Question

figure 7.13 shows a toy. it's lower part is a hemisphere and the upper part is a cone. find the volume and the surface area of the toy from the measure shown.

Answer 1

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Answer 2

$ANSWER :$

For the conical part,

Radius(r)=3cm

Height(h)=4cm

Let l be the slant height of the conical part,

$\\ {l}^{2} = {r}^{2} + {h}^{2} \\ \\ {l}^{2} = {3}^{2} + {4}^{2} \\ \\ {l}^{2} = 9 + 16 \\ \\ {l}^{2} = 25 \\ \\ \sqrt{ {l}^{2} } = \sqrt{25}.......(taking \: square \: roots \: on \: both \: sides) \\ \\ l = 5cm$

Therefore,

slant height (l) =5cm

Now,

For hemisphere,

Radius (r) =3cm

$\\ Volume \: of \: the \: toy = Volume \: of \: the \: Cone + Volume \: of \: Hemisphere \\ \\ Volume \: of \: the \: toy = \frac{1}{3} \pi \: {r}^{2} h + \frac{2}{3} \pi \: {r}^{3} \\ \\ Volume \: of \: the \:toy = \frac{1}{3} \pi \: {r}^{2} (h + 2r)\\ \\ Volume \: of \: the \: toy = \frac{1}{3} \times 3.14 \times 3 \times 3 + (4 + 2 \times 3) \\ \\ Volume \: of \: the \: toy = 3.14 \times 3(4 + 6).......(cancelling \: one \: of \: the \: 3 \: by \: \frac{1}{3} \: and \: multiplying \: 2 \: and \: 3 \: in \: bracket) \\ \\ Volume \: of \: the \: toy = 3.14 \times 3 \times 10 \\ \\ Volume \: of \: the \: toy = 3.14 \times 30 \\ \\ Volume \: of \: the \: toy = 94.2 \: {cm}^{3}$

Now,

$Surface \: area \: of \: toy = curved \: Surface \: area \: of \: the \: cone + curved \: Surface \: area \: of \: the \: hemisphere \\ \\ Surface \: area \: of \: the \: toy = \pi \: rl + 2\pi \: {r}^{2} \\ \\ Surface \: area \: of \: the \: toy = \pi \: r \: (l + 2r) \\ \\ Surface \: area \: of \: the \: toy = 3.14 \times 3(5 + 2 \times 3) \\ \\ Surface \: area \: of \: the \: toy = 3.14 \times 3 (5 + 6).......(multiplying \: 2 \: and \: 3) \\ \\ Surface \: area \: of \: the \: toy = 3.14 \times 3 \times 11 \\ \\ Surface \: area \: of \: the \: toy = 3.14 \times 33 \\ \\ Surface \: area \: of \: the \: toy = 103.62 {cm}^{2}$

Therefore,

the Volume of the toy=94.2 cm^3

&

the Surface area of the toy=103.62cm^3.