Question

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Answer 1

Answer:

Given Question :-

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of the cube and the toy. Also find the total surface area of the toy.

$\huge \mathbb \pink{ANSWER}$

GIVEN :-

A solid toy is in the form of a hemisphere surmounted by a right circular cone.

The height of the cone is 4 cm.

The diameter of the base is 8cm.

TO FIND :-

The volume of the toy.

If a cube circumscribes the toy, then find the difference of the volumes of the cube and the toy.

The total surface area of the toy.

Formula Used :-

$\longmapsto \boxed{ \purple{ \bf \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2}h }}$

$\longmapsto \boxed{ \purple{ \bf \:CSA_{(cone)} = \pi \: rl }}$

$\longmapsto \boxed{ \purple{ \bf \: Volume_{(hemisphere)} = \dfrac{2}{3} \pi \: {r}^{3} }}$

$\longmapsto \boxed{ \purple{ \bf \: CSA_{(hemisphere)} = 2\pi \: {r}^{2} }}$

$\longmapsto \boxed{ \purple{ \bf \:Volume_{(cube)} = {(side)}^{3} }}$

where,

r = radius

h = height of cone

l = slant height of cone

CALCULATION :-

1. VOLUME OF TOY

Given that

Radius of cone, r = 4 cm

Height of cone, h = 4 cm

Radius of hemisphere, r = 4 cm

So,

$\pink{\rm :\implies\:Volume_{(toy)} = Volume_{(cone)} + Volume_{(hemisphere)}}$

$\rm :\implies\:Volume_{(toy)} = \dfrac{1}{3} \pi \: {r}^{2}h + \dfrac{2}{3} \pi \: {r}^{3}$

$\rm :\implies\:Volume_{(toy)} = \dfrac{1}{3} \pi \: {r}^{2}(h + 2r)$

$\rm :\implies\:Volume_{(toy)} = \dfrac{1}{3} \times \dfrac{22}{7} \times {4}^{2} \times (4 + 8)$

$\rm :\implies\:Volume_{(toy)} = \dfrac{1}{ \cancel3} \times \dfrac{22}{7} \times 16 \times \cancel{12} \: \: \: ^{4}$

$\rm :\implies\: \boxed{ \red{ \bf \: Volume_{(toy)} = \dfrac{1408}{7} \: {cm}^{3} }}$

2. SURFACE AREA OF TOY

Given that

Radius of cone, r = 4 cm

Height of cone, h = 4 cm

Radius of hemisphere, r = 4 cm

So,

Slant (l) height of cone is given by

$\longmapsto \boxed{ \green{ \bf \: {l}^{2} = {h}^{2} + {r}^{2} }}$

$\rm :\implies\: {l}^{2} = {4}^{2} + {4}^{2}$

$\rm :\implies\: {l}^{2} = 16 + 16$

$\rm :\implies\: {l}^{2} = 32$

$\rm :\implies\:l = \sqrt{32} = 4 \sqrt{2} \: cm$

$\rm :\implies\:l \: = \: 4 \times 1.414$

$\rm :\implies\:l \: = \: 5.656 \: cm$

Now,

Surface Area of toy is given by

$\rm :\implies\: \pink{ \bf \: Surface Area_{(toy)} =CSA_{(cone)} + CSA_{(hemisphere)} }$

$\rm :\implies\:Surface Area_{(toy)} = \pi \: rl \: + \: 2\pi \: {r}^{2}$

$\rm :\implies\:Surface Area_{(toy)} = \pi \: r(l \: + \: 2r)$

$\rm :\implies\:Surface Area_{(toy)} = \dfrac{22}{7} \times 4 \times (5.656 + 8)$

$\rm :\implies\:Surface Area_{(toy)} = \dfrac{22}{7} \times 4 \times 13.656$

$\longmapsto \boxed{ \green{ \bf \:Surface Area_{(toy)} = 171.68 \: {cm}^{2} }}$

Now,

According to statement,

The toy is circumscribes by a cube.

So,

Edge of the cube = height of cone + height of hemisphere

So,

Edge of cube, x = 4 + 4 = 8 cm

$\rm :\implies\: \pink{ \bf \: Volume_{(cube)} = {x}^{3} }$

$\rm :\implies\:Volume_{(cube)} = {8}^{3}$

$\longmapsto \boxed{ \green{ \bf \: Volume_{(cube)} = 512 \: {cm}^{3} }}$

Now,

Difference in the volume of cube and volume of toy is

$\bull\purple{ \bf \: Volume_{(difference)} = Volume_{(cube)} - Volume_{(toy)}}$

$\rm :\implies\:Volume_{(difference)} =512 - \dfrac{1408}{7}$

$\rm :\implies\:Volume_{(difference)} =\dfrac{3584 - 1408}{7}$

$\rm :\implies\:Volume_{(difference)} =\dfrac{2176}{7}$

$\longmapsto \boxed{ \green{ \bf \:Volume_{(difference)} =310.86 \: {cm}^{3} }}$

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Answer 2

Answer:

Given Question :-

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of the cube and the toy. Also find the total surface area of the toy.

\huge \mathbb \pink{ANSWER}ANSWER

GIVEN :-

A solid toy is in the form of a hemisphere surmounted by a right circular cone.

The height of the cone is 4 cm.

The diameter of the base is 8cm.

TO FIND :-

The volume of the toy.

If a cube circumscribes the toy, then find the difference of the volumes of the cube and the toy.

The total surface area of the toy.

Formula Used :-

\longmapsto \boxed{ \purple{ \bf \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2}h }}⟼

Volume

(cone)

=

3

1

πr

2

h

\longmapsto \boxed{ \purple{ \bf \:CSA_{(cone)} = \pi \: rl }}⟼

CSA

(cone)

=πrl

\longmapsto \boxed{ \purple{ \bf \: Volume_{(hemisphere)} = \dfrac{2}{3} \pi \: {r}^{3} }}⟼

Volume

(hemisphere)

=

3

2

πr

3