Alan

As follows:

See https://web2.0calc.com/questions/let-be-an-angle-such-that-p-is-the-terminal-point-of_1#5

The angle \(\theta\) goes anti-clockwise from the positive x-axis around to the red line.  The angle \(\theta/2\) would go half-way (again in an anti-clockwise direction) from the positive x-axis towards the red line.  It would therefore get to point Q.

As follows:

\(\frac{1}{\sqrt2}+\frac{3}{\sqrt{8}}+\frac{6}{\sqrt{32}}\\\\ \frac{1}{\sqrt2}+\frac{3}{2\sqrt{2}}+\frac{6}{4\sqrt{2}}\\\\ \frac{1}{\sqrt2}(1+\frac{3}{2}+\frac{6}{4})\\\\ \frac{\sqrt2}{2}4\\\\ 2\sqrt2 \)

The radius of the inner circle is just half of the side length of the square, so you should still be able to calculate its area!

Presumably by "outer circle" you mean a circumscribing circle that just touches the four corners of the square.

If the area of the square is 196 decimeters squared, then the side length is \(\sqrt{196}\) or 14 decimeters, which means the diagonal of the square is \(14\sqrt2\)decimeters, which means half the diagonal is \(7\sqrt2\) decimeters.  Since the radius of the circumscribing circle is equal to half the square diagonal, I'm sure you can work out the circle area from here.

-1*1/(1+1) = -0.5

This graph should help (the other two roots are complex)

Alternatively:

\(81^n=9^{2n}\)

\(9*27*27*9=9*3*9*3*9*9=9^5\)

Hence \(9^{2n}=9^5\)  so \(2n=5\)  or \(n=5/2\)