sphere is 3.5 cm, then volume can be easily found as the formula of volume of sphere is given by,

Volume = 4πr³/3

⇒ Volume =

⇒ Volume =

⇒ Volume = 179.67

Thus, volume of sphere is 179.67 cm³.

Given: Total surface area of cone = 616 cm²

According to the question,

Slant height of cone = 3 (radius of the base of cone)

Let slant height of cone = l

And let radius of the base of cone = r

Then, l = 3r …(i)

Total surface area of cone is given by,

TSA = πrl + πr²

Substituting TSA = 616 cm²,

616 = πr(3r) + πr² [Using equation (i)]

⇒ 616 = 3πr² + πr²

⇒ 616 = 4πr²

⇒

⇒

⇒

⇒

⇒ r² = 49

⇒ r = √49

⇒ r = 7

Putting r = 7 in equation (i), we get

l = 3 × 7 = 21

Thus, slant height of cone is 21 cm.

Given: A well of cylindrical shape.

Inner diameter = 4.20 m

⇒ Inner radius (r) = 4.20/2 = 2.1 m

Height of well (h) = 10 m

To find inner surface area of the wall, we’ll have to apply curved surface area of cylinder:

CSA = 2πrh

Substituting values in the above equation, we get

⇒ CSA = 132 m²

Plastering of the well will be done on the wall only, so we will be using curved surface area of the well for finding cost of plastering the well.

If cost of plastering 1 m² = Rs. 52

Then cost of plastering 132 m² = 132 × 52

= Rs. 6864

Thus, inner surface area of the well is 132 m² and cost of plastering it is Rs. 6864.

The road roller is cylindrical in shape as described.

Given: length of cylindrical road roller (h) = 2.1 m

Diameter of the road roller = 1.4 m

⇒ Radius of the cylindrical road roller (r) = 1.4 /2 = 0.7 m

The figure is given below:

When the cylindrical road roller will roll on the ground, the only portion of the cylinder that will touch the ground is the curved surface of cylindrical road roller.

Thus, if we find curved surface area of cylinder, we can know area of ground levelled.

Curved surface area of cylinder is given by,

CSA = 2πrh

Where h = height of cylinder when placed vertical or length of the cylinder when placed horizontal

Substituting values in above equation, we get

⇒

⇒ CSA = 9.24

This implies that, area of ground leveled in 1 rotation = 9.24 m²

Then, area of ground leveled in 500 rotations = 9.24 × 500 = 4620 m²

If cost of levelling 7 m² ground = Rs. 7

Then cost of levelling 4620 m² ground = 4620 × 7 = Rs. 32340

Thus, area of the ground that was levelled by the road roller is 4620 m² and cost of levelling it is Rs. 32340.