So first of all we should know that the rectangular prism is a cuboid.
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Given :-
- heigth = 9 in.
- Length = 5 in.
- Width = 4 in.
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To find:-
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Solution:-
We know:-
[tex] \bigstar \boxed{ \rm volume \: of \: cuboid = length \times width \times height}[/tex]
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So:-
[tex] \dashrightarrow \sf volume \: of \: cuboid = length \times width \times height \\ [/tex]
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[tex] \dashrightarrow \sf volume \: of \: cuboid = 9 \times 5 \times 4 \\ [/tex]
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[tex] \dashrightarrow \sf volume \: of \: cuboid = 45 \times 4 \\ [/tex]
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[tex] \dashrightarrow \bf volume \: of \: cuboid = 180 {in}^{3} \\ [/tex]
Therefore option C is correct .
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know more:-
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[tex]\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}[/tex]
