Question

Diagonal AC is drawn in square ABCD to create a 45°-45°-90° triangle. Other45°-45°-90° triangles of different sizes are formed by drawing the altitudefrom the right angle. Find GN.

Answer 1

First let's calculate BE using Pythagorean theorem:

$\begin{gathered} AB^2=BE^2+AE^2 \\ (8\sqrt[]{3})^2=x^2+x^2 \\ 64\cdot3=2x^2^{} \\ x^2=32\cdot3 \\ x^2=96 \\ x=4\sqrt[]{6} \end{gathered}$

Now, let's find EG using the same formula:

$\begin{gathered} BE^2=BG^2+EG^2 \\ (4\sqrt[]{6})^2=y^2+y^2 \\ 2y^2=16\cdot6 \\ y^2=16\cdot3 \\ y=4\sqrt[]{3} \end{gathered}$

Finally, using the same formula to find GN, we have:

$\begin{gathered} EG^2=EN^2+GN^2 \\ (4\sqrt[]{3})^2=z^2+z^2 \\ 2z^2=16\cdot3 \\ z^2=8\cdot3 \\ z=2\sqrt[]{6} \end{gathered}$

So the length of GN is 2√6.