To solve this equation, first note that a line drawn from the origin to a vertex of the equilateral triangle will bisect the angle of the vertex. Furthermore, the length of this line is equal to the radius:

That this creates in turn is a 30-60-90 right triangle. Recall that the ratio of the sides of a 30-60-90 triangle is given as:

\(\displaystyle (1:\sqrt{3}:2)\)

Therefore, the length of the \(\displaystyle 60^{\circ}\) side can be found to be

\(\displaystyle \frac{10}{2}\sqrt{3}=5\sqrt{3}\)

This is also one half of the base of the triangle, so the base of the triangle can be found to be:

\(\displaystyle b=2\cdot 5\sqrt{3}=10\sqrt{3}\)

Furthermore, the length of the \(\displaystyle 30^{\circ}\) side is:

\(\displaystyle \frac{10}{2}=5\)

The vertical section rising from the origin is the length of the radius, which when combined with the shorter section above gives the height of the triangle:

\(\displaystyle h=5+10=15\)

The area of a triangle is given by one half the base times the height, so we can find the answer as follows:

\(\displaystyle Area=\frac{1}{2}bh=\frac{1}{2}(10\sqrt{3})(15)=75\sqrt{3}\)

All sides of an equilateral triangle are equal, so all sides of this triangle equal 4.  

Area = 1/2 base * height, so we need to calculate the height: this is easy for an equilateral triangle, since you can bisect any such triangle into two identical 30:60:90 triangles.

The ratio of lengths of a 30:60:90 triangle is 1:√3:2. The side of the equilateral triangle is 4, and we divided the base in half when we bisected the triangle, so that give us a length of 2, so our triangle must have sides of 2, 4, and 2√3; thus we have our height.

One of our 30:60:90 triangles will have a base of 2 and a height of 2√3. Half the base is 1, so 1 * 2√3 = 2√3.

We have two of these triangles, since we divided the original triangle, so the total area is 2 * 2√3 = 4√3.

You can also solve for the area of any equilateral triangle by applying the formula (s²√3)/4, where s = the length of any side.

To find the area of an equilateral triangle, notice that it can be divided into two \(\displaystyle 30-60-90\) triangles:

The ratio of sides in a \(\displaystyle 30-60-90\) triangle is \(\displaystyle 1-\sqrt{3}-2\), and since the triangle is bisected such that the \(\displaystyle 30\) degree side is \(\displaystyle \frac{s}{2}\), the \(\displaystyle 60\) degree side, the height of the triangle, must have a length of \(\displaystyle \frac{\sqrt{3}}{2}s\).

The formula for the area of the triangle is given as:

\(\displaystyle Area =\frac{1}{2}base*height\)

So the area of an equilateral triangle can be written in term of the lengths of its sides as:

\(\displaystyle \frac{1}{2}s\frac{\sqrt{3}s}{2}=\frac{\sqrt{3}}{4}s^2\)

For this particular triangle, since \(\displaystyle s=1\), its area is equal to \(\displaystyle \frac{\sqrt{3}}{4}\).

\(\displaystyle \frac{\sqrt{3}}{4}< \frac{1}{2}\)

If the relation between ratios is hard to visualize, realize that \(\displaystyle \frac{\sqrt{4}}{4}=\frac{2}{4}=\frac{1}{2}\)

The area of an equilateral triangle is \(\displaystyle \frac{s^2\sqrt{3}}{4}\).

So let's set-up an equation to solve for \(\displaystyle s\). 

\(\displaystyle \frac{s^2\sqrt{3}}{4}=6\sqrt{3}\) Cross multiply.

\(\displaystyle s^2\sqrt{3}=24\sqrt{3}\) 

The \(\displaystyle \sqrt{3}\) cancels out and we get \(\displaystyle s^2=24\).

Then take square root on both sides and we get \(\displaystyle 2\sqrt{6}\). To find height, we need to realize by drawing a height we create \(\displaystyle 2\) \(\displaystyle 30-60-90\) triangles. 

The height is opposite the angle \(\displaystyle 60\). We can set-up a proportion. Side opposite \(\displaystyle 60\) is \(\displaystyle \sqrt{3}\) and the side of equilateral triangle which is opposite \(\displaystyle 90\) is \(\displaystyle 2\).

\(\displaystyle \frac{h}{\sqrt{3}}=\frac{2\sqrt{6}}{2}\) Cross multiply.

\(\displaystyle 2\sqrt{18}=2h\) Divide both sides by \(\displaystyle 2\)

\(\displaystyle h=\sqrt{18}\) 

We can simplify this by factoring out a \(\displaystyle \sqrt{9}\) to get a final answer of \(\displaystyle 3\sqrt{2}\). 

This problem requires a bit of creative thinking (unless you have memorized the fact that an equilateral triangle always has an area equal to its side length times \(\displaystyle \sqrt{3}\).

Consider the equilateral triangle:

Since this kind of triangle is a species of isoceles triangle, we know that we can drop down a height from the top vertex. This will create two equivalent triangles, one of which will look like:

This gives us a 30-60-90 triangle. We know that for such a triangle, the ratio of the side across from the 30-degree angle to the side across from the 60-degree angle is:

\(\displaystyle \frac{1}{\sqrt{3}}\)

We can also say, given our figure, that the following equivalence must hold:

\(\displaystyle \frac{1}{\sqrt{3}} = \frac{4}{x}\)

Solving for \(\displaystyle x\), we get:

\(\displaystyle x = 4\sqrt{3}\)

Now, since \(\displaystyle \sqrt{3} < \sqrt{4}\), we know that \(\displaystyle \sqrt{3}\) must be smaller than \(\displaystyle 2\). This means that \(\displaystyle 4\sqrt{3} < 4 * 2\) or \(\displaystyle 4\sqrt{3} < 8\). Quantity B is larger than quantity A.

If the perimeter of our equilateral triangle is \(\displaystyle 27\), each of its sides must be \(\displaystyle \frac{27}{3}\) or \(\displaystyle 9\). This gives us the following figure:

Since this kind of triangle is a species of isoceles triangle, we know that we can drop down a height from the top vertex. This will create two equivalent triangles, one of which will look like:

This gives us a 30-60-90 triangle. We know that for such a triangle, the ratio of the side across from the 30-degree angle to the side across from the 60-degree angle is:

\(\displaystyle \frac{1}{\sqrt{3}}\)

Therefore, we can also say, given our figure, that the following equivalence must hold:

\(\displaystyle \frac{1}{\sqrt{3}} = \frac{4.5}{x}\)

Solving for \(\displaystyle x\), we get:

\(\displaystyle x = 4.5\sqrt{3}\)

Now, since \(\displaystyle \sqrt{3} < \sqrt{4}\), we know that \(\displaystyle \sqrt{3}\) must be smaller than \(\displaystyle 2\). This means that \(\displaystyle 4.5\sqrt{3} < 4.5 * 2\) or \(\displaystyle 4.5\sqrt{3} < 9\)

Therefore, quantity B is larger than quantity A.

The internal angles of a triangle must add up to 180. 180 - 75 -60= 45. 

To find the perimeter, add up the sides, here 5 + 12 + 13 = 30. To find the area, multiply the two legs together and divide by 2, here (5 * 12)/2 = 30.

If B is a midpoint of AC, then we know AB is 12. Moreover, triangles ACE and ABD share angle DAB and have right angles which makes them similar triangles. Thus, their sides will all be proportional, and BD is 4. ¹/₂bh gives us ¹/₂ * 12 * 4, or 24.

Area = 1/2(base)(height). You could use Pythagorean theorem to find the height or, if you know the special right triangles, recognize the 5-12-13. The area = 1/2(12)(5) = 30.