Pythagoras in Flächen

d = a . $$\sqrt[]{2}$$

d = $$\frac{a}{\sqrt[]{2}}$$

d = $$\sqrt[]{a\ .\ a}$$

d = $$\sqrt[]{a\ +\ a}$$

a = $$\frac{d}{\sqrt[]{2}}$$

a = $$\frac{d}{2}$$

a = $$\frac{d^2}{2}$$

d = $$\sqrt[]{a^2\ +\ b^2}$$

d = $$\sqrt[]{a\ +\ b}$$

d = $$\sqrt[]{a^2\ +\ b}$$

d = $$\sqrt[]{a\ +\ b^2}$$

a = $$\sqrt[]{d^2\ -\ b^2}$$

b = $$\sqrt[]{d^2\ -\ a^2}$$

a = $$\sqrt[]{d\ -\ b}$$

b = $$\sqrt[]{d\ -\ a}$$

a = $$\sqrt[]{h^2\ +\ \left(\frac{c}{2}\right)^2}$$

a = $$\sqrt[]{h^2\ +\ c^2}$$

a = $$\sqrt[]{h^2\ +\ \left(2.c\right)^2}$$

h = $$\sqrt[]{a^2\ -\ \left(\frac{c}{2}\right)^2}$$

$$\frac{c}{2}\ =\ \sqrt[]{a^2\ -\ h^2}$$

h = $$\sqrt[]{a^2\ -\ c^2}$$

c = $$\sqrt[]{a^2\ -\ \left(\frac{h}{2}\right)^2}$$

h = $$\frac{a}{2}.\ \sqrt[]{3}$$

h = $$\frac{a\ .\ \sqrt[]{3}}{2}$$

h = $$\frac{a}{2}$$

h = $$\sqrt[]{a^2\ +\ a^2}$$

a = $$\frac{2\ h}{\sqrt[]{3}}$$

a = $$\frac{h}{\sqrt[]{3}}$$

a = 2 . h . $$\sqrt[]{3}$$

$$a\ =\ \sqrt[]{\left(\frac{e}{2}\right)^2\ +\left(\frac{f}{2}\right)^2}$$

a = $$\sqrt[]{e^2\ +\ f^2}$$

a = $$\sqrt[]{\frac{e^2}{2}\ +\ \frac{f^2}{2}}$$

$$a\ =\ \sqrt[]{\left(\frac{e}{2}\right)^2\ -\left(\frac{f}{2}\right)^2}$$

$$\left(\frac{e}{2}\right)\ =\ \sqrt[]{a^2\ -\left(\frac{f}{2}\right)^2}$$

$$\left(\frac{f}{2}\right)\ =\ \sqrt[]{a^2\ -\left(\frac{e}{2}\right)^2}$$

e = $$\sqrt[]{a^2\ -f^2}$$

f = $$\sqrt[]{a^2\ -e^2}$$

$$f\ =\ \sqrt[]{a^2\ -\left(\frac{e}{2}\right)^2}$$