Length of a Curve (Semicirle) using integration

Question

I understand the concept of this problem, but need a little help deciphering the algebra

anyway, here goes;

consider a semicircle of radius $a$ given by the equation $$f(x)=\sqrt{a^{2}-x^{2}}$$

The arc length of a curve is given by the relationship $$L=\int^a_{-a}{\sqrt{1+f^{\prime}(x)^{2}}}$$

$$f^{\prime}(x)=\frac{1}{2}(a^{2}-x^{2})^{-1/2}*2x=\frac{x}{\sqrt{a^{2}-x^{2}}}$$

now since a semicircle has a vertical tangent at points $-a$ and $a$ we cant take the whole length of the semicircle, Instead we can draw a right triangle from the origin to a point on the curve, where the adjacent side of the triangle is $\frac{a}{\sqrt{2}}$ and the hypotenuse of the triangle is 1/8th of a full circle or 1/4 of the semicirle

Question 1: Why does this work, how did my textbook realize that the adjacent side of the triangle is $\frac{a}{\sqrt{2}}$, and how do they know this will be 1/8 of the full circle?

anyway therefore 1/4 of the arc length of the semicircle is; $$\frac{1}{4}L=\int_0^{\frac{a}{\sqrt{2}}}\sqrt{1+\bigg(\frac{x}{\sqrt{a^{2}-x^{2}}}\bigg)^{2}}dx$$

$$L=4\int_0^{\frac{a}{\sqrt{2}}}\sqrt{1+\frac{x^2}{a^{2}-x^{2}}}dx$$

finding the common denominator

$$=4\int_0^{\frac{a}{\sqrt{2}}}\sqrt{\frac{a^2-x^2+x^2}{a^{2}-x^{2}}}dx=4\int_0^{\frac{a}{\sqrt{2}}}\frac{a}{\sqrt{a^2-x^2}}dx$$

Thus the length of the curve of a semicircle is

$$=4a\int_0^{\frac{a}{\sqrt{2}}}\frac{dx}{\sqrt{a^2-x^2}}$$

$$=4a\bigg(sin^{-1}\frac{x}{a}\bigg)^{a/\sqrt{2}}_0$$

$$=\pi*a$$

So my confusion remains about the limits of integration, what did the authors use to identify that 1/8 of a the length of a semicircle of radius a is from x=0 to $x=\frac{a}{\sqrt{2}}$?

Answer 1

See this graphic:

So, in this picture, if the opposite of the triangle is $\frac{a}{\sqrt{2}}$, then $\theta$ would be equal to $\frac{\pi}{4}$ (because it is a $1,1,\sqrt{2}$ right triangle). By definition, the arc that corresponds to $\theta = \frac{\pi}{4} $ has a length equal to $\frac{1}{8}$ of a full circle. ($\frac{ \pi / 4 }{2 \pi} = \frac{1}{8}$)

Then, the complementary angle of $\theta$, $\phi$ would also be equal to $ \frac{\pi}{4} $, and its corresponding arc would also have a length equal to $\frac{1}{8}$ of a full circle. That arc, highlighted in red, corresponds to $x$ values from $0$ to $\frac{a}{\sqrt{2}}$ in the graph.