Question

If the radii of the circular ends of a conical bucket are 28cm and 7cm ,whose height is 45cm. Find the capacity of the bucket

Answer 1

Answer:

Capacity of the bucket : 48510 cm3.

Given:

➛The radii of the circular ends of a conical bucket are 28cm and 7cm.

➛Height is 45cm .

To Find:

The capacity of the bucket .

Solution:

We are given,

➛The radii of the circular ends of a conical bucket are 28cm and 7cm.

➛Height is 45cm i.e. h=45 cm.

Let, r and R be the radii of top and bottom end respectively.

⛬ r = 28 cm. and R = 7 cm.

We know,

The volume/capacity of the conical bucket is:

${\boxed{\rm{v = \dfrac{1}{3} \pi ( {r}^{2} + rR + {R}^{2} ) \times h}}}$

$v = \frac{1}{3} \pi( {28}^{2} + (28 \times 7) + {7}^{2} ) \times 45$

$v = \frac{1}{ 3} \times \frac{22}{7} \times 1029 \times 45$

$v = 22 \times 147 \times 15$

$v = 48510 {cm}^{3}$

Hence capacity of the bucket : 48510 cm³

Answer 2

Given ,

The radii of the circular ends of a conical bucket are 28 cm and 7 cm

We know that , the volume of frustum of cone is given by

$\sf \large \fbox{Volume = \frac{1}{3} \pi h \{( {R)}^{2} + {(r)}^{2} + Rr) \}}$

Thus ,

$\sf \mapsto Volume = \frac{1}{3} \times \frac{22}{7} \times 45 \{{(28)}^{2} + {(7)}^{2} + 28 \times 7 \} \\ \\ \sf \mapsto Volume = \frac{22 \times 15 \times 1029}{7} \\ \\\sf \mapsto Volume = \frac{339570}{7} \\ \\\sf \mapsto Volume = 48510 \: \: {cm}^{3}$

$\therefore \sf \underline{The \: volume \: or \: capacity \: of \: bucket \: is \: 48510 \: {cm}^{3} }$