Question

The perimeter of a room is 44 m. If hight and length is 12 m & 6m respectively .find the valume​

Answer 1

Given:-

• Perimeter of a Cuboid room is 44 m.
• Height of a Cuboid room is 12 m.
• Length of a Cuboid room is 6 m.

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To find:-

• Volume of the Cuboid room.

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Solution:-

Firstly,

• we have to find out the breadth.

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☆Formula used:-

${\dag}\:{\underline{\boxed{\sf{\purple{Perimeter_{(rectangle)} = 2(length + breadth)}}}}}$

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$\tt\longmapsto{44 = 2(l + b)}$

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$\tt\longmapsto{44 = 2(6 + b)}$

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$\tt\longmapsto{44 = 12 + 2b}$

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$\tt\longmapsto{2b = 44 - 12}$

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$\tt\longmapsto{2b = 32}$

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$\tt\longmapsto{b = \dfrac{32}{2}}$

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$\tt\longmapsto{b = 16\: m}$

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Hence,

• the breadth of the Cuboid room is 16 m.

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Then,

• finding the Volume of the Cuboid room.

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☆Formula used:-

${\dag}\:{\underline{\boxed{\sf{\purple{Volume_{(Cuboid)} = (length \times breadth \times height)}}}}}$

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$\tt\longmapsto{Volume_{(Cuboid)} = (l \times b \times h)}$

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$\tt\longmapsto{Volume_{(Cuboid)} = (6 \times 16 \times 12)}$

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$\tt\longmapsto{Volume_{(Cuboid)} = 96 \times 12}$

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$\sf\longmapsto{\boxed{\red{Volume_{(Cuboid)} = 1152\: m^3}}}$

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Hence,

• the Volume of Cuboid room is 1152 m³.

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★Formulas related to SA and 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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★More to know :-

$\sf{Area\;of\;Rectangle\;=\;Length\;\times\;Breadth}$

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$\sf{Area\;of\;Square\;=\;(Side)^{2}}$

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$\sf{Area\;of\;Triangle\;=\;\dfrac{1}{2}\;\times\;Base\;\times\;Height}$

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$\sf{Area\;of\;Parallelogram\;=\;Base\;\times\;Height}$

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$\sf{Area\;of\;Circle\;=\;\pi r^{2}}$

$\sf{Perimeter\;of\;Rectangle\;=\;2\;\times\;(Length\;+\;Breadth)}$

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$\sf{Perimeter\;of\;Rectangle\;=\;4\;\times\;(Side)}$

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$\sf{Perimeter\;of\;Circle\;=\;2\pi r}$