Question

The dimensions of the a metallic cuboidal box are 100cm × 80cm × 64 cm . It is melted and recast into a cube . Find the surface area of the cube

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Answer 1

Given :

• Dimensions of a metallic cuboidal box is 100 cm × 80 cm × 64 cm.
• It is melted and recasted into a cube.

To Find :

• The surface area of the cube.

Solution :

Here, dimensions of a metallic cuboidal box are -

• Length = 100 cm
• Breadth = 80 cm
• Height = 64 cm

Let the side of the cuboid be a,

We know,

• Volume of cuboid = l × b × h

Substituting values,

$\implies \: \: \tt \: a = l \times b \times h \\ \implies \tt \: a = 100 \times 80 \times 64 \: cm \\ \implies \bf \: 512000$

$\implies \tt \: \sqrt[3]{512000} \\ \implies \bf \: 80 \: cm$

• So, the side of the cuboid = 80 cm.

A.T.Q :

$\star$ Surface area of cube = 6a²

Plugging the values,

$\implies \tt \: 6 \times {80}^{2} \\ \implies \tt 6 \times 6400 \\ \implies \underline{ \boxed{ \bf{38400 \: {cm}^{2} }}}$

Therefore,

• The surface area of cube is 38400 cm².

$\large \underline{ \boxed{ \purple{ \tt{Surface \: area_{(Cube)} = \pmb{ 38400 \: cm^2}}}}}$

Answer 2

Answer:

$\underline{\underline{\bigstar{\textbf{\textsf{\: Given\::-}}}}}$

• ➽ The dimensions of the a metallic cuboidal box are 100cm × 80cm × 64 cm .
• ➽ It is melted and recast into a cube .

$\begin{gathered}\end{gathered}$

$\underline{\underline{\bigstar{\textbf{\textsf{\: To Find\::-}}}}}$

• ➽ The surface area of the cube

$\begin{gathered}\end{gathered}$

$\underline{\underline{\bigstar{\textbf{\textsf{\: Concept\::-}}}}}$

• ➽ Here the concept of Volume of Cuboid has been used. We are given that the dimensions of the a metallic cuboidal box are 100cm × 80cm × 64 cm. Firstly we will find the side of cuboidal box and then after finding side we will find the volume of cubical box by using the formula os Volume of Cube.

$\begin{gathered}\end{gathered}$

$\underline{\underline{\bigstar{\textbf{\textsf{\: Using Formula\::-}}}}}$

$\dag{\underline{\boxed{\pmb{\sf{\red{Volume \: of \: Cuboid = l \times b \times h}}}}}}$

$\dag{\underline{\boxed{\pmb{\sf{\red{Surface \: area \: of \: Cube=6{a}^{2}}}}}}}$

Where

• ➽ L = Lenght
• ➽ B = Breadth
• ➽ H = Height
• ➽ A = Side

$\begin{gathered}\end{gathered}$

$\underline{\underline{\bigstar{\textbf{\textsf{\: Solution\::-}}}}}$

★ Here, Dimensions of a metallic cuboid :

• ➽ Lenght = 100 cm
• ➽ Breadth = 80 cm
• ➽ Height = 64 cm

$\begin{gathered}\end{gathered}$

★ Now, Let the side of cuboid be a.

$\quad{: \implies{\sf{Volume \: of \: Cuboid = l \times b \times h}}}$

• Substuting the values

$\quad{ : \implies{\sf{ a= 100 \times80 \times 64}}}$

$\quad{ : \implies{\sf{ a=512000 \: {cm}^{3} }}}$

$\quad{ : \implies{\sf{ a= \sqrt[3]{51200} }}}$

$\quad{ : \implies{\sf{ a= \sqrt[3]{80 \times 80 \times 80} }}}$

$\quad{ : \implies{\sf{ a= 80 }}}$

${\dag{\underline{\boxed{\pmb{\sf{\red{ a= 80 \: cm }}}}}}}$

∴ The side of Cuboid is 80 cm.

$\begin{gathered}\end{gathered}$

★ Now,Finding the surface area of Cube.

$\quad{: \implies{\sf{Surface \: area \: of \: Cube=6{a}^{2}}}}$

• Substuting the values

$\quad{: \implies{\sf{Surface \: area \: of \: Cube=6{(80)}^{2}}}}$

$\quad{: \implies{\sf{Surface \: area \: of \: Cube=6{(80 \times 8)}}}}$

$\quad{: \implies{\sf{Surface \: area \: of \: Cube=6{(6400)}}}}$

$\quad{: \implies{\sf{Surface \: area \: of \: Cube=6 \times 6400}}}$

$\quad{: \implies{\sf{Surface \: area \: of \: Cube=38400 \: {cm}^{2} }}}$

$\dag{\underline{\boxed{\pmb{\sf{\red{Surface \: area \: of \: Cube=38400 \: {cm}^{2}}}}}}}$

∴ Surface area of Cube is 38400 cm².

$\begin{gathered}\end{gathered}$

$\underline{\underline{\bigstar{\textbf{\textsf{\:Learn More\::-}}}}}$

$\;\small\sf{\leadsto\;\;\;TSA\;of\;Cylinder\;=\;2\pi rh\;+\;2\pi r^{2}}$

$\;\small\sf{\leadsto\;\;\;CSA\;of\;Cylinder\;=\;2\pi rh}$

$\;\small\sf{\leadsto\;\;\;TSA\;of\;Cone\;=\;\pi rL\;+\;\pi r^{2}}$

$\;\small\sf{\leadsto\;\;\;CSA\;of\;Cone\;=\;\pi rL}$

$\;\small\sf{\leadsto\;\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

$\;\small\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;Length\;\times\;Breadth\;\times\;Height}$

$\;\small\sf{\leadsto\;\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\;\times\;\pi r^{2}h}$

$\;\small\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\;\times\;\pi r^{3}}$

$\;\small\sf{\leadsto\;\;\;Volume\;of\;Sphere\;=\;\dfrac{4}{3}\;\times\;\pi r^{3}}$

$\;\small\sf{\leadsto\;\;\;Volume\;of\;Hollow\;Sphere\;=\;\pi (R^{2}\;-\;r^{2})h}$

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