CBSE 10th Maths Chapter 13 - Surface Areas and Volumes-Exercise-13.4

Solution:

Radius ($r_1$) of the upper base = $\dfrac{4}{2}$ = 2 cm

Radius ($r_2$) of lower the base = $\dfrac{2}{2}$ = 1 cm

Height = 14 cm

Now, the capacity of glass = Volume of the frustum of the cone

So, Capacity of glass = ($\dfrac{1}{3}$)×π×h($r_1^{2}+r_2^{2}+r_1r_2$)

= ( $\dfrac{1}{3}$)×π×(14)($2^{2}+1^{2}$+ (2)(1))

∴ The capacity of the glass = 102×($\dfrac{2}{3}) cm^{3}$

Solution:

Given,

Slant height (l) = 4 cm

Circumference of upper circular end of the frustum = 18 cm

∴ 2π$r_1$ = 18

Or, $r_1$ = 9/π

Similarly, the circumference of the lower end of the frustum = 6 cm

∴ 2π$r_2$ = 6

Or, $r_2$ = 3/π

Now, the surface area of the frustum = π($r_1+r_2$) × l

= π(9/π+3/π) × 4

= 12×4 = 48 cm$^{2}$

Solution:

Given,

For the lower circular end, radius ($r_1$) = 10 cm

For the upper circular end, radius ($r_2$) = 4 cm

Slant height (l) of frustum = 15 cm

Now,

The area of material to be used for making the fez = CSA of frustum + Area of the upper circular end

CSA of frustum = π($r_1+r_2$)×l

= 210π

And, the Area of the upper circular end = π$r_2^{2}$

= 16π

The area of material to be used for making the fez = 210π + 16π = $\dfrac{(226 x 22)}{7}$ = 710 $\dfrac{2}{7}$

∴ The area of material used = 710 $\dfrac{2}{7}$ cm^{2}

Solution:

Given,

$r_1$ = 20 cm,

$r_2$ = 8 cm and

h = 16 cm

∴ Volume of the frustum = ($\dfrac{1}{3}$)×π×h($r_1^{2}+r_2^{2}+r_1r_2$)

= ($\dfrac{1}{3}$)×3.14 ×16($(20)^{2}+(8)^2+(20)(8)$)

= ($\dfrac{1}{3}$) ×3.14 ×16(400 + 64 + 160) = 10449.92 cm$^{3}$ = 10.45 lit

It is given that the rate of milk = Rs. 20/litre

So, the cost of milk = 20×volume of the frustum

= 20×10.45

= Rs. 209

Now, the slant height will be

l=$\sqrt{h^[2}+(r_1-r_2)^{2}}$ =$\sqrt{16^{2}+(20-8)^{2}}$ =$\sqrt{16^{2}+12^{2}}$

l = 20 cm

So, CSA of the container = π$(r_1+r_2)$×l

=$\dfrac{314}{100}(20+8) \times 20cm^{2}$

= 1758.4 cm$^{2}$

Hence, the total metal that would be required to make the container will be = 1758.4 + (Area of the bottom circle)

= 1758.4+πr$^{2}$ = 1758.4+π(8)$^{2}$

= 1758.4+201 = 1959.4 cm$^{2}$

∴ Total cost of metal = Rs. ($\dfrac{8}{100}$) × 1959.4 = Rs. 157

Solution:

The diagram will be as follows

Consider AEG

$\dfrac{EG}{AG}$=$tan30^{\circ}$ EG=$\dfrac{10}{\sqrt{3}}cm$ =$\dfrac{10\sqrt{3}}{3}$ In$ \triangleABD,$ $\dfrac{BD}{AD}$=$tan30^{\circ}$ BD=$\dfrac{20}{\sqrt{3}}$ =$\dfrac{20\sqrt{3}}{3}cm$

Radius ($r_1$) of upper end of frustum = $\dfrac{(10\sqrt{3})}{3}$ cm

Radius ($r_2$) of lower end of container = $\dfrac{(20\sqrt{3})}{3}$ cm

Height ($r_3$) of container = 10 cm

Now,

$\therefore$ Volume of the frustum = ($\dfrac{1}{3}$)×π×h($r_1^{2}+r_2^{2}+r_1r_2$)

=$\dfrac{1}{3}$xπx10[($\dfrac{(10\sqrt{3})}{3})^{2}+(\dfrac{(20\sqrt{3})}{3})^{2}+\dfrac{(10\sqrt{3})(20\sqrt{3})}{3 \times 3}$]

Solving this, we get

Volume of the frustum = $\dfrac{22000}{9} cm^{3}$

The radius (r) of wire = ($\dfrac{1}{16}$)×($\dfrac{1}{2}$) = $\dfrac{1}{32} $cm

Now,

Let the length of the wire be “l”.

The volume of wire = Area of cross-section x Length

= (π$r^{2}$)xl

= π($\dfrac{1}{32})^{2}$x l

Now, Volume of frustum = Volume of wire

$\dfrac{22000}{9}$ = $(\dfrac{22}{7}$)x($\dfrac{1}{32})^{2}$x l

Solving this, we get,

l = 7964.44 m