Chapter 12: Surface Areas and Volumes

We know that,

Volume of cube = a³, where a = side of cube

According to the question,

Side of first cube, a₁ = 3 cm

Side of second cube, a₂ = 4 cm

Side of third cube, a₃ = 5 cm

Let us assume that the side of cube recast from melting these cubes = a

We know that the total volume of the 3 cubes will be the same as the volume of the newly formed cube,

Volume of new cube = (volume of 1^st + 2^nd + 3^rd cube)

⇒ a³ = a₁³ + a₂³ + a₃³

⇒ a³ = (3)³ + (4)³ + (5)³

⇒ a³ = 27 + 64 + 125 = 216

⇒ a = 6 cm

Therefore, side of cube so formed is 6 cm.

Volume of cuboid = lbh, where, l = length, b = breadth and h = height

Cuboidal lead:

Length, l = 9 cm

Breadth, b = 11 cm

Height, h = 12 cm

Volume of lead = 9(11)(12) = 1188 cm³

Volume of sphere = 4/3πr³, where r = radius of sphere

Spherical shots,

Diameter = 3 cm

Radius, r = 1.5 cm

Volume of one shot = 4/3 × 22/7 × (1.5)³ = 99/7 cm³

Hence, the number of bullets that can be made from lead = 84.

According to the question,

The bucket is in the form of frustum of a cone.

We know that,

Volume of frustum of a cone = 1/3 πh(r₁² + r₂² + r₁r₂), where, h = height, r₁ and r₂ are the radii(r₁ > r₂)

For bucket,

Volume of bucket = 28.490 L

1 L = 1000 cm³

Volume of bucket = 28490 cm³

Radius of top, r₁ = 28 cm

Radius of bottom, r₂ = 21 cm

Let the height = h.

Substituting these values in the equation to find the volume of bucket,

We have.

Volume of bucket = 1/3 πh[28² + 21² + 28(21)]

28490 = 1/3 × 22/7 × h (784 + 441 + 588) = 22/7 × h × 1813

⇒ h = (28490 × 21) / (22 × 1813)

⇒ h = 15

According to the question,

Height of cone = OM = 12 cm

The cone is divided from mid-point.

Hence, let the mid-point of cone = P

OP = PM = 6 cm

From △OPD and △OMN

∠POD = ∠POD [Common]

∠OPD = ∠OMN [Both 90°]

Hence, by the Angle-Angle similarity criterion

We have,

△OPD ~ △OMN

And

Similar triangles have corresponding sides in equal ratio,

So, we have,

PD/MN = OP/OM

PD/8 = 6/12

PD = 4cm

For First part i.e. cone

Base Radius, r = PD = 4 cm

Height, h = OP = 6 cm

We know that,

Volume of cone for radius r and height h, V = 1/3 πr²h

Volume of first part = 1/3 π(4)²6 = 32π

For second part, i.e. Frustum

Bottom radius, r₁ = MN = 8 cm

Top Radius, r₂ = PD = 4 cm

Height, h = PM = 6 cm

We know that,

Volume of frustum of a cone = 1/3 πh(r₁² + r₂² + r₁r₂) , where, h = height, r₁ and r₂ are radii, (r₁ > r₂₎

Volume of second part = 1/3 π(6)[8² + 4² + 8(4)]

= 2π(112) = 224π

Therefore, we get the ratio,

Volume of first part : Volume of second part = 32π : 224π = 1 : 7

Let the side of one cube = a

Surfaces area of resulting cuboid = 2(Total surface area of a cube) – 2(area of single surface)

We know that,

Total surface area of cube = 6a² , where a = side of cube

⇒ Surfaces area of resulting cuboid = 2(6a²) – 2(a²) = 10a²

Also,

According to the question,

Volume of cube = 64 cm³

Volume of cube = a³

64 = a³

a = 4 cm

Therefore,

Surface area of resulting cuboid = 10a² = 10(4)² = 160 cm²

From the figure, we get,

Volume of remaining solid = volume of cube – volume of cone

For Cube

Side, a = 7 cm

We know that,

Volume of cube = a³, where a = side of cube

Volume of cube = (7)³ = 343 cm³

For cone

Radius, r = 3 cm

Height, h = 7 cm

Volume of cone = 1/3 πr²h = 1/3 π(3)²7 = 3 × (22/7) × 7 = 66 cm³

Volume of remaining solid = volume of cube – volume of cone

= 343 – 66

= 277 cm³

According to the question,

We get the figure given below,

We know that,

Total surface area of shape formed = Curved area of first cone + Curved surface area of second cone

Since, both cones are identical,

We have,

Total surface area of shape formed = Curved area of first cone + Curved surface area of the second cone

= 2(Surface area of cone)

We also know that,

Surface area of cone = πrl, where r = radius and l = slant height

And, the total Surface area of shape so formed = 2πrl

Given in the question that,

Radius, r = 8 cm

Height, h = 15 cm

Therefore,

Area = Curved area of first cone + Curved surface area of the second cone

= 2 (Surface area of the cone)

= 2 × πrl

= 2 × π × r × √(r² + h²)

= 2 × (22/7) × 8 × √(8² + 15²)

= 50.28 × √289

= 854.85 cm²

= 855 cm² (approx)

Hence, the surface area of shape so formed is 855cm².