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,

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