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:18:18

Trig challenge problem: area of a hexagon

Video transcript

let a equal zero zero and B equal lowercase B to be points on the coordinate plane let ABCDE F be a convex equilateral hexagon so convex means that it's not concave a concave hexagon would look like this so that's two sides three four five six this would be a concave hexagon so it's going to be popped out and all the sides are going to be equal so it's an equilateral hexagon they're not telling us that it's a regular hexagon so we don't know that all of the angles are going to be the same but all the sides will be the same such that F a B is equal to hundred twenty degrees then they show us a bunch of sides that are parallel to each other and then the y-coordinates of its vertices are distinct elements of the set 0 2 4 6 8 10 the area of the hexagon can be written in the form M square roots of n where m and n are positive integers and n is not divisive divisible by the square of any prime that's just a fancy way of saying that we've simplified this radical as much as possible find M plus n so really the first part let's just make sure we can visualize this hexagon so let me draw we know one point one vertex for short 0 0 so let me draw my x-axis that is my x-axis right over there and then my y-axis my y-axis would look like that y-axis we know that the vertex a sits at the point 0 0 that is vertex a now we know that all of the vertices have Y coordinates that are either 0 2 4 6 8 10 and they are distinct members of the set which mean no tour the vertices say share the same y-coordinate so they're not going to be on the same horizontal line so let me draw these horizontal lines we the x-axis is 0 then you have y is equal to 2 then you have 4 then you have 6 and then you have 8 and then you have 10 up here now V we already know so first of all we've already used up at the zero for a a is already using up the 0 B uses up the 2 they tell us that the y-coordinate of B is 2 the y-coordinate of B is 2 so we use that as well and let me see if I can draw B over here let me see if I can draw it sits on this horizontal someplace and we the the D the hexagon has side length s where we don't know what that length is but they're all the same so let's just call this s and it's going to help me think about it now that I know that since this equilateral lateral hexagon all of the sides are going to be the same length and so we're going to go out here we're going to go out here to the coordinate b comma - we don't know what B is but that is our vertex B now F is the other vertex that is connected to a F cannot sit on this horizontal cannot sit on Y is equal to two it can't sit on Y is equal to six because then this distance would be super far clearly much further than this distance over here or actually you could have that but then you would have you wouldn't be able to draw really a convex hexagon so Y the next vertex is just going to have to sit on this horizontal so it's going to be s away so it's going to be s away maybe it will be something like that something like that so let me draw it so that is the next vertex that is vertex F right because we're going ABCDEF and then back to a fair enough now what about vertex C well vertex C can't be on the four horizontal so it's going to have to be on the six horizontal so vertex C is going to have to be someplace like that someplace like that that's vertex C and once again that length is s this length is s now what about vertex E can't be on the sixth horizontal already taken up by vertex C so the four and the six already taken up so it has to be at that at the eight horizontal and so this is length s and we also know that we're going back to the origin now so this is vertex E right here we know that we're going back to the origin on to the origin we're going back to the same x-value this is going to be on the y-intercept and the reason why we know that is this is length s and this is length s and they both had to both of these diagonals travel the same they travel the same vertical distance this base is four this base is four so you can kind of view this as two right triangles both of them have base four and hypotenuse s and so they share this side right over here so they both this one goes out to the left that distance and then this one's going to have to come back that distance now by the same logic over here this guy is going to have to come back so he's we can now use the ten coordinate the ten y at the ten y horizontal or the y coordinate of ten that's the only one we haven't used yet for D and since we came out when we had a diagonal of length s traveling four up this time same logic we had a diagonal of length s it traveled up four over here and it moved out this distance when we go back in the other direction and traveling up four you're going to go back in the same direction so this is going to be directly on top of B so this the coordinate for D is actually going to be B comma 10 the y-coordinate here is 10 and there we have our hexagon we're done drawing our actual hexagon and all this parallel line information they told us AV is parallel to de so a B is parallel to de and this is kind of obvious here BC is parallel to EF BC is parallel to EF BC is parallel to EF and then they say CD is parallel to FA so CD is parallel to FA and the way that we drew it it looks pretty clear that that is the case now we need to find the area we need to find the area of this hexagon and it seems like a good starting point would be to figure out what s is and to figure out what s is it's really going to be a function of how much we've inclined this thing so let's let's draw and you can see that this isn't an equiangular hexagon that this is kind of skewed it's kind of we distorted it a little bit but all the sides are the same length so let's just call this theta let's call that angle right over there theta and then they tell us that angle FA b is 120 degrees F a B is 120 degrees that is 120 degrees so this angle over here on the left this angle over here on the left is going to be 180 minus 120 minus theta so 180 minus 120 is 60 so this angle over here is 60 - theta now the reason why I did that because we have some information we know that we traveled up for over here and we know that we traveled up to over here and maybe we can use that information to solve for s because s is the hypotenuse of both of these right triangles that I just constructed let me draw them so this right triangle right over here I could draw it like this I could draw it like this so I have s I have theta and I have two that's this right triangle right over here this right triangle looks like this it looks like this this is this angle is 60 minus theta and this height over here is 4 so let's see what we can do to solve for s this triangle on the left or it was on the right over here this triangle says if you take the sine of theta the sine of theta is equal to the opposite over the hypotenuse it's equal to 2 over s this triangle tells us that the sine and remember this hypotenuse over here is also s the sine of 60 minus theta the sine of 60 minus theta is equal to 4 over s and if we want to set these equal to each other we could multiply this guy by 2 on both sides so you could say 2 sine of theta is equal to 4 over s sine of 60 minus theta is also equal to 4 over s so we can set them equal to each other so we have 2 sine of theta - we have 2 sine - sine of theta is equal to sine of 60 minus theta and then we could use some of our trig identities we know the sine of a minus B is the same thing the sine of a minus B this is equal to the sine of a times the cosine of B or I should say theta in this case so the sine of 60 minus theta - this is just a standard trig identity ethical is the different sum and difference identity - cosine of 60 minus cosine of 60 times the sine of theta times the sine of theta and all this is equal to 2 sine of theta 2 sine of theta well sine of 60 degrees this is square root of three over two square root of three over two cosine of 60 degrees is 1/2 is 1/2 so we could add 1/2 sine theta to both sides of this and what are we going to get so we're going to add 1/2 sine theta then this guy is going to go away and then you add 1/2 sine theta to 2 sine of theta which is really 4 halves sine of theta so that's going to be 5 halves sine of theta so I'm just adding 1/2 sine theta to this so it's five five halves sine of theta is equal to square root of 3 over 2 cosine of theta square root of 3 over 2 cosine of theta right I added 1/2 sine theta to both sides of this to get this and I can multiply both sides by two just to simplify it so I get five sine of theta five sine of theta is equal to the square root of 3 cosine of theta now I want to use the identity sine squared of theta plus cosine squared of theta is equal to one so let me just square both sides and that also help us with this radical so we'll get 25 sine squared of theta is equal to square root of 3 squared is 3 instead of writing cosine squared of theta let's just write that's 1 minus sine squared of theta right cosine squared theta is 1 minus sine squared of theta I just squared both sides let me just write what I just did I just squared both sides and so we get 25 sine squared theta is equal to 3 minus 3 sine squared theta we can add 3 sine squared theta to both sides we get 28 sine squared theta is equal to 3 or that the sine squared of theta homestretch is equal to 3 over 28 or we could even write that's sine of theta sine of theta is equal to the square root of 3 over 28 so it's equal to the square root of 3 over 28 now we can simplify that 28 is 4 times 7 we could take it out but that's good enough for now that'll that maybe you know we'll simplify it later if we have to sometimes these are easier to deal with so let's see over so we have we have the sine of theta now we can relate that actually to s over here we know that we know that before I messed with this thing we know that the sine of theta is equal to 2 over s or that s over 2 is equal to 1 over sine of theta or that s is equal to 2 over the sine of theta well we know what sine of theta is this is square root of 3 over 28 so s is equal to 2 divided by sine of theta that's like multiplying by the inverse of sine of theta so that's 2 times the square root of 28 over 3 28 over 3 so that is we figured out our s 2 times this thing over here now given that we know an S let's see how we can figure out the area well what immediately pops out is that we have this triangle over here that has height or I should say maybe its base if you view it sideways its base is 8 its base is 8 and this distance right over here we should be able to figure out we should be able to figure out using the Pythagorean theorem because we know that this distance right over here is 4 we know that distance is 4 we know that this distance the hypotenuse is s so we could call this the height of it right over here we could say that H squared H squared plus 4 squared plus 16 is equal to the hypotenuse squared is equal to s squared s squared s is this thing over here so if we want to square s it becomes 4 times 28 4 times 28 over 3 and we just subtract 16 from both sides so H is equal to 4 times 28 over 3 minus if I want to write 16 over 3 or if I wonder at 16 is something over 3 that's going to be minus 48 minus 48 over 3 and let's see I don't want to multiply I don't want to have to multiply 4 times the 28 so we can write 48 as 4 times 12 is 4 times 12 so this numerator is going to be 4 times 8-12 over remember that's H squared I should say H squared is going to be 4 times 28 minus 12 over 3 which is equal to 4 times 16 over 3 which is equal to 64 over 3 that's H squared so H is going to be the square root of that which is 8 over the square root of 3 so H right over here is 8 over the square root of 3 so if I want to find the area of this whole thing over it well first let's find the area of this small thing right over here that's just going to be H times 4 so it's going to be well I could do it either way but let's just say this is H times 4 times 1/2 so it's going to be 2 times so the area of this triangle let me do it in blue right here this triangles area is going to be H which is 8 over the square root of 3 times 4 times 4 times 1/2 so this guy right over here is just going to be 2 times 8 over the square root of 3 or that's going to be 16 over the square root of 3 so this guy over here is 16 over the square root of 3 so that is 16 over the square root of 3 now we have a bunch of that we have this guy and now this guy is going to have the exact same area right over there and then you have this guy who's going to have the exact same area again same exact logic same exact logic same base same height they're actually congruent so you have four of these triangles you're going to multiply by 4 if you want the area of these all this area that I've already shaded in so 4 times this where it's 64 over where it's 64 over the square root of 3 now the only area we have left to figure out is the area of this parallelogram the area of the parallelogram in the middle now we know the base of the parallelogram the base of this parallelogram is 8 the base of the parallelogram is 8 we just have to figure out its height we just have to figure out its height and once again we can use the Pythagorean theorem so I'll call this I don't know we already used H well I'll use H again but this is you have to remember this is a different height over here this base over here is of length two two and it's hard to read now so we can now write we can now write that H squared plus four plus two squared is equal to is equal to s squared now we already figured out what s squared was in the past it's four times 28 over three four times 28 over three let's subtract 4 from both sides so you subtract four there so minus 12 over three and now let's see if 12 is the same thing as 4 times 3 so this is this is equal to 4 times 28 4 times 28 minus 3 so that's 4 times 25 over 3 which is equal to 100 over 3 that's H squared so this H is going to be equal to the square root of this which is 10 over the square root of 3 this is 10 over the square root of 3 so this distance right over here is 10 over the square root of 3 so if we want the area of this parallelogram it's going to be that height times the base of the parallelogram so the parallelogram is going to be 8 times 10 square roots of 3 or 80 80 square roots of 3 80 oh no let me be very careful let me be very careful this was 10 over this is 10 over the square root of 3 is this height so the whole area of this parallelogram is 8 times 10 over the square root of 3 it's 80 over the square root of 3 so our entire area now our if we add everything together if we add everything together we have 64 square roots of 3 for these four triangles plus 80 over square root of 3 so let's add it together so we have 80 over square roots of 3 for the parallelogram plus 64 over square root of 3 for the triangle parts and this is equal to 144 over the square root of 3 we can rationalize the denominator so times the square root of 3 over the square root of 3 and in the denominator now we're going to get a 3 144 over 3 is going to be what that's for eight right three times forty 123 times eight is twenty-four so it's going to be 48 square roots of three for the area of our entire our entire hexagon and so we have it in the form forty eight square roots of three so if you want to find M plus n it's 48 plus 3 which is 51 that was a tiring problem I just my brain started to fry near the end of it I had trouble keeping track of things anyway hopefully you enjoyed that