Question

A toy is in the shape of a cone mounted on a hemisphere of the same base radius if the volume of a toy is 231 cm cube and its diameter is 7 cm find the height of the toy

Answer 1

⏹It's given that:-

✔️Vol of toy=231 cm³

✔️Diameter of hemisphere= 7 cm

Remember :-

⭕️cone and hemisphere have equal radius because they are laid exactly on top of each other.....................................(1)

so from (1) we can say that

✔️radius of hemisphere = radius of cone = 3.5 cm

✔️Height of hemisphere= radius of hemisphere= 3.5 cm

⭕️Let H be height [TOY]

⭕️Let h be height [CONE]

therefore...

✔️H= height of cone+ height of hemisphere

H = h + r

H = h + 3.5

✔️volume of toy = vol of cone + vol of hemisphere

✔️Vol of toy
= 1/3πr²h + 2/3πr³
=πr²/3(h+2r)

=(22/7) × (3.5)² × (1/3) (h+2 × 3.5)=231

=( 22 × 3.5 × 3.5) / 7 (h+7)= 231 × 3

(231 × 3 × 7) / (3.5× 3.5 × 22) = (h+7)

= (3×3)/(0.5×0.5×2)=h+7

= 900 /50=h+7
= 90/5
= 18

=18=h+7

h=18-7

h= 11 cm

✔️Height of toy = h+r

= 11+3.5

= 14.5

⭕️So height of the toy = 14.5 cm⭕️

Answer 2

SOLUTION

A toy is in the shape of a cone mounted on a hemisphere of same base radius.

Volume of toy= 231cm³

Base diameter of toy= 7cm

Base radius of toy= 7/2cm= 3.5cm

Volume of hemisphere= 2/3πr³

$= > \frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 {cm}^{3} \\ = > 89.8 3 {cm}^{3}$

Volume of cone

=)Volume of hemisphere-Volume of toy

=) 231- 89.83cm³

=) 141.17cm³

Volume of cone is given by 1/3πr²h

(where h is the height of the cone)

$= > \frac{1}{3} \times \pi {r}^{2} h = 141.17 {cm}^{3} \\ = > \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times h = 141.17 {cm}^{3} \\ = > h = 141.17 \times 3 \times \frac{7}{22} \times \frac{1}{3.5} \times \frac{1}{3.5} \\ = > h = 11cm$

Height of the cone= 11cm

Height of toy,

=)Height of cone +Height of hemisphere

=) 11cm + 3.5cm

=) 14.5cm

[Height of hemisphere=Radius of hemisphere]

Height of toy is 14.5cm.

Hope it helps ☺️