Question

An 'ice-cream seller used to sell different kinds and different shapes of ice-cream like
rectangular shaped with one end hemispherical, cone-shaped and rectangular brick, etc. One
day a child came to his shop and purchased an ice-cream which has the following shape: ice-
cream cone as the union of a right circular cone and a hemisphere that has the same
(circular) base as the cone. The height of the cone is 9 cm and the radius of its base is 2.5
cm: By reading the above-given information, find the following:
i, The volume of the ice-cream without hemispherical end.
ii. The volume of the ice-cream with a hemispherical end.​

Answer 1

Answer:

1.For that you have to use formula of volume of cone - 1/3 πr²h and you get the answer 19.64 or you can right it 20 cm²

2. For that you have to take volume of cone and volume of hemisphere [ 1/3 πr²h + 2/3πr³]

Step-by-step explanation:

1. volume of cone - 1/3πr²h

= 1/3 × 22/7 × 2.5² × 9

= (22× 6.25) ÷ 7

= 19.64 or 20cm³

2. volume of cone + volume of hemispherical end

= 1/3πr²h + 2/3πr³

= 1/3 × 22/7 × 2.5² × 9 + 2/3 × 22/7 × 2.5³

= 137.5/7 + 687.5/21

= (412.5 + 687.5) ÷ 21

= 1100 ÷ 21

= 52.3 cm³

Answer 2

Answer:

(i) Volume without hemispherical end = 58.92 cm³.

(ii) Volume with hemispherical end = 91.66 cm³.

Step-by-step explanation:

Given, height of cone = 9 cm

           radius of cone = 2.5 cm

(i) The volume of ice-cream without hemispherical end is

   Volume of cone = $\frac{1}{3}\pi r^{2}h$

                               $=\frac{1}{3}*\frac{22}{7}*2.5*2.5*9$

                               $=58.92 cm^{3}$

(ii) The volume of ice-cream with a hemispherical end is

    Required volume = Volume of cone + volume of hemisphere

                                  $=\frac{1}{3}\pi r^{2}h + \frac{2}{3} \pi r^{3}$

                                  $=\frac{1}{3}*\frac{22}{7}*2.5*2.5*9 + \frac{2}{3}*\frac{22}{7}*2.5*2.5*2.5$

                                  $=58.92 + 32.74$

                                  $=91.66cm^{3}$