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30-60-90 Triangles
- Thread starter ATS
- Start date
I know that the Short leg would be 7Square root of 3.
I know that the long side is side x square root of 3,
The length of the longer leg is equal to the length of the shorter leg multiplied by the square root of 3.
and hypotenuse is 2 times the short side.
You stated what the length of the shorter leg is, so just double its length now
to get the length of the hypotenuse.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
A equilateral triangle has all sides of equal length (hence the name) and, so, all angles the same measure. Since the measures of the angles in any triangle must sum to 180 degrees, if they are all equal, they must each be 60 degrees. If you bisect an angle, continuing that bisector to the opposite side, you divide the equilateral triangle into two triangles in which one angle is 60 degrees and another angle is 60/2= 30 degrees. The third angle is, because, again, the measures of the angles sum to 180 degrees, 180- (30+ 60)= 180- 90= 90 degrees.
That is, we have constructed two "30-60-90" right triangles. If the original equilateral triangle has side length "s", then one side of each triangle (the hypotenuse) still has length s. By symmetry, one of the legs (opposite the bisected ngle) has length s/2. By the Pythagorean theorem, the third side has length \(\displaystyle \sqrt{s^2- s^2/4}= \sqrt{3s^2/4}= \frac{\sqrt{3}}{2}s\). \(\displaystyle \frac{\sqrt{3}}{2}\) is clearly larger than \(\displaystyle \frac{\sqrt{1}}{2}= \frac{1}{2}\) and so \(\displaystyle \frac{\sqrt{3}}{2}s\) is the length of the longer leg.
In this problem you are given that the length of the longer leg is 21: \(\displaystyle \frac{\sqrt{3}}{2}s= 21\) so that \(\displaystyle s= \frac{2}{\sqrt{3}}(21)= \frac{42}{\sqrt{3}}\). Rationalize the denominator by multiplying both numerator and denominator by \(\displaystyle \sqrt{3}\): \(\displaystyle \frac{42\sqrt{3}}{3}= 14\sqrt{3}\). That is the length of a side of the original equilateral triangle and so the length of the hypotenuse of the 30-60-90 right triangle. The shorter leg is half of that, \(\displaystyle 7\sqrt{3}\).
That is, we have constructed two "30-60-90" right triangles. If the original equilateral triangle has side length "s", then one side of each triangle (the hypotenuse) still has length s. By symmetry, one of the legs (opposite the bisected ngle) has length s/2. By the Pythagorean theorem, the third side has length \(\displaystyle \sqrt{s^2- s^2/4}= \sqrt{3s^2/4}= \frac{\sqrt{3}}{2}s\). \(\displaystyle \frac{\sqrt{3}}{2}\) is clearly larger than \(\displaystyle \frac{\sqrt{1}}{2}= \frac{1}{2}\) and so \(\displaystyle \frac{\sqrt{3}}{2}s\) is the length of the longer leg.
In this problem you are given that the length of the longer leg is 21: \(\displaystyle \frac{\sqrt{3}}{2}s= 21\) so that \(\displaystyle s= \frac{2}{\sqrt{3}}(21)= \frac{42}{\sqrt{3}}\). Rationalize the denominator by multiplying both numerator and denominator by \(\displaystyle \sqrt{3}\): \(\displaystyle \frac{42\sqrt{3}}{3}= 14\sqrt{3}\). That is the length of a side of the original equilateral triangle and so the length of the hypotenuse of the 30-60-90 right triangle. The shorter leg is half of that, \(\displaystyle 7\sqrt{3}\).