Chapter 12: Surface Areas and Volumes

Given that,

Diameter of cylinder = 10 cm

So, the radius of the cylinder (r) = 10/2 cm = 5 cm

∴ Length of wire in completely one round = 2πr = 3.14×5 cm = 31.4 cm

It is given that diameter of wire = 3 mm = 3/10 cm

∴ The thickness of the cylinder covered in one round = 3/10 m

Hence, the number of turns (rounds) of the wire to cover 12 cm will be

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds

40 x 31.4 cm = 1256 cm

Radius of the wire = 0.3/2 = 0.15 cm

The volume of wire = Area of the cross-section of wire × Length of wire

= π(0.15)²×1257.14

= 88.898 cm³

We know,

Mass = Volume × Density

= 88.898×8.88

= 789.41 gm

Draw the diagram as follows:

Let us consider the ABA

Here,

AS = 3 cm, AC = 4 cm

So, Hypotenuse BC = 5 cm

We have got 2 cones on the same base AA’ where the radius = DA or DA’

Now, AD/CA = AB/CB

By putting the value of CA, AB and CB, we get,

AD = 2/5 cm

We also know,

DB/AB = AB/CB

So, DB = 9/5 cm

As, CD = BC-DB,

CD = 16/5 cm

Now, the volume of the double cone will be

Solving this, we get

V = 30.14 cm³

The surface area of the double cone will be

= 52.75 cm²

Given that the dimension of the cistern = 150 × 120 × 110

So, volume = 1980000 cm³

Volume to be filled in cistern = 1980000 – 129600

= 1850400 cm³

Now, let the number of bricks placed to be “n”

So, the volume of n bricks will be = n×22.5×7.5×6.5

Now, as each brick absorbs one-seventeenth of its volume, the volume will be

= n/(17)×(22.5×7.5×6.5)

For the condition given in the question,

The volume of n bricks has to be equal to the volume absorbed by n bricks + the volume to be filled in the cistern

Or, n×22.5×7.5×6.5 = 1850400+n/(17)×(22.5×7.5×6.5)

Solving this, we get

n = 1792.41

From the question, it is clear that

Total volume of 3 rivers = 3×[(Surface area of a river)×Depth]

Given,

Surface area of a river = [1072×(75/1000)] km

And,

Depth = (3/1000) km

Now, volume of 3 rivers = 3×[1072×(75/1000)]×(3/1000)

= 0.7236 km³

Now, the volume of rainfall = total surface area × total height of rain

= 0.7280 km³

For the total rainfall to be approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to the volume of 3 rivers.

But, 0.7280 km³ = 0.7236 km³

So, the question statement is true.

Given,

Diameter of the upper circular end of the frustum part = 18 cm

So, radius (r₁) = 9 cm

Now, the radius of the lower circular end of the frustum (r₂) will be equal to the radius of the circular end of the cylinder

So, r₂ = 8/2 = 4 cm

Now, height (h₁) of the frustum section = 22 – 10 = 12 cm

And,

Height (h₂) of cylindrical section = 10 cm (given)

Now, the slant height will be-

Or, l = 13 cm

Area of tin sheet required = CSA of frustum part + CSA of the cylindrical part

= π(r₁+r₂)l+2πr₂h₂

Solving this, we get

Area of tin sheet required = 782 4/7 cm²

Consider the diagram

Let ABC be a cone. From the cone, the frustum DECB is cut by a plane parallel to its base. Here, r₁ and r₂ are the radii of the frustum ends of the cone, and h is the frustum height.

Now, consider the ΔABG and ΔADF,

Here, DF||BG

So, ΔABG ~ ΔADF

Now, by rearranging, we get

The total surface area of the frustum will be equal to the total CSA of the frustum + the area of the upper circular end + the area of the lower circular end

= π(r₁+r₂)l+πr₂²+πr₁²

∴ Surface area of frustum = π[(r₁+r₂)l+r₁²+r₂²]

Consider the same diagram as the previous question.

Now, approach the question in the same way as the previous one and prove that

ΔABG ~ ΔADF

Again,

Now, rearrange them in terms of h and h₁

The total volume of the frustum of the cone will be = Volume of cone ABC – Volume of cone ADE

= (⅓)πr₁²h₁ -(⅓)πr₂²(h₁ – h)

= (π/3)[r₁²h₁-r₂²(h₁ – h)]

Now, solving this, we get

∴ The volume of frustum of the cone = (⅓)πh(r₁²+r₂²+r₁r₂)