Respuesta :

Hrishii

Answer:

See Explanation

Step-by-step explanation:

  • For cement block:

  • Given is a square cement block, thus, it's width and height would be equal.

  • -> length (l)= h, width (w)= 2x, height (h) = 2x

  • [tex]V_{cement \: block} = lwh[/tex]

  • [tex]\implies V_{cement \: block} = h(2x)(2x)[/tex]

  • [tex]\implies V_{cement \: block} = 4x^2h.....(1)[/tex]

  • For cylinder:

  • Radius (r) = 2x/2 = x, height (h) = h

  • [tex]V_{cylinder} = \pi r^2 h[/tex]

  • [tex]\implies V_{cylinder} = \pi x^2 h.....(2)[/tex]

  • Subtract (2) from (1), we find:

  • [tex]V_{cement \: block}-V_{cylinder} = 4x^2h- \pi x^2 h[/tex]

  • [tex] V(cement \: block\: without\: cylinder)= x^2h(4- \pi)[/tex]

  • Thus proved

  • To solve 6.2.1 and 6.2.2 plug the given values in [tex]x^2h(4- \pi)[/tex] and calculate.

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