(a) A circle with radius 8 inches has crircumference \(\displaystyle C = 2\pi r = 2\pi \cdot 8 = 16\pi\) inches. An arc 10 inches long is \(\displaystyle \frac{10}{16\pi } = \frac{5}{8\pi }\) of that circle. \(\displaystyle \frac{5}{8\pi } \times 360 ^{\circ } = \frac{225}{\pi } ^{\circ }\), the degree measure of this arc.

(b) A circle with radius 10 inches has crircumference \(\displaystyle C = 2\pi r = 2\pi \cdot 10 = 20\pi\) inches. An arc 12 inches long is \(\displaystyle \frac{12}{20\pi }=\frac{3}{5\pi }\) of that circle. \(\displaystyle \frac{3}{5\pi } \times 360 ^{\circ } = \frac{216}{\pi } ^{\circ }\), the degree measure of this arc.

(a) is the greater quantity.

Since 

\(\displaystyle 7^{2} + 12 ^{2} = 49 + 144 = 169 = 13 ^{2}\),

the triangle is a right triangle with right angle \(\displaystyle \angle ACB\).

\(\displaystyle \angle ACB\) is an inscribed angle on the circle, so the arc it intercepts is a semicricle. Therefore, \(\displaystyle \widehat{ACB}\) is also a semicircle, and it measures \(\displaystyle 180 ^{\circ }\).

The measure of an arc intercepted by an inscribed angle of a circle is twice that of the angle. Therefore, \(\displaystyle y = 2 \cdot 44 = 88 < 90\)

\(\displaystyle \odot A\) has twice the radius of \(\displaystyle \odot B\), so \(\displaystyle \odot A\) has four times the area of \(\displaystyle \odot B\). This means that for a sector of \(\displaystyle \odot A\) to have the same area as a sector of \(\displaystyle \odot B\), the central angle of the latter sector must be four times that of the former sector. This makes (b) greater than (a), which is only twice that of the former sector.

The measure of an inscribed angle of a circle is one-half that of the arc it intercepts. Therefore, \(\displaystyle x = \frac{1}{2} \cdot 110 = 55\).

Remember that the angle for a sector or arc is found as a percentage of the total \(\displaystyle 360\) degrees of the circle.  The proportion of \(\displaystyle x\) to \(\displaystyle 360\) is the same as \(\displaystyle 19.4\) to the total circumference of the circle.

The circumference of a circle is found by:

\(\displaystyle C = 2*\pi*r\)

For our data, this means:

\(\displaystyle C = 2*8.7*\pi=17.4\pi\)

Now we can solve for \(\displaystyle x\) using the proportions:

\(\displaystyle \frac{x}{360} = \frac{19.4}{17.4\pi}\)

Cross multiply:

\(\displaystyle 17.4x\pi=6984\)

Divide both sides by \(\displaystyle 17.4\pi\):

\(\displaystyle x=127.76300259239047\)

Therefore, \(\displaystyle x\) is \(\displaystyle 127.76\)˚.

Remember that the angle for a sector or arc is found as a percentage of the total \(\displaystyle 360\) degrees of the circle.  The proportion of \(\displaystyle x\) to \(\displaystyle 360\) is the same as \(\displaystyle 9.31\) to the total circumference of the circle.

The circumference of a circle is found by:

\(\displaystyle C = 2*\pi*r\)

For our data, this means:

\(\displaystyle C = 2*9.1*\pi=18.2\pi\)

Now we can solve for \(\displaystyle x\) using the proportions:

\(\displaystyle \frac{x}{360} = \frac{9.31}{18.2\pi}\)

Cross multiply:

\(\displaystyle 18.2x\pi=3351.6\)

Divide both sides by \(\displaystyle 18.2\pi\):

\(\displaystyle x=58.61798980953807\)

Therefore, \(\displaystyle x\) is \(\displaystyle 58.62\)˚.

Remember that the angle for a sector or arc is found as a percentage of the total \(\displaystyle 360\) degrees of the circle.  The proportion of \(\displaystyle x\) to \(\displaystyle 360\) is the same as \(\displaystyle 4.5\pi\) to the total area of the circle.

The area of a circle is found by:

\(\displaystyle A = \pi * r^2\)

For our data, this means:

\(\displaystyle A = 7.5^2\pi = 56.25\pi\)

Now we can solve for \(\displaystyle x\) using the proportions:

\(\displaystyle \frac{x}{360} = \frac{2.81\pi}{56.25\pi}\)

Cross multiply:

\(\displaystyle 56.25x\pi=1011.6\pi\)

Divide both sides by \(\displaystyle 56.25\pi\):

\(\displaystyle x=17.984\)

Therefore, \(\displaystyle x\) is \(\displaystyle 17.98\)˚.

Remember that the angle for a sector or arc is found as a percentage of the total \(\displaystyle 360\) degrees of the circle.  The proportion of \(\displaystyle x\) to \(\displaystyle 360\) is the same as \(\displaystyle 4.5\pi\) to the total area of the circle.

The area of a circle is found by:

\(\displaystyle A = \pi * r^2\)

For our data, this means:

\(\displaystyle A = 6^2\pi = 36\pi\)

Now we can solve for \(\displaystyle x\) using the proportions:

\(\displaystyle \frac{x}{360} = \frac{4.5\pi}{36\pi}\)

Cross multiply:

\(\displaystyle 36x\pi=1620\pi\)

Divide both sides by \(\displaystyle 36\pi\):

\(\displaystyle x=45\)

Therefore, \(\displaystyle x\) is \(\displaystyle 45\)˚.

\(\displaystyle \bigtriangleup ABC\) has angles of degree measure 30 and 60; the third angle must measure 90 degrees, making \(\displaystyle \bigtriangleup ABC\) a right triangle.

For the sake of simplicity, let \(\displaystyle BC = 1\); the reasoning is independent of the actual length. The smaller leg of a 30-60-90 triangle has length equal to \(\displaystyle \frac{\sqrt{3}}{3}\) times that of the longer leg; this is about

\(\displaystyle AB = \frac{\sqrt{3}}{3} \approx \frac{1.7}{3} \approx 0.57\)

 The area of a right triangle is half the product of its legs, so 

\(\displaystyle A = \frac{1}{2} \cdot 1 \cdot 0.57 = 0.5 \cdot 0.57 \approx 0.285\)

Also, if \(\displaystyle BC = 1\), then the orange semicircle has diameter 1 and radius \(\displaystyle \frac{1}{2}\). Its area can be found by substituting \(\displaystyle r = \frac{1}{2}\) in the formula:

\(\displaystyle A = \frac{1}{2} \cdot \pi r^{2}\)

\(\displaystyle = \frac{1}{2} \cdot \pi \left ( \frac{1}{2} \right ) ^{2}\)

\(\displaystyle = \frac{1}{2} \cdot \pi \cdot \left ( \frac{1}{4 } \right )\)

\(\displaystyle = \frac{1}{8} \cdot \pi\)

\(\displaystyle \approx 0. 125 \cdot 3.14\)

\(\displaystyle \approx 0. 3925\)

The orange semicircle has a greater area than \(\displaystyle \bigtriangleup ABC\)