The hypotenuse in an isosceles right triangle is equal to $\sqrt{2}$ times the length of a leg.

The legs of an isosceles right triangle are the sides that are opposite the $45^\circ$ angles. So in the given triangle, the legs have length $\dfrac{3\sqrt2}{2}.$

Consequently, the hypotenuse has a length of

\[ \sqrt{2}\cdot\dfrac{3\sqrt{2}}{2} = \dfrac{\sqrt{2}\cdot3\sqrt{2}}{2}=\dfrac{3\cdot 2}{2}=3. \]

The hypotenuse in an isosceles right triangle is equal to $\sqrt2$ times the length of a leg.

The legs of an isosceles right triangle are the sides that are opposite the $45^\circ$ angles. So in the given triangle, the legs have length $6\sqrt2.$

Consequently, the hypotenuse has a length of

\[ \sqrt{2}\cdot 6\sqrt{2} = 6\cdot 2 = 12. \]

The hypotenuse in a $45^\circ$-$45^\circ$-$90^\circ$ triangle is equal to $\sqrt{2}$ times the length of a leg.

Here, we know that the hypotenuse has length $\dfrac{3}{4}.$ So, if the length of a leg is $a,$ then we must have

\begin{align} \dfrac{3}{4} &= \sqrt2 \cdot a \\[3pt] a &= \dfrac{3}{4\sqrt2}. \end{align}

To remove the square root from the denominator, we can multiply both the numerator and denominator by $\sqrt2,$ as follows:

\begin{align} a &= \dfrac{3\cdot \sqrt2}{4\sqrt2 \cdot \sqrt2} \\[5pt] &= \dfrac{3\sqrt2}{4\cdot 2} \\[5pt] &= \dfrac{3\sqrt2}{8} \end{align}

Therefore, the length of each leg is $\dfrac {3\sqrt{2}}{8}.$

The hypotenuse in a $45^\circ$-$45^\circ$-$90^\circ$ triangle is equal to $\sqrt{2}$ times the length of a leg.

Here, we know that the hypotenuse has length $4\sqrt{2}.$ So, if the length of a leg is $a,$ then we must have \begin{align} 4\sqrt{2} &= \sqrt2 \cdot a \\[5pt] a &= \dfrac{4\sqrt{2}}{\sqrt2}\\[5pt] a &= 4. \end{align}

Therefore, the length of each leg is $4.$

Because the triangle is isosceles, the legs must have the same length. Let $a$ be the length of each leg. So, we have

\[ AB=BC = a. \]

Moreover, because the triangle is a right isosceles triangle, the hypotenuse must be equal to $\sqrt2$ times the length of a leg. Therefore,

\[ AC = AB\cdot \sqrt2 = a\sqrt2. \]

Finally, since the perimeter is $p=14+7\sqrt2,$ we have

\begin{align*} \require{cancel} p &= AB + BC + AC \\[3pt] 14 + 7\sqrt2 &= a + a + a\sqrt{2} \\[3pt] 14 + 7\sqrt2 &= 2a + a\sqrt{2} \\[3pt] 7(2 + \sqrt2) &= a(2 + \sqrt{2}) \\[3pt] 7\cancel{(2 + \sqrt2)} &= a\cancel{(2 + \sqrt{2})} \\[3pt] 7 & = a. \end{align*}

Therefore, the length of each leg is $a=7,$ and the length of the hypotenuse is

\[ AC = a\sqrt2 = 7\sqrt2. \]

Because the triangle is isosceles, the legs must have the same length. Let $a$ be the length of each leg. So, we have

\[ AB=BC = a. \]

Moreover, because the triangle is a right isosceles triangle, the hypotenuse must be equal to $\sqrt2$ times the length of a leg. Therefore,

\[ AC = AB\cdot \sqrt2 = a\sqrt2. \]

Finally, since the perimeter is $p=12 + 6\sqrt{2},$ we have \begin{align*} \require{cancel} p &= AB + BC + AC \\[3pt] 12 + 6\sqrt{2}&=a + a + a\sqrt{2}\\[3pt] 6(2 + \sqrt{2})&=2a + a\sqrt{2}\\[3pt] 6(2 + \sqrt{2})&=(2 + \sqrt{2})a\\[3pt] 6\cancel{(2 + \sqrt{2})}&=\cancel{(2 + \sqrt{2})}a\\[3pt] 6&=a. \end{align*}

Therefore, the measure of a leg is $6.$

A diagram of the given situation is shown below.

Notice that $\triangle{ACD}$ is a $45^\circ$-$45^\circ$-$90^\circ$ triangle since $AD=CD$ and $m\angle DAC=90^\circ.$

The hypotenuse in an isosceles right triangle is equal to $\sqrt{2}$ times the length of a leg.

Here, the length of the hypotenuse $\overline{AC}$ is $5\sqrt{2}\,\mathrm{mm}.$ So the length of the leg $\overline{AD}$ can be found by solving the following equation: \begin{align} AC & = \sqrt{2} \cdot AD\\[5pt] 5\sqrt{2} & = \sqrt{2} \cdot AD\\[5pt] AD & = \dfrac{5\sqrt{2}}{\sqrt{2}}\\[5pt] AD & = 5 \,\mathrm{mm} \end{align}

Finally, the perimeter $p$ is given by

\[ {p}=4\cdot AD= 4\cdot 5 = 20\,\mathrm{mm}.\]

Since the perimeter is $p=8,$ the length of a single side is \[ AB = \dfrac{p}{4} = \dfrac{8}{4} = 2\,\textrm{mm}. \]

Now, consider $\triangle{ABC},$ which is an isosceles right triangle since $\angle{ABC} = 90^\circ$ and $AB=BC.$

The hypotenuse in an isosceles right triangle is equal to $\sqrt{2}$ times the length of a leg.

Here, the length of a leg is $AB = 2,$ so the length of the hypotenuse is

\[ AC = \sqrt{2}\cdot AB = \sqrt{2}\cdot 2 = 2\sqrt{2}\,\textrm{mm}. \]

Therefore, the diagonal of the square $ABCD$ has a length of $AC=2\sqrt2\,\textrm{mm}.$