The equation of a circle on the coordinate plane is 

\(\displaystyle x^{2} + y^{2} = r^{2}\) 

where \(\displaystyle r\) is the radius. Therefore, 

\(\displaystyle r^{2}= 49\)

and 

\(\displaystyle r = \sqrt{49} = 7\).

The circumference of a circle is \(\displaystyle 2 \pi\) times is radius, which here would be 

\(\displaystyle C= 2 \pi \cdot 7 = 14 \pi\)

The figure is a sector of a circle with radius 8; the sector has degree measure \(\displaystyle 180 ^{\circ } - 45 ^{\circ }= 135 ^{\circ }\). The length of the arc is 

\(\displaystyle A = \frac{ 135^{\circ }}{360 ^{\circ }} \cdot 2\pi r\)

\(\displaystyle A = \frac{ 135^{\circ }}{360 ^{\circ }} \cdot 2\pi \cdot 8\)

\(\displaystyle A = \frac{3}{8} \cdot 16\pi\)

\(\displaystyle A = 6\pi\)

The equation of a circle on the coordinate plane is 

\(\displaystyle x^{2} + y^{2} = r^{2}\) ,

where \(\displaystyle r\) is the radius. Therefore, 

\(\displaystyle r ^{2} = 75\)

and 

\(\displaystyle r = \sqrt{ 75} = \sqrt{25} \times \sqrt{ 3} = 5 \sqrt{ 3}\).

The circumference of a circle is \(\displaystyle 2 \pi\) times is radius, which here would be

\(\displaystyle C = 2 \pi \times 5 \sqrt{3} = 10\pi \sqrt{3}\).

The figure below shows \(\displaystyle \angle AOB\), which matches this description, along with its chord \(\displaystyle \overline{AB}\) and triangle bisector \(\displaystyle \overline{OM}\). 

We will concentrate on \(\displaystyle \Delta AOM\), which is a 30-60-90 triangle.

Chord \(\displaystyle \overline{AB}\) has length 24, so \(\displaystyle \overline{AM}\) has length half this, or 12.

By the 30-60-90 Theorem, 

\(\displaystyle OM = AM \div \sqrt{3}= 12 \div \sqrt{3}= \frac{12}{ \sqrt{3}}\)

and 

\(\displaystyle AO = 2 \cdot OM = 2 \cdot \frac{12}{ \sqrt{3}} = \frac{24}{ \sqrt{3}} = \frac{24 \cdot \sqrt{3}}{ \sqrt{3} \cdot \sqrt{3} } =\frac{24 \cdot \sqrt{3}}{ 3 } = 8\sqrt{3}\)

This is the radius, so the circumference is

\(\displaystyle C = 2 \pi r = 2 \pi \cdot 8\sqrt{3}= 16 \pi \sqrt{3}\)

If a (perpendicular) radius of the inscribed circle is constructed to the triangle, and a radius of the circumscribed circle is constructed to a neighboring vertex, a right triangle is formed. By symmetry, it can be shown that this is a 30-60-90 triangle, and, subsequently,

\(\displaystyle BO = 2 \cdot AO\)

If we let \(\displaystyle r = AO\), the circumference of the inscribed circle is \(\displaystyle C =2 \pi r\).

Then \(\displaystyle BO = 2r\), and the circumference of the circumscribed circle is \(\displaystyle 2 \pi \cdot 2r = 4 \pi r = 2C\).

The ratio of the circumferences is therefore 2 to 1.

If a right triangle is inscribed inside a circle, then the arc intercepted by the right angle is a semicircle, making the hypotenuse of triangle a diameter. 

The length of the hypotenuse of this triangle can be calculated using the Pythagorean Theorem:

\(\displaystyle d = \sqrt {18^{2}+24^{2}} = \sqrt{324+576} = \sqrt{900} = 30\)

This is the diameter, also, so the circumference is \(\displaystyle C = \pi d = 30 \pi\).

If a right triangle is inscribed inside a circle, then the arc intercepted by the right angle is a semicircle, making the hypotenuse of triangle a diameter. 

The length of the shorter leg of a 30-60-90 triangle is that of the longer leg divided by \(\displaystyle \sqrt{3}\), so the shorter leg will have length

\(\displaystyle 7\frac{1}{2} \div \sqrt{3} = \frac{15}{2} \times \frac{1}{\sqrt{3}} = \frac{15}{2} \times \frac{\sqrt{3}}{3} =\frac{5\sqrt{3}}{2}\)

The hypotenuse will have length twice that of its short leg, so the hypotenuse of this triangle will have twice this length, or 

\(\displaystyle 2 \times \frac{5\sqrt{3}}{2} = 5\sqrt{3}\)

This is the diameter, so multiply this by \(\displaystyle \pi\) to get the circumference:

\(\displaystyle 5\sqrt{3} \cdot \pi = 5 \pi \sqrt{3}\)

An equilateral triangle of perimeter 60 has sidelength one-third of this, or 20. 

Construct this triangle and its inscribed circle, as well as a radius to one side - which, by symmetry, is a perpendicular bisector - and a segment to one of that side's endpoints:

Each side of the triangle has measure 20, so \(\displaystyle AB = 10\). Also, the triangle formed by the segments, by symmetry, is a 30-60-90 triangle. Therefore, 

\(\displaystyle BO = \frac{AB}{\sqrt{3}} = \frac{10}{\sqrt{3}} = \frac{10\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}} = \frac{10 \sqrt{3}}{3}\)

which is the radius of the circle. The cricumference of the circle is \(\displaystyle 2 \pi\) times this, or 

\(\displaystyle \frac{10 \sqrt{3}}{3} \times 2 \pi = \frac{20 \pi \sqrt{3}}{3}\)

An equilateral triangle of perimeter 84 has sidelength one-third of this, or 28. 

Construct this triangle and its circumscribed circle, as well as a perpendicular bisector to one side and a radius to one of that side's endpoints:

Each side of the triangle has measure 28, so \(\displaystyle AB = 14\). Also, the triangle formed by the segments, by symmetry, is a 30-60-90 triangle. Therefore, by the 30-60-90 Theorem,

\(\displaystyle BO = \frac{AB}{\sqrt{3}} = \frac{14}{\sqrt{3}} = \frac{14\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}} = \frac{14 \sqrt{3}}{3}\)

and \(\displaystyle AO = 2 \cdot BO = 2 \cdot \frac{14 \sqrt{3}}{3}= \frac{28 \sqrt{3}}{3}\).

This is the radius, so the circumference is \(\displaystyle 2 \pi\) times this, or 

\(\displaystyle \frac{28 \sqrt{3}}{3} \times 2 \pi = \frac{56 \pi \sqrt{3}}{3}\)

The figure below shows \(\displaystyle \angle AOB\), which matches this description, along with its chord \(\displaystyle \overline{AB}\):

By way of the Isosceles Triangle Theorem, \(\displaystyle \Delta AOB\) can be proved equilateral, so \(\displaystyle AB = AO= 9\). This is the radius, so the circumference is

\(\displaystyle C = 2 \pi r = 2 \pi \cdot 9 = 18 \pi\)