Chapter 1 - 1.6 - Other Types of Equations - 1.6 Exercises - Page 130: 108

$a=\sqrt {\frac{C^2}{2\pi^2}-b^2}$

$C=2\pi \sqrt {\frac{a^2+b^2}{2}}$ $\frac{C}{2\pi}=\sqrt {\frac{a^2+b^2}{2}}$ Squaring both sides of the equation. $\frac{C^2}{4\pi^2}=\frac{a^2+b^2}{2}$ Multiplying both sides of the equation by $2$ $\frac{C^2}{2\pi^2}=a^2+b^2$ $\frac{C^2}{2\pi^2}-b^2=a^2$ $a=\sqrt {\frac{C^2}{2\pi^2}-b^2}$