First, find $\displaystyle CB$.

Since $\displaystyle \overline{DE}$ is an altitude of $\displaystyle \Delta CDB$ from its right angle to its hypotenuse, 

$\displaystyle \Delta DEC \sim \Delta BED$

$\displaystyle \frac{BE}{DE}= \frac{ED}{EC}$

$\displaystyle \frac{BE}{6}= \frac{6}{12}$

$\displaystyle BE= \frac{6}{12} \cdot 6 = 3$

$\displaystyle CB =BE + EC = 3+ 12 = 15$

$\displaystyle \Delta DEC \sim \Delta ABC$ by the Angle-Angle Postulate, so 

$\displaystyle \frac{AB}{DE} = \frac{CB}{CE}$

$\displaystyle \frac{AB}{6} = \frac{15}{12}$

$\displaystyle AB= \frac{15}{12} \cdot 6$

$\displaystyle AB = 7\frac{1}{2}$

By the Pythagorean Theorem,

$\displaystyle x^{2}+ 15^{2} = (x+12)^{2}$

$\displaystyle x^{2}+ 225 = x^{2}+24x+144$

$\displaystyle 225 = 24x+144$

$\displaystyle 24x= 81$

$\displaystyle x= 3\frac{3}{8}$

The arcs intercepted by a right angle are both semicircles, so hypotenuse $\displaystyle \overline{AC}$ shares its endpoints with two semicircles. This makes $\displaystyle \overline{AC}$ a diameter of the circle, and $\displaystyle AC = 2r = 2 \cdot 26 = 52$.

By the Pythagorean Theorem,

$\displaystyle BC = \sqrt{(AC)^{2} - (AB)^{2}} = \sqrt{52^{2} -20^{2}} = \sqrt{2,704-400} = \sqrt{2,304} = 48$

$\displaystyle \Delta ABC \sim \Delta DEF$. Corresponding angles of similar triangles are congruent, so since $\displaystyle \angle B$ is a right angle, so is $\displaystyle \angle E$. 

The hypotenuse $\displaystyle \overline{DF}$ of $\displaystyle \Delta DEF$ is twice as long as leg $\displaystyle \overline{DE}$; by the $\displaystyle 30 ^{\circ } - 60^{\circ } -90^{\circ }$ Theorem, $\displaystyle m \angle D = 60 ^{\circ }$. Again, by similiarity, 

$\displaystyle m \angle A = m \angle D = 60 ^{\circ }$.

To solve this problem, we must utilize the Pythagorean Theorom, which states that:

$\displaystyle a^{2}+b^{2}=c^{2}$

We know that the base is $\displaystyle 7$, so we can substitute $\displaystyle 7$ in for $\displaystyle a$.  We also know that the height is $\displaystyle 4$, so we can substitute $\displaystyle 4$ in for $\displaystyle b$.

$\displaystyle 7^{2}+4^{2}=c^{2}$

Next we evaluate the exponents:

$\displaystyle 7^{2}=49$

$\displaystyle 4^{2}=16$

Now we add them together:

$\displaystyle 49+16=65$

Then, $\displaystyle 65=c^{2}$.

$\displaystyle 65$ is not a perfect square, so we simply write the square root as  $\displaystyle \sqrt{65}$.

$\displaystyle c=\sqrt{65}$

To solve this problem, we are going to use the Pythagorean Theorom, which states that $\displaystyle a^{2}+b^{2}=c^{2}$.

We know that this particular right triangle has a base of $\displaystyle 3$, which can be substituted for $\displaystyle a$, and a height of $\displaystyle 9$, which can be substituted for $\displaystyle b$. If we rewrite the theorom using these numbers, we get:

$\displaystyle 3^{2}+9^{2}=c^{2}$

Next, we evaluate the expoenents:

$\displaystyle 3^{2}=9$

$\displaystyle 9^{2}=81$

$\displaystyle 9+81=c^{2}$

$\displaystyle 9+81=90$

Then, $\displaystyle 90=c^{2}$.

To solve for $\displaystyle c$, we must find the square root of $\displaystyle 90$. Since this is not a perfect square, our answer is simply $\displaystyle c=\sqrt{90}$.

According to the Pythagorean Theorem, the equation for the hypotenuse of a right triangle is $\displaystyle a^{2} + b^{2}=c^{2}$. Plugging in the sides, we get $\displaystyle 5^{2}+8^{2}=c^{2}$. Solving for $\displaystyle c$, we find that the hypotenuse is $\displaystyle \sqrt{89}$:

$\displaystyle 25+64=c^2$

$\displaystyle 89=c^2$

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let $\displaystyle c=$ hypotenuse and $\displaystyle s=$ side length.

$\displaystyle c^2=s^2+s^2\Rightarrow c^2=(2t)^2+(2t)^2\Rightarrow c^2=8t^2\Rightarrow c=\sqrt{8t^2}=2\sqrt{2}t$

Given that a rectangle has all right angles, drawing a diagonal will create a right triangle the legs are each 6 feet and 8 feet. 

We know that in a 3-4-5 right triangle, when the legs are 3 feet and 4 feet, the hypotenuse will be 5 feet. 

Given that the legs of this triangle are twice as long as those in the 3-4-5 triangle, it follows that the hypotense will also be twice as long. 

Thus, the diagonal in through the rectangle creates a 6-8-10 triangle. 10 is therefore the length of the diagonal.