Math, asked by CheeseCakeMouse, 5 hours ago

find the volume of the following solid figures using the steps provided.use π = 3.14. Express your answer to the nearest hundreds​

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Answers

Answered by Aryan0123
18

Answer :-

Volume of sphere = 904.32 cubic inches

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Step-by-step explanation :-

Analyzing the question,

Given:

  • Radius of sphere = 6 inches
  • Value of π = 3.14

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To find:

Volume of sphere = ?

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Solution:

For finding the volume of sphere we use this formula:

 \blue{ \star} \:  \:  \pink{ \sf{Volume \: of \: sphere =  \dfrac{4}{3} \pi {r}^{3} }} \\  \\

 \implies \sf{Volume =  \dfrac{4}{3} \times 3.14 \times (6)^{3}  } \\  \\

 \implies \sf{Volume =  \dfrac{4}{3} \times 3.14 \times 216 } \\  \\

 \implies \sf{Volume =  \dfrac{4}{ \cancel{3}}  \times 3.14 \times  \cancel{216} \:  \:  ^{72} } \\  \\

 \implies \sf{Volume = 4 \times 3.14 \times 72} \\  \\

 \implies \sf{Volume = 904.32 \: \:  in {}^{3} } \\  \\

\therefore \red{\boxed{\bf{Volume \: of \: sphere = 904.32 \: \: in^{3}}}}

Answered by Anonymous
72

Answer:

\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{ Given:}}}}}}}\end{gathered}

  • ★ Radius of Sphere = 6 inches

\begin{gathered}\end{gathered}

\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{ To Find:}}}}}}}\end{gathered}

  • ★ Volume of Sphere

\begin{gathered}\end{gathered}

\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{ Using Formula:}}}}}}}\end{gathered}

\begin{gathered}\dag{\underline{\boxed{\sf{ \sf{Volume \: of \: sphere = \dfrac{4}{3} \pi{r}^{3}}}}}}\end{gathered}

Where

  • ★ π = 3.14
  • ★ r = radius

\begin{gathered}\end{gathered}

\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{ Diagram:}}}}}}}\end{gathered}

\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\qbezier(-2.3,0)(0,-1)(2.3,0)\qbezier(-2.3,0)(0,1)(2.3,0)\thinlines\qbezier (0,0)(0,0)(0.2,0.3)\qbezier (0.3,0.4)(0.3,0.4)(0.5,0.7)\qbezier (0.6,0.8)(0.6,0.8)(0.8,1.1)\qbezier (0.9,1.2)(0.9,1.2)(1.1,1.5)\qbezier (1.2,1.6)(1.2,1.6)(1.38,1.9)\put(0.2,1){\bf 6in}\end{picture}

  • See this diagram from website Brainly.in.

\begin{gathered}\end{gathered}

\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{ Solution:}}}}}}}\end{gathered}

\begin{gathered}\quad{:\implies{\sf{Volume \: of \: sphere = \bf\dfrac{4}{3} \pi{r}^{3}}}}\end{gathered}

  • Substituting the values

\begin{gathered}\quad{:\implies{\sf{Volume \: of \: sphere = \bf\dfrac{4}{3} \times 3.14 \times {(6)}^{3}}}}\end{gathered}

\begin{gathered}\quad{:\implies{\sf{Volume \: of \: sphere = \bf\dfrac{4}{3} \times 3.14 \times {(6 \times 6 \times 6)}}}}\end{gathered}

\begin{gathered}\quad{:\implies{\sf{Volume \: of \: sphere = \bf\dfrac{4}{3} \times 3.14 \times {(216)}}}}\end{gathered}

\begin{gathered}\quad{:\implies{\sf{Volume \: of \: sphere = \bf\dfrac{4}{\cancel{3}}\times 3.14 \times {\cancel{216}}}}}\end{gathered}

\begin{gathered}\quad{:\implies{\sf{Volume \: of \: sphere = \bf{4}\times 3.14 \times 72}}}\end{gathered}

\begin{gathered}\quad{:\implies{\sf{Volume \: of \: sphere = \bf{904.32 \:  {in}^{3} }}}}\end{gathered}

\begin{gathered}\quad\dag{\underline{\boxed{\sf{\purple{Volume \: of \: sphere ={904.32 \:  {inches}^{3}}}}}}}\end{gathered}

  • Henceforth,The Volume of Sphere is 904.32 inches³.

\begin{gathered}\end{gathered}

\begin{gathered}{\underline{\underline{\maltese{\large{\textsf{\textbf{ Additional Information :}}}}}}}\end{gathered}

Formulas related to SA & Volume :

\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}

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