Question

please help

see attached picture​

Answer 1

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

• $V_{cement \: block} = lwh$

• $\implies V_{cement \: block} = h(2x)(2x)$

• $\implies V_{cement \: block} = 4x^2h.....(1)$

• For cylinder:

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

• $V_{cylinder} = \pi r^2 h$

• $\implies V_{cylinder} = \pi x^2 h.....(2)$

• Subtract (2) from (1), we find:

• $V_{cement \: block}-V_{cylinder} = 4x^2h- \pi x^2 h$

• $V(cement \: block\: without\: cylinder)= x^2h(4- \pi)$

• Thus proved

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