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

By the 30-60-90 Theorem, 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 \(\displaystyle \frac{11}{\sqrt{3}}\); the hypotenuse will have length twice this length, or 

\(\displaystyle 2 \cdot \frac{11}{\sqrt{3}} = \frac{22}{\sqrt{3}}\).

The diameter of the circle is therefore \(\displaystyle \frac{22}{\sqrt{3}}\); the radius is half this, or \(\displaystyle \frac{11}{\sqrt{3}}\). The area of the circle is therefore

\(\displaystyle A = \pi r^{2} = \pi \cdot \left ( \frac{11}{\sqrt{3}} \right ) ^{2} = \frac{121}{3}\cdot \pi = \frac{121 \pi}{3}\)

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

By way of the Isoscelese Triangle Theorem, \(\displaystyle \Delta AOB\) can be proved a 45-45-90 triangle with hypotenuse 15. By the 45-45-90 Theorem, its legs, each a radius, have length that can be determined by dividing this by \(\displaystyle \sqrt{2}\):

\(\displaystyle A O = \frac{AB }{\sqrt{2}} = \frac{15}{\sqrt{2}}\)

The area is therefore

\(\displaystyle A = \pi r^{2} = \pi \cdot \left ( \frac{15}{\sqrt{2}} \right )^{2} = \pi \cdot \frac{225}{2} =\frac{225 \pi}{2}\)

An equilateral triangle of perimeter 72 has sidelength one-third of this, or 24. 

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 24, so \(\displaystyle AB = 12\). Also, the triangle formed by the segments, by symmetry, is a 30-60-90 triangle. Therefore, 

\(\displaystyle BO = \frac{AB}{\sqrt{3}} = \frac{12}{\sqrt{3}} = \frac{12\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}} = \frac{12 \sqrt{3}}{3} =4\sqrt{3}\)

which is the radius of the circle. The area of this circle is 

\(\displaystyle A = \pi r^{2} = \pi (4 \sqrt{3})^{2} = \pi \cdot 4 ^{2} (\sqrt{3})^{2} = 16\cdot 3 \cdot \pi= 48\pi\)

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

\(\displaystyle d = \sqrt {10^{2}+24^{2}} = \sqrt{100+576} = \sqrt{676} = 26\)

The radius is half this, or 13, so the area is

\(\displaystyle A = \pi r^{2} = \pi \cdot 13 ^{2} = 169 \pi\)

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 15, so \(\displaystyle AM = \frac{1}{2} AB = \frac{1}{2} \cdot 15 = \frac{15}{2}\)

By the 30-60-90 Theorem, 

\(\displaystyle OM = AM \div \sqrt{3}= \frac{15}{2} \div \sqrt{3}= \frac{15}{2\sqrt{3}}\)

and 

\(\displaystyle AO = 2 \cdot OM = 2 \cdot \frac{15}{2\sqrt{3}} = \frac{15}{ \sqrt{3}}\)

This is the radius, so the area is

\(\displaystyle A = \pi r^{2} = \pi \cdot \left ( \frac{15}{ \sqrt{3}} \right )^{2}= \frac{225}{3} \pi = 75 \pi\)

The formula for finding the area of a circle is \(\displaystyle \pi r^{2}\). In this formula, \(\displaystyle r\) represents the radius of the circle.  Since the question only gives us the measurement of the diameter of the circle, we must calculate the radius.  In order to do this, we divide the diameter by \(\displaystyle 2\).

\(\displaystyle \frac{15}{2}=7.5\)

Now we use \(\displaystyle 7.5\) for \(\displaystyle r\) in our equation.

\(\displaystyle \pi (7.5)^{2}=176.7146 \: in^{2}\)

First, solve for radius:

\(\displaystyle r=\frac{d}{2}=\frac{6}{2}=3\)

Then, solve for area:

\(\displaystyle A=r^2\pi=3^2\pi=9\pi\)

The area of a circle can be calculated using the formula:

\(\displaystyle Area=\frac{\pi d^2}{4}\),

where \(\displaystyle d\) is the diameter of the circle, and \(\displaystyle \pi\) is approximately \(\displaystyle 3.14\).

\(\displaystyle Area=\frac{\pi d^2}{4}=\frac{\pi\times 4^2}{4}=4\pi \Rightarrow Area\approx 4\times 3.14\Rightarrow Area\approx 12.56 \ cm^2\)

The area of a circle can be calculated using the formula:

\(\displaystyle Area=\frac{\pi d^2}{4}\),

where \(\displaystyle d\)  is the diameter of the circle and \(\displaystyle \pi\) is approximately \(\displaystyle 3.14\).

\(\displaystyle Area=\frac{\pi (4t)^2}{4}=\frac{16\pi t^2}{4}=4\pi t^2 \Rightarrow Area\approx 4\times 3.14\times t^2\)

\(\displaystyle \Rightarrow Area\approx 12.56t^2\)