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# Trig challenge problem: area of a triangle

## Video transcript

triangle ABC is a right triangle with AC is equal to seven BC is equal to 24 and a right angle at C so let's just try to draw that let's draw then let's draw the right angle maybe we'll have to use some coordinate so let's draw the right angle at the origin so let's say that this is C right over here this is C let's the AC is equal to seven so a will put over here this distance over here is going to be seven and then we have the hypotenuse of our triangle and then this can be B and this distance over here they tell us that BC is equal to 24 all right now point M is the midpoint of a B so point M is right over here let me do that in a different color point M is the midpoint it's the midpoint of a B so this distance is equal to that distance and D is on the same side of a B so D is on the same side of a BSc so C is on this side I guess you call it the left the bottom left-hand side of a B so that ad is equal to BD is equal to 15 so D is going to be someplace over here it's going to be equal distance between a and B obviously all of the points equal distance between a and B are going to sit on a line that looks something like that this is the midpoint of a V so D is going to sit right over here and it's 15 away from both a and B so it would look something like that that distance over there is 15 and then this distance this distance over here is also going to be 15 given that the area of triangle C DM so the area of triangle C DM this is D so C D M is the triangle that they want to think about given that the area of triangle C DM may be expressed as M times the square root of n over P where m and n P are positive integers and M and P are relatively prime so that just means you can't simplify it and n is not divisible by the square of any prime so kind of n is you've simplified the radical as much as possible find M plus n plus P so we essentially need to find the area the area of this green triangle of C DM over here and so let's see what we can do to figure it out so we could try some of the coordinates for some of these points so this point right over here a is going to be its x value is going to be 7 I could draw the coordinate just so you see what I'm doing that could be the x-axis and then this side is on the y-axis so the coordinate for a would be 7 comma 0 coordinate for C would be 0 comma 0 and coordinate for B would be 24 sorry would be 0 24 0 24 so the coordinate for M would just be the average of B and a so the coordinate for M the average of 0 & 7 is 7 halves and the coordinate for the y coordinate the average of 24 and 0 is 12 that's fair enough now let's see what we can figure out about the side so we know this is a right triangle so our gut reaction is always to use the Pythagorean theorem we know this side and that side so if we wanted to figure out a B we could just say that we just know that 24 squared plus 7 squared is equal to is equal to a B is equal to a B squared and 24 squared is 576 plus 49 is equal to a B squared and let's see 576 plus 49 if it was plus 50 would get us to 5 20 will give us a 626 but it's one less than that so 625 so 625 is equal to a B squared so a B is going to equal 25 so a B is equal to 25 so this distance this the distance of this big hypotenuse here is 25 or half of the distance this is going to be 25 over 2 from V to M from M to a is also going to be 25 over 2 now the other thing we know the other thing we know is that em right over here em right over here the triangle see ma is an isosceles triangle how do we know that well em if you look at its x-coordinate its x-coordinate is directly in between the x-coordinates for C and a seven-halves it's the average this is seven this is zero it's M coordinate is right over there's directly above the midpoint of this base over here so this is going to be an isosceles triangle it's symmetric you could flip the triangle over so this length this length and this seems useful because this is kind of the base of the area that of the truck the base of the triangle we care about we could kind of view this as the base of CDM this is also going to be 25 over 225 over 2 it's an isosceles triangle this is going to be the same as that because we're symmetric around this right over here so let's see we know one side of this triangle let's see if we could figure out this side over here it seems pretty straightforward this is going to be a right triangle right over here because this line from D to M is going to be perpendicular to a B all of the points that are equidistant between a and B are going to be on a line that is perpendicular that is perpendicular to a B so this is going to be a right triangle so we can figure out DM using the Pythagorean theorem again we get 25 over 2 squared 25 over 2 squared plus DM squared plus DM squared plus DM squared is going to be equal to 15 squared is going to equal to the hypotenuse of this triangle so it's going to equal 225 and so what do we get we get DM squared DM squared is equal to 225 minus 625 over 4 so let's put this over 4 so 225 over 4 or 225 with fours the denominators the same thing is 900 900 over 4 and in the last video actually figuratively said 900 over 4 was 125 obviously boneheaded mistake but the numbers show up again here so 225 is obscene 100 over 4 and then we're going to subtract from that 625 over 4 minus 625 over 4 and this is equal to C in our numerator let's see we have 900 minus 625 would be 300 minus 20 it's going to be 200 275 over four and so DM is going to be equal to the square root of this so it's equal to the square root of 275 over four 275 is 25 times 11 because 25 times 12 would be 300 so 25 times 11 over 4 so this is going to be equal to 5 square roots of 11 over 2 so DM is 5 square roots of 11 over 2 now if we could just figure out so what we need to do here is figure out the height of this triangle right over here if we could figure out the height of that triangle where we're pretty much done 1/2 the base times the height but we don't know let's see we could figure that if we know the law of cosines we could do something over there if we know the sine of this angle right over here if we know the sine of that angle the sine of that angle is going to be this height over the side we just figured out so if we could figure out the sine of that angle then we'd be done or will be very close to being done but it's not any obvious way but one thing we can do if you look at this bigger triangle over here so let me highlight it so if you look at triangle B M C let me draw a triangle VMC B well I'll just draw it on this drawing right over here I don't want to get too overloaded with triangle B M C right over here we know this side is 25 over 2 we know this side over here is 25 over 2 we know this side over here is 24 and we know this so what we want to do is figure out the sine of this angle right here the sine of theta of angle C M D now that's hard to do but what we can do is use the law of cosines of theta plus 90 so let's let me let me redraw our triangle so triangle B cm I could redraw like this I could redraw it's actually an isosceles triangle so we have b c m and what is this angle right over here it's going to be our theta that we care about plus 90 degrees so this angle right here is Theta theta plus 90 degrees and this over here is 24 this is 25 over 2 and this is 25 over 2 and so using this we could use the law of cosines to figure out to figure out what theta actually is here so let's do that we're going to use a little bit of trig identities but law of cosines so we get the opposite to the angle squared so we get 24 squared 24 squared is equal to 25 over 2 squared 25 over 2 squared plus 25 over 2 squared plus 25 over 2 squared minus 2 times 25 over 2 times 25 over 2 times the cosine of this angle times let me scroll over the cosine of theta theta plus theta plus 90 degrees now you're saying hey Sal you know this has it in terms of cosine of theta plus 90 degrees how can we figure out the sine of theta that's what we actually care about to figure out the area of this triangle directory to figure out the height of this triangle and to do that you just have to make the realization we know the trig identity we know the trig identity that the cosine of theta is equal I won't use theta because I don't to overload theta cosine of X is equal to sine of 90 minus X so the cosine of theta plus 90 degrees so the cosine of theta plus 90 degrees is going to be equal to the sine of let me put in parenthesis 90 minus whatever Z or 90 minus theta minus 90 which is equal to the 90s cancel out sine of negative theta and we know sine of negative theta is equal to the negative sine of theta so this over here simplifies - this is the negative sign of theta so we could write the sign of theta here and then put the negative out here and this becomes a positive so what does this simplify to we have 24 squared which is 576 576 is equal to let's see we have I'll just I won't skip any steps here so we have 25 squared plus 25 squared this is 2 times 25 this is 2 2 times 25 over 2 squared squared plus 2 times this is 25 over 2 squared again 2 times 25 over 2 squared times sine of theta now we just have to solve for sine of theta so this is going to be equal to well this is 576 is equal to 2 times 25 over 2 squared I'm just factoring that out 1 plus sine of theta 1 plus sine of theta or we can just divide both sides of the equation by this you actually let me just simplify it this thing over here is 625 over 4 but then we're going to multiply that by 2 so this thing over here is 625 over 2 so let's divide both sides of this by 625 over 2 and we get 576 x times 2 over 625 just multiplying both sides by the inverse is equal to when you multiply both sides the inner side this askyou cancels out is equal to 1 plus sine of theta or sine of theta we just subtract 1 from both sides we get sine of theta is equal to 576 times - let's see 76 times 2 is 152 plus a thousand so it's 1152 over 625 that's that part there minus 1 but instead of 1 let's say 625 over 625 so minus 625 and this is equal to I'll have to do a little math on the side so 1152 minus 6 and 25 you get a 12 there this becomes a 4 12 minus 5 is 7 4 minus 2 is 2 11 minus 6 is 5 so this is equal to 527 over 625 now you might not realize it but we're in the homestretch now let's let's draw let me draw this CDM triangle the one that is really the focus of the problem and let me draw it a little bit differently let me draw it let me draw it slightly differently so now we know some very interesting things about CDM so this is C this is D and this is M we know this side over here we figured out is 5 halves square root of n' 5 over 2 times the square root of n' that's the length of DM we also know that cm right over here is 25 over 2 we also know that sine of theta here we also know that sine of theta is equal to 527 over 625 that's what we just figured out and we can use that to figure out the height the height of this triangle because we know that sine is is the opposite over the hypotenuse if we draw a right triangle right here so sine of theta is which is 527 over 625 is equal to the opposite is equal to the height of this triangle is equal to the height of this triangle over the hypotenuse over 5 over 2 square roots of 11 so we can multiply both sides of this by 5 square roots of 5 over 2 square roots 11 and we get the height of the triangle is equal to 527 over 625 times 5 over 2 square root of 11 and let's see we can divide 6 25 divided by 5 is 125 so you get a 1 here and you have a 125 over here so this is equal to 527 square roots 11 over 125 times 2 which is 200 250 that's the height now what's the area of the triangle it's 1/2 base times height area is equal to one half base which is 25 over 225 over two times the height which we just figured out is 527 square roots of 11 over 250 let's see 25 divide the numerator by 25 divide the denominator by 25 you get a 10 there so this is equal to 527 square roots of 11 over 2 times 2 times 10 which is 40 so that's our area and the whole problem they didn't want us to find the area they want us to find M plus n plus P M plus n plus P so they want us to find essentially 527 plus 11 plus 45 27 plus 11 plus 40 so 527 plus 11 is 5 let me make sure I do this 538 and then plus 40 is 578 and we're done