3. The volume of a cuboidal box isx’ + 14x’ + 52x + 21 cubic units. Its height isx + 7 units. What is
its base area?

3. The volume of a cuboidal box isx’ + 14x’ + 52x + 21 cubic units. Its height isx + 7 units. What is
its base area?

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2 thoughts on “3. The volume of a cuboidal box isx’ + 14x’ + 52x + 21 cubic units. Its height isx + 7 units. What is<br />its base area?<br /><br”

  1. Step-by-step explanation:

    • Length of cuboid = (2x + 2) units

    • Breadth of cuboid = (2x – 2) units

    ➪Height of cuboid = (2x – 2) units

    ➪Volume of cuboid is

    [tex]✵\sf\begin{gathered} Volume = Length \times Breadth \times Height\\\\ Volume = (\,2x + 2)\, \times (\,2x – 2)\, \times (\,2x – 2)\,\\\\ Volume = (\,(\,2x)\,^2 – (\,2)\,^2)\ \times (\,2x – 2)\,\\\\ Volume = (\,4x^2 – 4)\, \times (\,2x – 2)\, \\\\ Volume = 8x^3 – 8x^2 – 8x + 8\end{gathered}[/tex]

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  2. Answer:

    Step-by-step explanation:

    Length of cuboid = (2x + 2) units

    Breadth of cuboid = (2x – 2) units

    ➪Height of cuboid = (2x – 2) units

    ➪Volume of cuboid is

    \begin{gathered}✵\sf\begin{gathered} Volume = Length \times Breadth \times Height\\\\ Volume = (\,2x + 2)\, \times (\,2x – 2)\, \times (\,2x – 2)\,\\\\ Volume = (\,(\,2x)\,^2 – (\,2)\,^2)\ \times (\,2x – 2)\,\\\\ Volume = (\,4x^2 – 4)\, \times (\,2x – 2)\, \\\\ Volume = 8x^3 – 8x^2 – 8x + 8\end{gathered}\end{gathered}

    Volume=Length×Breadth×Height

    Volume=(2x+2)×(2x−2)×(2x−2)

    Volume=((2x)

    2

    −(2)

    2

    ) ×(2x−2)

    Volume=(4x

    2

    −4)×(2x−2)

    Volume=8x

    3

    −8x

    2

    −8x+8

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