Question

a solid is in the form of a right circular cylinder with a hemisphere shape at one end and a cone at the other end. their common diameter is 4.2 cm and heights of the cylindrical and conical portion are 12 cm and 7cm. respectively find the volume of the solid toy (π = 22/7)​

Answer 1

Step-by-step explanation:

•SOLUTION:-

$\bf •The \: shape \: of \: the \: toy \: is \: attached \: above⇭$

$\bf➢ Radius \: of \: the \: hemisphere =2.1cm$

$\bf \: ➢ Radius \: of \: cylinder =2.1cm$

$\bf ➢Radius \: of \: the \: base \: of \: the \: cone =2.1 cm$

$\bf➢Height \: of \: the \: cylinder(H)=12cm$

$\bf➢Height \: of \: the \: cone(h)=7cm$

⇨Volume of the given toy =(volume of the hemisphere + volume of the cylinder + volume of the cone)

$\bf•Volume \: of \: hemisphere = \frac{2}{3} \pi r³$

$\bf•Volume \: of \: cylinder= \pi r²H$

$\bf • Volume \: of \: cone= \frac{1}{3} \pi r²h$

$\bf➦Now ,Volume \: of \: the \: toy=$

$\bf \: \: \: \: \: \: \: \: \: \: = \huge( \small \frac{2}{3} \pi r {}^{3} + \pi r {}^{2} H+ \frac{1}{3} \pi r²h \huge) \small cm {}^{3}$

$\: \: \: \: \: \: \: \: \: \: \bf = \huge[ \small \frac{2}{3} \pi×(2.1)³+ \pi×(2.1)²w×12+ \frac{1}{3} \pi×(2.1)² \huge] \small cm³$

$\: \: \: \: \: \: \: \: \: \: \bf = \frac{1}{3} \pi \times (2.1) {}^{2} \big [ \small2×(2.1)+3×12+7 \big ] \small \: cm {}^{3}$

$\: \: \: \: \: \: \: \: \: \: \bf = \huge[ \small \frac{1}{3} \pi×(2.1)²×47.2 \huge] \small \: cm³$

$\: \: \: \: \: \: \: \: \: \: = \bf \huge( \small \frac{1}{ \cancel{3}} × \frac{22}{ \cancel{7}} × \frac{ {\cancel{21 }\: \:\cancel{3 }}}{10} \times \frac{21}{10} \times \frac{472}{10} \huge) \small \: cm {}^{3}$

$\bf \: \: \: \: \: \: \: \: \: \: = 218.064 \: cm {}^{3}$

➠Hence,the volume of the given toy is 218.064 cm³

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Answer 2

Step-by-step explanation:

Diameter of base of conical part=4.2cm

Diameter of base of cylindrical part=4.2cm

Diameter of hemispherical part=4.2m

Therefore radius=2.1cm

Height of conical part, h

1

=7cm

Height of cylindrical part, h

2

=12cm

Volume of toy=volume of cone +volume of cylinder+volume of hemisphere

=

3

1

πr

2

h

1

+πr

2

h

1

+

3

2

πr

3

=πr

2

(

3

h

1

+h

2

+

3

2r

)

=

7

22

×2.1×2.1(

3

7

+12+

3

2×2.1

)

=

7

22

×2.1×2.1×15.73

=218.02cm

3