Question

important question ☺️​

Answer 1

Answer:

Mark a brain list

Step-by-step explanation:

We have

Edge of cube= 5 cm

Diameter of hemisphere= 4.2 cm

Clearly

Surface area of decorative block

= total surface area of the cube - base area of hemisphere + curved surface area of hemisphere

=(6×(edge)

2

−πr

2

+2πr

2

)cm

2

=(25×6+πr

2

)cm

2

={150+

7

22

×(2.1)

2

}cm

2

=163.86cm

2

Answer 2

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

$\tt The\: figure\: shows\: a \:decorative \:block\: which\: is\\\tt made \:up\: of \:two\: solids \:a \:cube\: and\: a\: hemisphere. \\\tt The\: base\: of \:the\: block \:is \:cube\: with \:edge \:5\:cm\:and\\\tt the\: hemisphere\:, fixed \:on\: top, \:has \:a\: diameter \:of\\\tt 4.2\:cm. \:Find \:the \:total\: surface\: area\: of\: the\: block.$

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

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

• $\bf Edge \:of \:the \:cube(a) = 5\:cm$
• $\bf Diameter \:of \:the \:hemisphere(d) = 4.2\:cm$

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

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

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

${\pink {\underline {\bf {\pmb {Radius \:of \:the \:hemisphere}}}}}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {r=\dfrac{d}{2}}}}}}}}$

• $\sf r=radius \:of \:the \:hemisphere$
• $\sf d=diameter\:of \:the \:hemisphere$

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

$\bf \implies r=\dfrac{4.2}{2}$

$\bf \implies r=\dfrac{\cancel{4.2}}{\cancel{2}}$

$\implies {\blue {\boxed {\boxed {\purple {\sf r=2.1}}}}}$

————————————————————————————

${\pink {\underline {\bf {\pmb {Total \:Surface \:Area \:of \:the\:block}}}}}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {TSA\:of\:cube=6{a}^{2}}}}}}}}$

• $\sf TSA=Total \:surface \:area \:of \:cube$
• $\sf a=edge \:of \:the \:cube$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {CSA\:of \:hemisphere=2 \pi{r}^{2}}}}}}}}$

• $\sf CSA=Curved\:surface \:area \:of \:hemisphere$
• $\sf r=radius \:of \:the \:hemisphere$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {Base\:area\:of \:hemisphere= \pi{r}^{2}}}}}}}}$

• $\sf r=radius \:of \:the \:hemisphere$

————————————————————————————

${\pmb {\orange {\boxed {\green {\Big(TSA\:of \:block\Big) =\Big(TSA\:of\:cube\Big) +\Big(CSA\:of \:hemisphere\Big)-\Big(Base\:area\:of \:hemisphere\Big)}}}}}$

${\boxed{\sf TSA\:of\:block =6{a}^{2}+2\pi{r}^{2}-\pi{r}^{2}}}$

$\bf \implies TSA\:of\:block =6{a}^{2}+\pi{r}^{2}$

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

$\bf \implies TSA\:of\:block =6{\big(5\big)}^{2}+\dfrac{22}{7}\times {\big(2.1\big)}^{2}$

$\bf \implies TSA\:of\:block =6\times 25+\dfrac{22}{7}\times 2.1\times 2.1$

$\bf \implies TSA\:of\:block= 150+\dfrac{22}{\cancel{7}}\times \cancel{2.1}\times 2.1$

$\bf \implies TSA\:of\:block= 150+22\times 0.3\times 2.1$

$\bf \implies TSA\:of\:block= 150+13.86$

$\implies {\blue {\boxed {\boxed {\purple {\mathfrak {TSA\:of\:block= 163.86\:{cm}^{2}}}}}}}$

${\underbrace {\red {\overline {\red {\underline {\red {\sf {\pmb {{\therefore}The\:Total \:Surface \:Area \:of \:the \:block \:is\:163.86\:{cm}^{2}}}}}}}}}}$

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

____________________________________________

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

$\sf Lateral \:surface \:area \:of \:cube =4{a}^{2}$

$\sf Total \:surface \:area \:of \:cube =6{a}^{2}$

$\sf Volume\:of \:cube ={a}^{3}$

$\sf Curved \:surface \:area \:of \:hemisphere =2\pi{r}^{2}$

$\sf Total \:surface \:area \:of \:hemisphere =3\pi{r}^{2}$

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