We derive the height formula from the area of the triangle formula:

\(\displaystyle A=\frac{(bh)}2{}\)

\(\displaystyle h=\frac{2A}{b}\)

\(\displaystyle =\frac{(2*21)}{3}\)

\(\displaystyle =14\)

The area of an equilateral triangle with sidelength \(\displaystyle s = 20\) is 

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

Using this area for \(\displaystyle A\) and 20 for \(\displaystyle b\) in the general triangle formula, we can obtain \(\displaystyle h\):

\(\displaystyle \frac{1}{2} bh = A\)

\(\displaystyle \frac{1}{2}\cdot 20 \cdot h =100\sqrt{3}\)

\(\displaystyle 10 \cdot h =100\sqrt{3}\)

\(\displaystyle 10 \cdot h \div 10 =100\sqrt{3}\div 10\)

\(\displaystyle h=10\sqrt{3}\)

The height of an upright equilateral triangle is the perpendicular distance from the center of its base to its top. We can imagine that this line cuts the equilateral triangle into two congruent right triangles whose height is half the length of the original base and whose hypotenuse is the original side length. In these two congruent triangles, their base, which is the height of the equilateral triangle, is the only unknown side length, so we can use the Pythagorean theorem to solve for it:

\(\displaystyle a^2+b^2=c^2\)

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

\(\displaystyle b^2=5^2-(\frac{5}{2})^2=\frac{75}{4}\)

\(\displaystyle b=\sqrt{\frac{75}{4}}=\frac{\sqrt{3*25}}{2}=\frac{5\sqrt{3}}{2}\)

We know the length of the side, therefore we can use the formula for the height in an equilateral triangle:

 \(\displaystyle h=s\frac{\sqrt{3}}{2}\), where \(\displaystyle s\) is the length of a side and \(\displaystyle h\) the length of the height.

Therefore, the final answer is \(\displaystyle \frac{3\sqrt{3}}{2}\).

To solve, we must use pythagorean's theorem given that we know the hypotenuse is \(\displaystyle 6\) and one side length is \(\displaystyle 3\) \(\displaystyle \left (\frac{1}{2} \textup{ of } 6 \right )\). Therefore:

\(\displaystyle c^2=a^2+b^2\Rightarrow 6^2=a^2+3^2\Rightarrow 27=a^2\)

\(\displaystyle a=\sqrt{27}=\sqrt{3^2*3}=3\sqrt3\)

The area is given, which will allow us to calculate the side of the triangle and hence we can also find the perimeter.

The area for an equilateral triangle is given by the formula 

\(\displaystyle \frac{s^{2}\sqrt{3}}{4}\), where \(\displaystyle s\) is the length of the side of the triangle.

Therefore, \(\displaystyle s=\sqrt{\frac{4a}{\sqrt{3}}}\), where \(\displaystyle a\) is the area.

Thus \(\displaystyle s=\sqrt{\frac{28}{\sqrt{3}}}\), and the perimeter of an equilateral triangle is three times the side, hence, the final answer is \(\displaystyle 3\sqrt{\frac{28}{\sqrt{3}}}\).

An equilateral triangle with a side length \(\displaystyle x\) has a perimeter \(\displaystyle P=3x\).

Given: 

\(\displaystyle x=9cm\), 

\(\displaystyle P=3(9cm)=27cm\). 

An equilateral triangle with a side length \(\displaystyle x\) has a perimeter \(\displaystyle P=3x\).

Given: 

\(\displaystyle x=17cm\), 

\(\displaystyle P=3(17cm)=51cm\). 

An equilateral triangle with a side length \(\displaystyle x\) has a perimeter \(\displaystyle P=3x\).

Given: 

\(\displaystyle x=21cm\), 

\(\displaystyle P=3(21cm)=63cm\). 

The trick to this one is to carefully put together what you are given.

We know that two of our angles are equal to 60 degrees. This means that the last angle is also 60 degrees. This make HFT an equilateral triangle.

Equilateral triangles always have equal sides and equal angles, so our last side has to be 7 light years as well.