Question

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same
radies. The total hecht of the toy is 15.5 cm. Find the total surface area of the toy​

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt A \:toy\: is\: in \:the\: form \:of\: a \:cone\: of\: radius\: 3.5\: cm\\\tt mounted\: on \:a \:hemisphere\: of \:same\: radius. \\\tt The \:total\: height \:of\: the \:toy\: is\: 15.5 \:cm.\\\tt Find\:the \:total\: surface\: area\: of \:the\: toy.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\sf \bf Radius \:of \:the \:cone=3.5\:cm$
• $\sf \bf Radius \:of \:the \:hemisphere=3.5\:cm$
• $\sf \bf Total \:height \:of \:the \:toy=15.5\:cm$

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

• $\sf \bf Total \:Surface \:Area \:of \:the \:toy(TSA)$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

${\pink{\underline {\sf (i)\:Height \:of \:the\:cone}}}$

${\blue {\mathfrak{Radius \:of \:the \:hemisphere \:is\:the \:height \:of \:hemisphere}}}$

${\orange {\boxed {\green{\sf Height\: of\: the \:cone=Total\: height \:of \:the \:toy - Height \:of \:hemisphere}}}}$

${\blue{\mathfrak{ Let \:the \: height \: of \: the \: cone \: be} \: {\mathcal{H}}}}$

${\underbrace {\overbrace {\orange {\bf Substitute \:the \:values}}}}$

$\sf \bf \implies H=15.5-3.5$

$\implies {\orange {\boxed {\boxed{\purple {\sf \bf H=12\:cm}}}}}$

————————————————————————————

${\pink{\underline {\sf (i)\:Slant\:height \:of \:the\:cone}}}$

${\blue {\boxed {\boxed {\boxed {\green {\sf \bf l=\sqrt{{H}^{2}+{r}^{2}}}}}}}}$

• $\sf l=slant \:height \:of \:the \:cone$
• $\sf H=height \:of \:the \:cone$
• $\sf r=radius \:of \:the \:cone$

${\underbrace {\overbrace {\orange {\bf Substitute \:the \:values}}}}$

$\sf \bf \implies l=\sqrt{{(12)}^{2}+{(3.5)}^{2}}$

$\sf \bf \implies l=\sqrt{144+12.25}$

$\sf \bf \implies l=\sqrt{156.25}$

$\implies {\orange {\boxed {\boxed{\purple {\sf \bf H=12.5\:cm}}}}}$

————————————————————————————

${\pink{\underline {\sf (iii)\:Total \:Surface \:Area \:of \:the \:toy}}}$

${\blue {\boxed {\boxed {\boxed {\green {\sf \bf TSA =3\pi {r}^{2}}}}}}}$

• $\sf TSA=Total \:surface \:area \:of \:hemisphere$
• $\sf r=radius \:of \:the \:hemisphere$

${\blue {\boxed {\boxed {\boxed {\green {\sf \bf TSA =\pi r (r+l)}}}}}}$

• $\sf TSA=Total \:surface \:area \:of \:cone$
• $\sf r=radius \:of \:the \:cone$
• $\sf l=slant \:height \:of \:the \:cone$

${\orange {\boxed {\green {\bf\Big(TSA \:of \:toy\Big) =\Big(TSA \:of \:hemisphere\Big) +\Big(TSA \:of \:cone\Big)}}}}$

$\sf \bf \implies TSA \:of \:toy=3\pi {r}^{2}+\pi r(r+l)$

${\underbrace {\overbrace {\orange {\bf Substitute \:the \:values}}}}$

$\sf \bf \implies TSA \:of \:toy=3\times\dfrac{22}{7}\times {\Big(3.5\Big)}^{2}+\dfrac{22}{7}\times 3.5\Big(3.5+12.5\Big)$

$\sf \bf \implies TSA \:of \:toy=3\times\dfrac{22}{7}\times 3.5\times 3.5+\dfrac{22}{7}\times 3.5\Big(16\Big)$

$\sf \bf \implies TSA \:of \:toy=3\times\dfrac{22}{\cancel{7}}\times \cancel{3.5} \times 3.5+\dfrac{22}{7}\times 56$

$\sf \bf \implies TSA \:of \:toy=3\times 22\times 0.5 \times 3.5+\dfrac{22}{\cancel{7}}\times \cancel{56}$

$\sf \bf \implies TSA \:of \:toy=3\times 22\times 0.5 \times 3.5+22\times 8$

$\sf \bf \implies TSA \:of \:toy= 115.5+176$

$\implies {\blue{\boxed {\boxed {\boxed{\purple {\mathfrak{TSA\:of \:toy=291.5\:{cm}^{2}}}}}}}}$

${\underbrace {\red{\underline{\red {\overline {\red{\sf \therefore The\:Total \:Surface \:Area \:of \:toy\:is\:291.5\:{cm}^{2}}}}}}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

_________________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf CSA\:of \:hemisphere = 2 \pi {r}^{2}$

$\sf TSA\:of \:hemisphere = 3 \pi {r}^{2}$

$\sf Volume \:of \:hemisphere =\dfrac{2}{3} \pi {r}^{3}$

$\sf CSA\:of \:cone=\pi r l$

$\sf TSA\:of \:cone=\pi r(r+l)$

$\sf Volume \:of \:cone =\dfrac{1}{3} \pi {r}^{2}h$

Answer 2

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