Question

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, their common diameter is 4.2 cm and height of
the cylindrical and conical portion are 12 cm and 7 cm respectively. Find the vol-
ume of the solid toy. (T1 = 22/7)​

Answer 1

Answer:

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Answer 2

•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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