Chapter 12, Problem 10GRFC

To determine

Find the value of a .

Answer to Problem 10GRFC

  $a=\pm 9$

Explanation of Solution

Given:

  $a^2+{40}^2={41}^2$

Calculation:

  $\begin{array}{l}{a}^2+{40}^2={41}^2\\ \Rightarrow a^2={41}^2-{40}^2\mathrm{..............}[subtract{40}^2\text{ from each side]}\\ \Rightarrow a^2=(41+40)(41-40)\mathrm{..}[a^2-b^2=(a+b)(a-b)\\ \Rightarrow a^2=81\\ \Rightarrow a=\pm \sqrt{81}\mathrm{.......................}[take\text{ square root each side]}\\ \Rightarrow a=\pm 9\end{array}$