Answer:
[tex]\qquad\boxed{ \sf{ \:Volume\:of\:wood\: \: = \: 27000 \: {cm}^{3} \: }} \\ \\ [/tex]
Step-by-step explanation:
External dimensions of wooden box are as follow
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:Length \: of \: wooden \: box=60 \: cm \qquad \: \\ \\& \qquad \:\sf \:Breadth \: of \: wooden \: box=45 \: cm\\ \\& \qquad \:\sf \:Height \: of \: wooden \: box=32 \: cm\end{aligned}} \qquad \: \\ \\ [/tex]
Now, as it is given that wood used in making the closed wooden box is 2.5 cm thick.
So, Internal dimensions of wooden box are as follow :
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:Length \: of \: wooden \: box=60 - 5 = 55 \: cm \qquad \: \\ \\& \qquad \:\sf \:Breadth \: of \: wooden \: box=45 - 5 = 40 \: cm\\ \\& \qquad \:\sf \:Height \: of \: wooden \: box=32 - 5 = 27\: cm\end{aligned}} \qquad \: \\ \\ [/tex]
So,
[tex]\sf \: Volume\:of\:wood\:used\:in\:making\:a\:box \\ \\ [/tex]
[tex]\sf \: = \: Outer\:volume- Inner\:volume \\ \\ [/tex]
[tex]\sf \: = \: 60 \times 45 \times 32 - 55 \times 40 \times 27 \\ \\ [/tex]
[tex]\sf \: = \: 86400 - 59400 \\ \\ [/tex]
[tex]\sf \: = \: 27000 \: {cm}^{3} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: Volume\:of\:wood\: \: = \: 27000 \: {cm}^{3} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{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}[/tex]
The volume of wood is 27000
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Verified answer
Answer:
[tex]\qquad\boxed{ \sf{ \:Volume\:of\:wood\: \: = \: 27000 \: {cm}^{3} \: }} \\ \\ [/tex]
Step-by-step explanation:
External dimensions of wooden box are as follow
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:Length \: of \: wooden \: box=60 \: cm \qquad \: \\ \\& \qquad \:\sf \:Breadth \: of \: wooden \: box=45 \: cm\\ \\& \qquad \:\sf \:Height \: of \: wooden \: box=32 \: cm\end{aligned}} \qquad \: \\ \\ [/tex]
Now, as it is given that wood used in making the closed wooden box is 2.5 cm thick.
So, Internal dimensions of wooden box are as follow :
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:Length \: of \: wooden \: box=60 - 5 = 55 \: cm \qquad \: \\ \\& \qquad \:\sf \:Breadth \: of \: wooden \: box=45 - 5 = 40 \: cm\\ \\& \qquad \:\sf \:Height \: of \: wooden \: box=32 - 5 = 27\: cm\end{aligned}} \qquad \: \\ \\ [/tex]
So,
[tex]\sf \: Volume\:of\:wood\:used\:in\:making\:a\:box \\ \\ [/tex]
[tex]\sf \: = \: Outer\:volume- Inner\:volume \\ \\ [/tex]
[tex]\sf \: = \: 60 \times 45 \times 32 - 55 \times 40 \times 27 \\ \\ [/tex]
[tex]\sf \: = \: 86400 - 59400 \\ \\ [/tex]
[tex]\sf \: = \: 27000 \: {cm}^{3} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: Volume\:of\:wood\: \: = \: 27000 \: {cm}^{3} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{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}[/tex]
Answer:
The volume of wood is 27000