In order to find the area of the triangle, we must first calculate the height of its altitude.  An altitude slices an equilateral triangle into two $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ triangles. These triangles follow a side-length pattern. The smallest of the two legs equals $\displaystyle x$and the hypotenuse equals $\displaystyle 2x$. By way of the Pythagorean Theorem, the longest leg or $\displaystyle h=x\sqrt{3}$.

Therefore, we can find the height of the altitude of this triangle by designating a value for $\displaystyle x$. The hypotenuse of one of the $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ is also the side of the original equilateral triangle.  Therefore, one can say

that $\displaystyle 2x=s=89cm$ and $\displaystyle x=44.5cm$.

$\displaystyle h=x\sqrt{3}$

$\displaystyle h=44.5cm*\sqrt{3}$

Now, we can calculate the area of the triangle via the formula

 $\displaystyle A=\frac{1}{2}bh$

$\displaystyle A= \frac{1}{2} * 89cm * 44.5cm*\sqrt{3}$

$\displaystyle A=3430 cm^{2}$

Now convert to meters.

$\displaystyle \rightarrow \frac{3430cm^{2}}{1}*\frac{1m}{100cm}*\frac{1m}{100cm}= 0.343m^{2}$

The formula for the area of a right triangle is $\displaystyle A = \frac{1}{2}bh$, where $\displaystyle b$ is the length of the triangle's base and $\displaystyle h$ is its height. Since the longest side is the hypotenuse, use the two smaller numbers given as sides for the base and height in the equation to calculate the area of Triangle A:

$\displaystyle A = \frac{1}{2}(6)(8)$

 $\displaystyle A = 24$

The formula for the area of an equilateral triangle is $\displaystyle A = \frac{s^2\sqrt{3}}{4}$, where $\displaystyle s$ is the length of each side. (Alternatively, you can divide the equilateral triangle into two right triangles and find the area of each). Triangle B's area is thus calculated as:

$\displaystyle A = \frac{(8)^2\sqrt{3}}{4}$

$\displaystyle A = \frac{64\sqrt{3}}{4}$

$\displaystyle A = 16\sqrt{3}$ 

To determine which of the two areas is greater without using a calculator, rewrite the areas of the two triangles with comparable factors. Triangle A's area can be expressed as $\displaystyle 24= 8\sqrt{3}\sqrt{3}$, and Triangle B's area can be expressed as $\displaystyle 16\sqrt{3}= (8)(2)\sqrt{3}$. Since $\displaystyle 2$ is greater than $\displaystyle \sqrt{3}$, the product of the factors of Triangle B's area will be greater than the product of the factors of Triangle A's, so Triangle B has the greater area.

Since the area of a triangle is 

$\displaystyle Area=\frac{1}{2}bh$

you need to find the height of the triangle first. Because of the 30-60-90 relationship, you can determine that the height is $\displaystyle 5.5\sqrt{3}$.

Then, multiply that by the base (11).

Finally, divide it by two to get 52.4.

The formula for the area of an equilateral triangle is:

$\displaystyle A=\frac{s^2\sqrt{3}}{4}$

Where $\displaystyle s$ is the length of the side

Plugging in our values, we get:

$\displaystyle A=\frac{s^2\sqrt{3}}{4}$

$\displaystyle A=\frac{(20m)^2\sqrt{3}}{4}=\frac{400m^2\sqrt{3}}{4}=100\sqrt{3}m^2$

The formula for the area of an equilateral triangle is:

$\displaystyle A=\frac{s^2\sqrt{3}}{4}$,

where $\displaystyle s$ is the length of the sides.

Plugging in our value, we get:

$\displaystyle A=\frac{(8cm)^2\sqrt{3}}{4}=\frac{64cm^2\sqrt{3}}{4}=16cm^2\sqrt{3}$

The formula for the perimeter of an equilateral triangle is:

$\displaystyle P = 3 (s)$

Plugging in our values, we get

$\displaystyle 21cm = 3 (s)$

$\displaystyle s = 7cm$

The formula for the area of an equilateral triangle is:

$\displaystyle A = \frac{s^2 \sqrt{3}}{4}$

Plugging in our values, we get

$\displaystyle A = \frac{(7cm)^2 \sqrt{3}}{4} = \frac{49cm^2 \sqrt{3}}{4}$

Note that an equilateral triangle has equal sides and equal angles. The question gives us the length of the base, 5, but doesn't tell us the height. 

If we split the triangle into two equal triangles, each has a base of 5/2 and a hypotenuse of 5. 

Therefore we can use the Pythagorean Theorem to solve for the height:

$\displaystyle a^2+b^2=c^2$

$\displaystyle (\frac{5}{2})^2+b^2=5^2$

$\displaystyle \frac{25}{4}+b^2=25$

$\displaystyle b^2=\frac{75}{4}$

$\displaystyle b=\sqrt{\frac{3\times 25}{2\times 2}}=\frac{5\sqrt{3}}{2}$

Now we can find the area of the triangle:

$\displaystyle Area=\frac{1}{2}\times base\times height=\frac{1}{2}\times 5\times \frac{5\sqrt{3}}{2}=\frac{25\sqrt{3}}{4}$

Recall that an equilateral triangle also obeys the rules of isosceles triangles.   That means that our triangle can be represented as having a height that bisects both the opposite side and the angle from which the height is "dropped."  For our triangle, this can be represented as:

Now, although we do not yet know the height, we do know from our 30-60-90 regular triangle that the side opposite the 60° angle is √3 times the length of the side across from the 30° angle. Therefore, we know that the height is 3√3.

Now, the area of a triangle is (1/2)bh. If the height is 3√3 and the base is 6, then the area is (1/2) * 6 * 3√3 = 3 * 3√3 = 9√(3).

The radius of the circle is 2, from the equation circumference $\displaystyle =2r\pi$. Each triangle is the same, and is equilateral, with side length of 2. The area of a triangle $\displaystyle =\frac{1}{2}b*h$

To find the height of this triangle, we must divide it down the centerline, which will make two identical 30-60-90 triangles, each with a base of 1 and a hypotenuse of 2. Since these triangles are both right traingles (they have a 90 degree angle in them), we can use the Pythagorean Theorem to solve their height, which will be identical to the height of the equilateral triangle.

$\displaystyle a^{2}+b^{2}=c^{2}$

We know that the hypotenuse is 2 so $\displaystyle 2^{2}=4$. That's our $\displaystyle c^{2}$ solution. We know that the base is 1, and if you square 1, you get 1.

Now our formula looks like this: $\displaystyle a^{2}+1=4$, so we're getting close to finding $\displaystyle a$.

Let's subtract 1 from each side of that equation, in order to make things a bit simpler: $\displaystyle a^{2}+0=3 \rightarrow a^{2}=3$

Now let's apply the square root to each side of the equation, in order to change $\displaystyle a^{2}$ into $\displaystyle a$: $\displaystyle a=\sqrt{3}$

Therefore, the height of our equilateral triangle is $\displaystyle \sqrt{3}$

To find the area of our equilateral triangle, we simply have to multiply half the base by the height: $\displaystyle \frac{2}{2}*\sqrt{3}=1*\sqrt{3}=\sqrt{3}$

The area of our triangle is $\displaystyle \sqrt{3}$

The altitude, $\displaystyle \dpi{100} h$, divides the equilateral triangle into two $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ right triangles and divides the bottom side in half.  

In a $\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}$ right triangle, the sides of the triangle equal $\displaystyle x$, $\displaystyle x\sqrt{3}$, and $\displaystyle 2x$. In these equations $\displaystyle x$ equals the length of the smallest side, which in our triangle is $\displaystyle \frac{s}{2}$ or $\displaystyle \frac{\sqrt{3}}{2}$.  

In this scenario:

$\displaystyle h=x\sqrt{3}$

and

$\displaystyle x=\frac{\sqrt{3}}2{}$

Therefore, 

$\displaystyle h=\frac{\sqrt{3}}{2}*\frac{\sqrt{3}}{1}$ 

$\displaystyle \rightarrow\frac{3}{2}$