Chapter MPS, Problem 6.8MPS

To determine

To Find: The width of the brick.

Answer to Problem 6.8MPS

  $6\sqrt{2}unit$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Length of the patio}=8ft\\ \text{Width of the patio}=6ft\\ \text{Combine area of patio and brick is }=120f{t}^2\\ \text{Width of the brick }=x\end{array}$

Formula Used:

  $\begin{array}{l}\text{Area of a rectangle}=length\times breadth\\ \text{Area of square}=sid{e}^2\end{array}$

Calculation:

  $\begin{array}{l}\text{Length of the patio}=8ft\\ \text{Width of the patio}=6ft\\ \text{Combine area of patio and brick is }=120f{t}^2\\ \text{Width of the brick }=x\\ \text{Area of a rectangle}=length\times breadth\\ \text{Area of square}=sid{e}^2\\ \text{⇒}\text{Area of patio}=\left(6\times 8\right)f{t}^2=48f{t}^2\\ \text{Area of brick}=x^2\\ \text{Combine area }=120f{t}^2\\ \text{⇒}120f{t}^2=48f{t}^2+x^2\\ \text{⇒}{x}^2=72f{t}^2\\ \text{⇒}x=6\sqrt{2}unit\end{array}$

Conclusion:

Thus the width of brick is $6\sqrt{2}unit$ .