An isosceles triangle, by definition, has two sides of equal length. Having the third side measure either 3 feet or 4 feet would make the triangle meet this criterion.

3 feet is equal to $\displaystyle 3 \times 12 = 36$ inches, and 4 feet is equal to $\displaystyle 4 \times 12 = 48$ inches. We choose 36 inches, since that, but not 48 inches, is a choice.

Because the triangles are similar, proportions can be used to solve for the length of the side:

$\displaystyle \frac{9}{3}=\frac{6}{x}$

Cross-multiply:

$\displaystyle 9x=18$

$\displaystyle x=2$

The base angles of an isosceles triangle are always equal. Therefore both base angles are $\displaystyle 42^{\circ}$.

Let $\displaystyle x=$ the measure of the third angle. Since the sum of the angles of a triangle is $\displaystyle 180^{\circ}$, we can solve accordingly:

$\displaystyle 42+42+x=180\Rightarrow x=180-84\Rightarrow x=96^{\circ}$

To find the missing side, use the Pythagorean Theorem $\displaystyle (a^2+b^2=c^2)$. Plug in (remember c is always the hypotenuse!) so that $\displaystyle 6^2+b^2=10^2$. Simplify and you get $\displaystyle 36+b^2=100.$ Subtract 36 from both sides so that you get $\displaystyle b^2=64.$Take the square root of both sides. B is 8.

By the Pythagorean Theorem,

$\displaystyle x^{2}+ (x+3)^{2} = 20^{2}$

$\displaystyle x^{2}+ x^{2}+6x+9= 400$

$\displaystyle 2x^{2}+6x-391 = 0$

By the Pythagorean Theorem,

$\displaystyle x^{2}+ (x+5)^{2} = (x+7)^{2}$

$\displaystyle x^{2}+ x^{2} +10x+25 = x^{2} +14x +49$

$\displaystyle 2x^{2} +10x+25 = x^{2} +14x +49$

$\displaystyle x^{2} -4x-24= 0$

First, find $\displaystyle AB$.

Since $\displaystyle \overline{DF }$ is an altitude of right $\displaystyle \Delta ADB$ to its hypotenuse, 

$\displaystyle \frac{FB}{FD}= \frac{FD}{AF}$

$\displaystyle \frac{FB}{8}= \frac{8}{6}$

$\displaystyle FB= \frac{8}{6} \cdot 8 = 10 \frac{2}{3}$

$\displaystyle AB = AF + FB = 6 + 10 \frac{2}{3} = 16 \frac{2}{3}$

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

$\displaystyle \frac{BC}{FD} = \frac{AB}{AF}$

$\displaystyle \frac{BC}{8} = \frac{16 \frac{2}{3}}{6}$

$\displaystyle BC = 16 \frac{2}{3}\cdot 8 \div 6 = 22\frac{2}{9}$

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,

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