Question

if a cone of height 8m has a curved surface area 188.4m^2 find the radius of its base and its volume also​

Answer 1

★$\large\bold{\underline{\sf{\pink{Question :-}}}}$

• If a cone of height 8m has a curved surface area 188.4m^2. Find the radius of its base and its volume also.

★$\large\bold{\underline{\sf{\red{Answer :-}}}}$

• Radius of cone is 6m and volume is 301.44cm³.

★$\large\bold{\underline{\sf{\blue{To\:Find :-}}}}$

• Radius(of the base) and volume of Cone.

★$\large\bold{\underline{\sf{\green{Given :-}}}}$

• Height of the Cone = 8m
• Curved Surface Area (CSA) = 188.4m²

★$\large\bold{\underline{\sf{\orange{Explanation :-}}}}$

$\\ { \sf {\: Let \: r \: is \: the \: radius \: and \: l \: is \: the}} \\ {\sf{\: slant \: height \: of \: the \: cone.}}$

${ \underline{ \underline{ \pink{ \sf{According \: to \: the \: question,}}}}}$

${ \underline{ \sf{Curved \: surface \: area \: of \: the \: cone = 188.4}}} \\ \\ \sf\implies\pi rl = 188.4 \\ \\ \sf\implies \pi r \sqrt{( {r }^{2} + {h}^{2})} = 188.4$

${ \small{\boxed{ \red{ \sf{since \: slant \: height \: l =\sqrt{( {r }^{2} + {h}^{2})}}}}}}$

$\sf \implies \: 3.14 \times r \sqrt{ {r }^{2} + {8}^{2} } = 188.4 \\ \\ \sf\implies \: r\sqrt{ {r }^{2} + 64 } = \dfrac{188.4}{3.14} \\ \\ \sf\implies \: r\sqrt{ {r }^{2} + 64 } = 60 \\ \\ \sf\implies \: {r}^{2} \times ( {r}^{2} + 64) = {60}^{2} \\ \\ \sf\implies {r}^{4} + 64 {r}^{2} = 3600 \\ \\ \sf\implies{r}^{4} + 64 {r}^{2} - 3600 = 0 \\ \\ \sf\implies( {r}^{2} - 36) \times ( {r}^{2} + 100) = 0 \\ \\ \sf\implies {r}^{2} = 36 \: and - 100$

${ \sf {\small{ \pink{Since , \: square \: of \: a \: number \: can \: not \: be \: negative.}}}}$

$\sf \: Hence, {r}^{2} - 100 \: is \: not \: possible$

$\\ \sf \: So, {r}^{2} = 36 \\ \\\sf\implies\: r=\pm6$

${ \pink{ \small{\sf {\: Since, \: radius \: can \: not \: be \: negative.}}}}$

$\\ \sf \: So, r = 6m$

${\boxed{ \red {\sf{Now \: volume \: of \: the \: cone = \dfrac{1}{3}\pi {r}^{2}h}}}}$

$\sf\longmapsto \cancel\dfrac{3.41 \times 36 \times 8}{3} \\ \\ \sf\longmapsto3.41 \times 12 \times 8 \\ \\ { \boxed{ \blue{\longmapsto{ \sf{301.44 {m}^{3} }}}}}$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{Question}}}}}}}$

• If a cone of height 8m has a curved surface area 188.4m^2 find the radius of its base and its volume also

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{Solution}}}}}}}$

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

• Height of the cone = h = 8m
• curved surface area = CSA = 188.4m^2

$\sf{\bold{\green{\underline{\underline{To\:find}}}}}$

• radius of the base
• volume of the cone

$\sf{\bold{\pink{\underline{\underline{solution}}}}}$

$\sf{\bold{\orange{\boxed{\boxed{CSA = \pi r l }}}}}$--------(i)

$\sf{\bold{\orange{\boxed{\boxed{l = \sqrt{(r^2 + h^2}}}}}}$--------(ii)

• from eq (i) and (ii)

$\sf{\bold{\orange{\boxed{\boxed{CSA =\pi r \sqrt{r^2 + h^2}}}}}}$

$\sf \longrightarrow\: 3.14 \times r \sqrt{ {r }^{2} + {8}^{2} } = 188.4$

$\sf\longrightarrow\: r\sqrt{ {r }^{2} + 64 } = \dfrac{188.4}{3.14}$

$\sf\longrightarrow \: r\sqrt{ {r }^{2} + 64 } = 60$

$\longrightarrow \: {r}^{2} \times ( {r}^{2} + 64) = {60}^{2}$

$\sf\longrightarrow{r}^{4} + 64 {r}^{2} = 3600$

$\sf\longrightarrow{r}^{4} + 64 {r}^{2} - 3600 = 0$

• middle term split

$\sf\longrightarrow r^4 + 100r^2 - 36r^2 - 3600 = 0$

$\sf\longrightarrow( {r}^{2} - 36) ( {r}^{2} + 100) = 0$

$\sf\longrightarrow {r}^{2} = 36 | r^2 = - 100$

$\sf r = \sqrt {36}$ | $\sf r= -\sqrt 100$

$\sf r = 6$ | $\sf r = - 10$

• radius cannot be negetive

therefore radius = 6 m

$\sf{\bold{\green{\boxed{\boxed{\dag radius = 6m }}}}}$

$\sf{\bold{\orange{\boxed{\boxed{volume = \dfrac{1}{3} \pi r^2 h}}}}}$

$\sf\longrightarrow volume = \dfrac{3.41 \times\cancel 36 \times 8}{\cancel 3}$

$\sf\longrightarrow volume = 3.41 \times 12 \times 8$

$\sf\longrightarrow volume = 301.44cm^2$

$\sf{\bold{\green{\boxed{\boxed{\dag volume = 301.44cm^2}}}}}$

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