Question

a solid cube has been cut into two cuboid of equal volumes.what is the ratio of total surface area of a cube and cuboid thus obtained??​

Answer 1

Let the side of the cube = x

Total surface area of cube :

= 6 × (side)²

= 6x²

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[ The cube is divided into two same cuboid, the height and length of the base Will be the same, but the length will be the half of the length of the cube]

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Height of cuboid = x

Base of cuboid = x

Length of the cuboid = x/2

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$\underline{\boxed{\rm TSA\: of\: cuboid = 2( lb+bh+hl)}}$

TSA = total surface area

l = length

b = base

h = height

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$\rm TSA =2\:\bigg\{ \bigg( \dfrac{x}{2}\times x \bigg)+\bigg(x\times x\bigg)+\bigg(x\times\dfrac{x}{2}\bigg)\bigg\}$

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$\rm TSA =2\bigg(\: \dfrac{x^2}{2} + x^2 + \dfrac{x^2}{2}\bigg)$

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$\rm TSA =\not2\bigg( \dfrac{x^2+2x^2+x^2}{\not2}\bigg)$

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$\rm TSA= 4x^2$

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Ratio of TSA of cube to TSA of cuboid :

$\rm\implies\dfrac{6x^2}{4x^2}$

$\rm\implies\dfrac{\not6\!\!\!\not x^2}{\not4\!\!\!\not x^2}$

$\rm \implies \dfrac{3}{2}$

Their ratio is 3:2