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Current time:0:00Total duration:9:06

Video transcript

what I want to do in this video is come up with a relationship between the area of a triangle and that triangles circumscribed circle or circum circle so before we even think about the circum circle let's just think about the area of the triangle so let's say the triangle let's say that the triangle looks something looks something like let's say that it looks something like that actually I don't wanna make it look isosceles so let me make it a little bit make let's look at let's make it so it doesn't Harbor it doesn't look like any particular type of triangle and let's call this triangle capital a capital B capital C that's the vertices and then the lengths of the side opposite capital a is lowercase a lowercase B over here and then lowercase C now we know how to calculate the area of this triangle if we know its height so we can drop an altitude right here we can drop an altitude right here and if this altitude has length H we know that the area we know that the area of a b c and writing ABC with the brackets around it means the area of triangle ABC is equal to 1/2 times the base which is lowercase b times the height times the height fair enough so we have an expression for the area let's see if we can somehow relate some of these things or the area to the radius of this triangles circumscribed circle and so a circumscribed circle is a circle that passes through all of the vertices of the triangle and every triangle has a circumscribed circle so let me try to draw it this is the hard part right over here so let me it might look something might look something like this that's fair enough that's close enough to a circle I think you get the general idea that is the circum circle for this triangle or this triangles circumscribed circumscribed circle let me label it this is this is that what we're looking arrow this is the circum circle circum circle for this trying now let's think about the center of that circumcircle sometimes referred to as the circumcenter so it looks like it would be sitting I don't know just eyeballing it right on this little B here so that is the circumcenter of the circle and let's draw a diameter through that circumcenter so let's draw a diameter from vertex B through that circumcenter so then we go there and then we just keep we just keep going over here and let's call this point over here let's call that point over there D now let's create a triangle with vertices a B and D so we can just draw another line right over here and we have triangle a B D now we proved in the geometry playlist and it's not actually a crazy proof at all that any inscribed draw any triangle that's inscribed in a circle where one of the sides of the triangle is the hype is that is a diameter of the circle then that is going to be a right triangle and that's the angle that is going to be 90 degrees is the angle opposite the diameter so this is the right angle right here and you can derive that pretty pretty straight forward you have this arc here you have an arc here that is 180 degrees that is 180 degrees because obviously this is the diameter and it subtends this inscribed angle and we've also proved that an inscribed angle that's subtended by an arc will be half of the arc length this is this is 180 degree arc so this will be a 90 degree angle so either way this is going to be 90 degrees over there now the other thing that we see is we have this arc right over here that I'm drawing in magenta the arc that goes from A to B well that arc subtends two different angles and are drawing it subtends this angle right over your angle ACB it subtends that right over there but it also subtends angle a B that's why we constructed it this way so it also subtends this so these two angles are going to be congruent they're both going to have half the degree measure of this arc over here because they're both inscribed angles subtended by the same exact arc now something interesting is popping up we have two triangles here we have triangle abd and triangle B and triangle B I don't know we could call this e and triangle B EC they have two angles and there's names they have a right angle and they have this magenta angle right angle in this magenta angle so their third angle must be the same I'll do it in yellow their third angle this angle must be congruent to that angle so they have three angles that are the same they must be similar triangles or the ratio between corresponding sides must be the same so we can use that information now to relate the length of this side which is really diameter it's 2 times the radius to the height of this smaller triangle and we know relationship between the height of the small triangle in the area and we are essentially in the homestretch so let's do that so these are two similar triangles so we know that the ratio we know that the ratio of C there we know that the ratio of C C to this diameter right over here but what's the length of the diameter the length of the diameter is 2 times the radius this is a radius and this is a radius right here we know that the ratio of C times C to 2 times the radius is going to be the same exact thing as the ratio of H as the ratio of H and we want to make sure we're using the same side to the hypotenuse of that triangle to the ratio of H to a and the way we figured that out the way we figured out we looked at corresponding side C and the hypotenuse are both quart are both the sides adjacent to this angle right over here so we have H and a over here so C is 2 2 R as H is 2 a or we could solve well we can do a lot of things so we could 1 we could just substitute we can solve for H over here and then substitute this and substitute an expression that has the area in for H actually let's just do that so let's say so if we use this first expression that we have for the area we could we can multiply both sides by 2 we can multiply both sides by 2 and divide both sides by B and divide both sides by B that cancels with that that cancels with that we get that H is equal to 2 times the area over B so we can rewrite this little relationship as C over 2 R is equal to H which is 2 times 2 times the area of our triangle 2 times our the area of our triangle over V and then all of that is going to be all of that is going to be over a or we could rewrite this second part right over here as 2 times the area 2 times the area over we're dividing by B and then dividing by a that's just the same thing as dividing by dividing by a B and so we can ignore this right here so we have C over 2 R is equal to 2 times the area over a B and now we can cross multiply we could have a B times C I'll do a new color just to ease the monotony a B times C a B times C a B times C is going to be equal to 2 R times 2 ABC so that's going to be 4 R times the area of our triangle I just cross multiply this times this is going to be equal to that times that and we know that all cross multiplication is is just multiplying both sides of the equation by 2r and multiplying both sides of the equation by a B so we did that on the left hand side we also have to do it on the right hand side to R and a B obviously that cancels with that that cancels with that so you get ABC is equal to 2r times 2 ABC or 4 R times the area of our triangle and now we're at the home stretch we divide both sides of this you divide both sides of this by 4 times the area 4 times 4 times the area and we're done this cancels with that that cancels with that and we have our relationship the radius the radius of what we could call the circum radius the radius of this triangle circumscribed circle is equal to the product of the sides of the triangle divided by four times four times the area four times the area of the triangle that's a pretty neat result