2. For the first one use $a^2+b^2=c^2$
in your case $(\frac{6}{2})^2+d^2=5^2$
find $d$ from here

3. Each corner is a right isosceles triangle. Let the lengths of the legs each by s:
\begin{aligned}
s^2 + s^2 &= a^2 \\
2s^2 &= a^2 \\
s^2 &= \frac{a^2}{2} \\
s &= \frac{a\sqrt{2}}{2}
\end{aligned}

(The above should look familiar. I did this in your other thread:http://www.mathhelpforum.com/math-he...eorem-2-a.html)

The length of a side of the square, x, would be s + a + s:
\begin{aligned}
x &= s + a + s \\
&= a + 2s \\
&= a + 2\left(\frac{a\sqrt{2}}{2}\right) \\
&= a + a\sqrt{2} \\
&= a(1 + \sqrt{2})
\end{aligned}

01