To find perimeter of an quilateral triangle, simply multiply the side length by $\displaystyle 3$. Thus,

$\displaystyle P=s\cdot3=2.1\cdot3=6.3$

To solve, simply multiply the side length by $\displaystyle 3$. Thus,

$\displaystyle P=3s=3*4=12$

To solve, simply multiply the side length by 3 since they are all equal. Thus,

$\displaystyle P=3*s=3*2=6$

An equilateral triangle has three congruent sides, and is also an equiangular triangle with three congruent angles that each meansure 60 degrees.

To find the height we divide the triangle into two special 30 - 60 - 90 right triangles by drawing a line from one corner to the center of the opposite side. This segment will be the height, and will be opposite from one of the 60 degree angles and adjacent to a 30 degree angle. The special right triangle gives side ratios of $\displaystyle x$, $\displaystyle x\sqrt{3}$, and $\displaystyle 2x$. The hypoteneuse, the side opposite the 90 degree angle, is the full length of one side of the triangle and is equal to $\displaystyle 2x$. Using this information, we can find the lengths of each side fo the special triangle.

$\displaystyle 8=2x\rightarrow 4=x\rightarrow 4\sqrt{3}=x\sqrt{3}$

The side with length $\displaystyle x\sqrt{3}$ will be the height (opposite the 60 degree angle). The height is $\displaystyle 4\sqrt{3}$ inches.

Draw a vertical line from the vertex. This will divide the equilateral triangle into two congruent right triangles. For the hypothenuse of one right triangle, the length will be $\displaystyle 2\:cm$. The base will have a dimension of $\displaystyle 1\:cm$.  Use the Pythagorean Theorem to solve for the height, substituting in $\displaystyle 2$ for $\displaystyle c$, the length of the hypotenuse, and $\displaystyle 1$ for either $\displaystyle a$ or $\displaystyle b$, the length of the legs of the triangle:

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

$\displaystyle 1^2+b^2=2^2$

$\displaystyle b^2=2^2-1^2$

$\displaystyle b^2 = 3$

$\displaystyle b=\sqrt3$

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 $\displaystyle 30-60-90$ triangles. Therefore, you can create a ratio to help you find $\displaystyle h$.

The ratio of $\displaystyle 6$ to $\displaystyle h$ is the same as the ratio of $\displaystyle 1$ to $\displaystyle \sqrt{3}$.

As an equation, this is written:

$\displaystyle \frac{6}{h}=\frac{1}{\sqrt{3}}$

Solving for $\displaystyle h$, we get: $\displaystyle h=6\sqrt{3}$

To calculate the height of an equilateral triangle, first draw a perpendicular bisector for the triangle. By definition, this splits the opposite side from the vertex of the bisector in two, resulting in two line segments of length $\displaystyle 4$ inches. Since it is perpendicular, we also know the angle of intersection is $\displaystyle 90^{\circ}$. 

So, we have a new right triangle with two side lenghts $\displaystyle 8$ and $\displaystyle 4$ for the hypotenuse and short leg, respectively. The Pythagorean theorem takes over from here:

$\displaystyle 8^2 - 4^2 = x^2$ ---> $\displaystyle x^2 = 48 = 4\sqrt{3}$

So, the height of our triangle is $\displaystyle 4\sqrt{3}$.

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:

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

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 $\displaystyle 30-60-90$ triangles. Therefore, you can create a ratio to help you find $\displaystyle h$.

The ratio of $\displaystyle 10$ to $\displaystyle h$ is the same as the ratio of $\displaystyle 1$ to $\displaystyle \sqrt{3}$.

As an equation, this is written:

$\displaystyle \frac{10}{h}=\frac{1}{\sqrt{3}}$

Solving for $\displaystyle h$, we get: $\displaystyle h=10\sqrt{3}$

Now, the area of the triangle is merely $\displaystyle \frac{1}{2}bh$.  For our data, this is: $\displaystyle \frac{1}{2}*20*10\sqrt{3}$ or $\displaystyle 100\sqrt{3}$.

Notice that this is the same as $\displaystyle \frac{b^2\sqrt{3}}{4}$.  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 $\displaystyle \frac{51}{3}$ or $\displaystyle 17$. 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 $\displaystyle 30-60-90$ triangles. Therefore, you can create a ratio to help you find $\displaystyle h$.

The ratio of $\displaystyle 8.5$ to $\displaystyle h$ is the same as the ratio of $\displaystyle 1$ to $\displaystyle \sqrt{3}$.

As an equation, this is written:

$\displaystyle \frac{8.5}{h}=\frac{1}{\sqrt{3}}$

Solving for $\displaystyle h$, we get: $\displaystyle h=8.5\sqrt{3}$

Now, the area of the triangle is merely $\displaystyle \frac{1}{2}bh$.  For our data, this is: $\displaystyle \frac{1}{2}*17*8.5\sqrt{3}$ or $\displaystyle 72.25\sqrt{3}$.