If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:6:59

Special right triangles proof (part 1)

CCSS.Math:

Video transcript

what I want to do in this video is discuss a special class of triangles called 30-60-90 triangles and I think you know why they're called this the measures of its angles are 30 degrees 60 degrees and 90 degrees and what we're going to prove in this video this tends to be a very useful result at least for a lot of what you see in a geometry class and then later on in trigonometry class is the ratios between the sides of a 30-60-90 triangle that if the hypotenuse has length x so that's remember the hypotenuse is opposite the 90-degree side if the hypotenuse has length X what we're going to prove is that the shortest side which is opposite the 30-degree side has length x over 2 and that the 60-degree side the 60-degree side or the side that's opposite the 60-degree angle I should say is going to be square root of 3 times the shortest side so square root of 3 times x over 2 that's going to be its length so that's what we're going to prove in this video and other videos we're going to apply this we're going to show this is actually pretty useful result now let's start with a triangle that we're very familiar with so let me draw ourselves an equilateral triangle so drawing the triangles is always the hard part so this is my best shot at an equilateral triangle let's call so let's call this a B C I'm just going to assume that I've constructed an equilateral triangle so triangle a b c is equilateral and if it's equilateral that that means all of its sides are equal and let's say that it has length and let's say equilateral with sides of length of length X so this is going to be X this is going to be X and this is going to be X we also know based on what we've seen from equilateral triangles before that all the measures of all of these angles are going to be 60 degrees so this is going to be 60 degrees this is going to be 60 degrees and then this is going to be 60 degrees now what I'm going to do is I'm going to drop an altitude from this top point right over here so I'm going to drop an altitude right down and by definition when I'm constructing an altitude it's going to intersect the base right here at right angle so that's going to be a right angle and then this is going to be a right angle and it's a pretty straightforward proof to show that this is not only not only is this an altitude not only is it perpendicular to the space but it's a pretty straightforward proof to show that it bisects the base and you could pause it if you like and prove it yourself but it really comes out of the fact that it's easy to prove that these two triangles are congruent so let me prove it for you so let's call this point D right over here so triangles abd and BD see they clearly both share this side so this side is common to both of them right over here and then we could do whatever that we have they this angle right over here is congruent to this angle over there this angle right over here is congruent to this angle over here and so if these two are congruent to each other then the third angle has to be congruent to each other so this angle right over here needs to be congruent to that angle right over there so these two are congruent and so you can use actually a variety of our of our congruence postulates we could say side-angle-side side-angle-side congruence we could use we could use angle side angle any of those to show that triangle triangle a B D is congruent is congruent to triangle C C BD and what that does for us and we could use R as I said we could use angle side angle or side angle side whatever we like to use for this what that does for us is it tells us that the corresponding sides of these triangles are going to be equal in particular in particular the length of in particular ad is going to be equal to CD ad is going to be equal to CD these are corresponding sides so these are going to be equal to each other and if we know that they're equal to each other and they add up to X remember this was an equilateral triangle of length X we know that this side right over here is going to be x over 2 we know this is going to be x over 2 not only do we know that but we also knew when we when we drop this altitude we showed that this angle has to be congruent to that angle and their measures have to add up to 60 so if two things are the same they add up to 60 this is going to be a 30 degrees and this is going to be 30 degrees so we've already shown one of the interesting parts of a 30-60-90 triangle that if the hypotenuse notice and I guess I didn't point this out by dropping this altitude I've essentially split this equilateral triangle into two 30-60-90 triangles 30-60-90 triangles and so we've already shown that if the side opposite the 90 degree side is X that the side opposite the 30-degree side is going to be x over two that's what we should showed right over here now we just have to come up with the third side the side that is opposite the 60-degree side the side that is opposite the 60-degree side right over there and let's call that let's call that length let's call well I'll just use I'll just use the letters that we already don't have here this is BD and we can just use the Pythagorean theorem right here BD squared plus this length right over here squared plus x over 2 squared is going to be equal to the hypotenuse squared so we get BD BD squared plus x over 2 squared this is just straight out of the Pythagorean theorem plus x over 2 squared is going to equal this hypotenuse squared it's going to equal x squared and just to be clear I'm looking at this triangle right here I'm looking at this triangle right over here on the right and I'm just applying the Pythagorean theorem this side squared plus this side squared is going to equal the hypotenuse squared and let's solve now for BD you get be d squared plus x squared over 4 x squared over 4 is equal to x squared is equal to x squared you could view this as 4 x squared over 4 that's the same thing obviously as x squared and then if we subtract if you distract one if you subtract 1/4 x squared from both sides or x squared over 4 from both sides you get B d squared is equal to 4x squared over 4 minus 2 x squared over 4 is going to be 3x squared over 4 so we're going to be 3x squared over 4 take the principal root of both sides you get BD is equal to is equal to the square root of 3 times X times X principal root of 3 is square root of 3 principal root of x squared is just x over the principal root of 4 which is two and BD is a side opposite the 60-degree side so we're done if this hypotenuse is X the side opposite the 30-degree side is going to be x over two and the side opposite the 60-degree side is going to be square root of three over two times X square root of 3x over two depending on how you want to view it