Math, asked by zeher007, 8 months ago

a solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other end there common diameter is 4.2cm and heights of the cylindrical and conical portions are 12 and 7 cm respectively. find volume

Answers

Answered by solankidivya465
0

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

Answered by Anonymous
1

•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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