Chapter SH, Problem 9.6.17EP

To determine

The area of a square is to be estimated as monomial.

  

Answer to Problem 9.6.17EP

The area of a square is to be estimated as monomial:

  $A={\left(6m^2{n}^3\right)}^2=36m^4{n}^6$

Explanation of Solution

Given:

  

Concept Used:

The area of a square is given by the following expression:

  $A=a^2$

Where,

Edge of square: $a$

The product rule:

  ${\left(ab\right)}^m=a^m{b}^m$

Powers of power property:

  ${\left(a^m\right)}^n=a^{mn}$

Calculation:

The edge of given square is $a=6m^2{n}^3$

The area of a square is given as:

  $\begin{array}{l}A=a^2\\ A={\left(6m^2{n}^3\right)}^2\end{array}$

So the number is: ${\left(6m^2{n}^3\right)}^2$

By using the product rule:

  ${\left(ab\right)}^m=a^m{b}^m$

Put the given values:

  $\begin{array}{l}{\left(6m^2{n}^3\right)}^2=6^2{\left(m^2\right)}^2{\left(n^3\right)}^2\\ {\left(6m^2{n}^3\right)}^2=36{\left(m^2\right)}^2{\left(n^3\right)}^2\end{array}$

Use powers of power property:

  ${\left(a^m\right)}^n=a^{mn}$

Put the given values:

  $\begin{array}{l}{\left(6m^2{n}^3\right)}^2=36{\left(m^2\right)}^2{\left(n^3\right)}^2\\ {\left(6m^2{n}^3\right)}^2=36m^{2\times 2}{n}^{3\times 2}\\ {\left(6m^2{n}^3\right)}^2=36m^4{n}^6\end{array}$

So, the volume of cube $A={\left(6m^2{n}^3\right)}^2=36m^4{n}^6$

Conclusion:

Hence, $A={\left(6m^2{n}^3\right)}^2=36m^4{n}^6$ .