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:

Now we can find the area of the triangle:

While you can very quickly compute the area of an equilateral triangle by using a shortcut formula, it is best to understand how to analyze a triangle like this for other problems. Let's consider this. The shortcut will be given below.

Recall that from any vertex of an equilateral triangle, you can drop a height that is a bisector of that vertex as well as a bisector of the correlative side. This gives you the following figure:

Notice that the small triangles within the larger triangle are both  triangles. Therefore, you can create a ratio to help you find .

The ratio of  to  is the same as the ratio of  to .

As an equation, this is written:

Solving for , we get: 

Now, the area of the triangle is merely .  For our data, this is:  or .

Notice that this is the same as .  This is a shortcut formula for the area of equilateral triangles.

Since an equilateral triangle is comprised of sides having equal length, we know that each side of this triangle must be  or . While you can very quickly compute the area of an equilateral triangle by using a shortcut formula, it is best to understand how to analyze a triangle like this for other problems. Let's consider this. The shortcut will be given below.

Recall that from any vertex of an equilateral triangle, you can drop a height that is a bisector of that vertex as well as a bisector of the correlative side. This gives you the following figure:

Notice that the small triangles within the larger triangle are both  triangles. Therefore, you can create a ratio to help you find .

The ratio of  to  is the same as the ratio of  to .

As an equation, this is written:

Solving for , we get: 

Now, the area of the triangle is merely .  For our data, this is:  or .

Draw out 3 radii and 3 lines to the corners of each triangle, creating 6 30-60-90 triangles.  

See that these 30-60-90 triangles can be used to find side length. 

Formula for side of equilateral triangle is

.

Now substitute the new equation that is in terms of R in for S.

Draw 3 radii, and then 3 new lines that bisect the radii. You get six 30-60-90 triangles.  

These triangles can be used ot find a side length .  

Using the formula for the area of an equilateral triangle in terms of its side, we get

The radius of the circle is 2, from the equation circumference . Each triangle is the same, and is equilateral, with side length of 2. The area of a triangle 

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.

We know that the hypotenuse is 2 so . That's our  solution. We know that the base is 1, and if you square 1, you get 1.

Now our formula looks like this: , so we're getting close to finding .

Let's subtract 1 from each side of that equation, in order to make things a bit simpler: 

Now let's apply the square root to each side of the equation, in order to change  into : 

Therefore, the height of our equilateral triangle is 

To find the area of our equilateral triangle, we simply have to multiply half the base by the height: 

The area of our triangle is 

From the given information, we can draw the following picture:

Since this is an equilateral triangle, all three of its angles will be congruent. All three of its sides will also be congruent.

The dotted line, representing the height, bisects not only the base of the triangle but also the apical angle. This means that the two smaller right triangles created by the dotted line have bases that are half the side length of the equilateral triangle with apical angle measures of . 

This creates two smaller 30/60/90 special right triangles. This problem can be solved in one of two ways:

1. You can use the derived side ratio for 30/60/90 triangles, , to solve for the length of one of the equilateral triangle's sides.

OR

2. You can use trig functions to solve for the length of one of the equilateral triangle's sides.

Using trig functions, we can use  as our angle of interest, with the length  and the hypotenuse being our mystery side.

Going back to SOH CAH TOA, we can determine that we will be using sine because we have the information available for the side opposite ("O") of our angle of interest and are looking for the hypotenuse ("H"). This means we will be using the "SOH" part of SOH CAH TOA—sine.

Substituting in the problem's values, we can solve for the length of the hypotenuse of one of the two smaller triangles, which is the length of one side of the equilateral triangle:

Remember that the problem asks for the triangle's perimeter. That means that we need to multiply the length of one of its edges by three to find the final answer:

For any triangle,  .

We know our height is  and our area is .

Therefore  so .

Since this triangle is equilateral, all of the sides (including the base) are equal.

Use the formula for area of a triangle to solve for the base:

In this case:

So the base is equal to 

Since this is an equilateral triangle, all of the sides and all of the angles would be equal (60 degrees each since the 3 angles must add up to 180 degrees).

1. The height splits the triangle's base in half and creates two right triangles. Create expressions for the length of the two unknown sides in this new triangle:

The value we are looking for is the side length so we'll make that   

The base of the new triangle is half the length of a side so we'll make that    

2. Use the Pythagorean Theorem to find the value of  or the side length:

In this case:

3. Multiply the side length you found by 3 to get the perimeter: