The width of a box is two-third of its length and height is one-third of its length. If the volume of the box is 384 cubic meter, find the dimensions of the box.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
The width of a box is two-third of its length and height is one-third of its length.
Let assume that
[tex]\sf \: Length\:of\:the\:box = x \: m \\ \\ [/tex]
So,
[tex]\sf \: Breadth\:of\:the\:box = \frac{2}{3} x \: m \\ \\ [/tex]
and
[tex]\sf \: Height\:of\:the\:box = \frac{1}{3} x \: m \\ \\ [/tex]
Further, given that the volume of the box is 384 cubic meter.
The width of a box is two-third of its length and height is one-third of its length. If the volume of the box is 384 cubic meter, find the dimensions of the box.
❒ Information provide tous ❒
➪ The width of a box is two-third of its length and height is one-third of its length.
➪ Volume of the box is 384 cubic metre.
❒ Assume ❒
[tex]\begin{gathered}\rm \: Length\:of\:the\:box = x \: m \\ \\ \end{gathered} [/tex]
Let'sSolve :
[tex]\begin{gathered}\rm \: Breadth\:of\:the\:box = \frac{2}{3} x \: m \\ \\ \end{gathered} [/tex]
And,
[tex]\begin{gathered}\ \ \ \: \rm Height\ \rm \ of\:the\:box = \frac{1}{3} x \: m \\ \\ \end{gathered} [/tex]
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Verified answer
Question :-
The width of a box is two-third of its length and height is one-third of its length. If the volume of the box is 384 cubic meter, find the dimensions of the box.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
The width of a box is two-third of its length and height is one-third of its length.
Let assume that
[tex]\sf \: Length\:of\:the\:box = x \: m \\ \\ [/tex]
So,
[tex]\sf \: Breadth\:of\:the\:box = \frac{2}{3} x \: m \\ \\ [/tex]
and
[tex]\sf \: Height\:of\:the\:box = \frac{1}{3} x \: m \\ \\ [/tex]
Further, given that the volume of the box is 384 cubic meter.
We know,
[tex]\sf \: Volume\:of\:the\:box = Length \times Breadth \times Height \\ \\ [/tex]
[tex]\sf \: x \times \dfrac{2x}{3} \times \dfrac{x}{3} = 384 \\ \\ [/tex]
[tex]\sf \: {x}^{3} = 384 \times \dfrac{9}{2} \\ \\ [/tex]
[tex]\sf \: {x}^{3} = 192 \times 9 \\ \\ [/tex]
[tex]\sf \: {x}^{3} = 4 \times 4 \times 4 \times 3 \times 3 \times 3 \\ \\ [/tex]
[tex]\sf \: {x}^{3} = {4}^{3} \times {3}^{3} \\ \\ [/tex]
[tex]\sf \: {x}^{3} = {12}^{3} \\ \\ [/tex]
[tex]\bf\implies \:x = 12 \: \\ \\ [/tex]
Hence,
[tex]\bf \: Length\:of\:the\:box = 12 \: m \\ \\ [/tex]
[tex]\bf \: Breadth\:of\:the\:box = \frac{2}{3} \times 12 = 8 \: m \\ \\ [/tex]
[tex]\bf \: Height\:of\:the\:box = \frac{1}{3} \times 12 = 4\: m \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \mathfrak{Additional\:Information}}}[/tex]
[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]
♣ Question :
The width of a box is two-third of its length and height is one-third of its length. If the volume of the box is 384 cubic meter, find the dimensions of the box.
❒ Information provide to us ❒
➪ The width of a box is two-third of its length and height is one-third of its length.
➪ Volume of the box is 384 cubic metre.
❒ Assume ❒
[tex]\begin{gathered}\rm \: Length\:of\:the\:box = x \: m \\ \\ \end{gathered} [/tex]
Let's Solve :
[tex]\begin{gathered}\rm \: Breadth\:of\:the\:box = \frac{2}{3} x \: m \\ \\ \end{gathered} [/tex]
And,
[tex]\begin{gathered}\ \ \ \: \rm Height\ \rm \ of\:the\:box = \frac{1}{3} x \: m \\ \\ \end{gathered} [/tex]
❒ Also given that ❒
The volume of the box is 384 cubic meter.
➪ We know :-
[tex] \colorbox{red}{ \boxed{\sf \: Volume\:of\:the\:box = Length \times Breadth \times Height }}[/tex]
[tex]\begin{gathered}\tt \: \: x \times \dfrac{2x}{3} \times \dfrac{x}{3} = 384 \\ \\ \end{gathered}[/tex]
[tex]\begin{gathered}\tt \: {x}^{3} = 384 \times \dfrac{9}{2} \\ \\ \end{gathered} [/tex]
[tex]\begin{gathered}\sf \: {x}^{3} = 192 \times 9 \\ \\ \end{gathered} [/tex]
[tex]\begin{gathered}\sf \: {x}^{3} = 4 \times 4 \times 4 \times 3 \times 3 \times 3 \\ \\ \end{gathered} [/tex]
[tex]\begin{gathered}\sf \: {x}^{3} = {4}^{3} \times {3}^{3} \\ \\ \end{gathered} [/tex]
[tex]\begin{gathered}\sf \: {x}^{3} = {12}^{3} \\ \\ \end{gathered}[/tex]
[tex]\bold {\implies x = 12 }[/tex]
Hence ,
[tex]➪ \: { \color {cyan}{\bf \: Length\:of\:the\:box = 12 \: m } }[/tex]
[tex]➪ \: { \color{pink}{\bf \: Breadth\:of\:the\:box = \frac{2}{3} \times 12 = 8 \: m }}[/tex]
[tex]➪ \: { \color{purple}{{ \bf \: Height\:of\:the\:box = \frac{1}{3} \times 12 = 4\: m }}}[/tex]
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[tex] \colorbox{red}{Answered \: by \green{★} 9218 \ {★}}[/tex]