Ask your own question, for FREE!
Mathematics 23 Online
OpenStudy (anonymous):

GEOMETRY QUESTION: if the radius of your circular pattern is 6 inches, calculate the perimeter of the triangle inscribed in the circle

OpenStudy (lncognlto):

|dw:1401280605819:dw|

OpenStudy (anonymous):

stop drawing my diagrams. ;-;

OpenStudy (anonymous):

you ninja.

OpenStudy (lncognlto):

LOL sowwies.

OpenStudy (anonymous):

@lncognlto so each radius is 6....

OpenStudy (lncognlto):

Yup, that is correct.

OpenStudy (anonymous):

@lncognlto how do i go about calculating the perimeter of the triangle from that point on?

OpenStudy (anonymous):

I would use the Law of Sines to determine the hypotenus

OpenStudy (lncognlto):

As this is an equilateral triangle, each of the inner angles is going to be 120 degrees. So then I would use the cosine rule to determine the length of one side of the triangle, and then multiply by three.

OpenStudy (lncognlto):

Well, the sin rule should work too, if you want.

OpenStudy (lncognlto):

*sine

OpenStudy (anonymous):

|dw:1401280934230:dw|

OpenStudy (anonymous):

so if thats the diagram then you set up SOHCAHTOA to work it out right @lncognlto

OpenStudy (lncognlto):

Yep, that should sort it. ^-^

OpenStudy (lncognlto):

You don't really need to break it down that much; you could take (sin 120)/x = (sin 30)/6 and solve for the side length from there.

OpenStudy (anonymous):

I got 10.392 as my final answer; does that sound right? @lncognlto

OpenStudy (lncognlto):

Well, that is certainly the length of one side of the triangle, yes, but it is not the entire perimeter of it...

OpenStudy (anonymous):

So the whole perimeter is 31.176? @lncognlto

OpenStudy (vishweshshrimali5):

Actually I don't think thats exactly right method.

OpenStudy (anonymous):

@vishweshshrimali5 what would be the right method?

OpenStudy (vishweshshrimali5):

I would prefer using the formula for circumradius

OpenStudy (vishweshshrimali5):

I think that may work

OpenStudy (lncognlto):

Yes, I am curious... but that was the answer I got too, @ontour

OpenStudy (anonymous):

@lncognlto i mean that got me what sounds like a reasonable answer

OpenStudy (anonymous):

@lncognlto also, how would i calculate the area of the triangle if it had the same radius measurements

OpenStudy (anonymous):

could someone please tell me how to calculate the area??

OpenStudy (lncognlto):

Sorry, I'm having a look... I'd do it like this, although this is not the only way:

OpenStudy (lncognlto):

|dw:1401281776818:dw|

OpenStudy (lncognlto):

I'd solve for x by Pythag, and then I would use the formula A = (1/2)bh for a triangle to get the area.

OpenStudy (anonymous):

My final answer worked out to be 23.3723 @lncognlto

OpenStudy (lncognlto):

I get double that as my answer: 46.74...

OpenStudy (anonymous):

How did you get it doubled? @lncognlto

OpenStudy (lncognlto):

10.39^2 - 5.195^2 = x^2 Therefore x = sqrt(10.39^2 - 5.195^2) = 8.998 Therefore A = (0.5)bh = (0.5)(10.39)(8.998) = 46.74....

OpenStudy (anonymous):

pythagorean theorem is a2+b2=c2 right?? @lncognlto

OpenStudy (lncognlto):

Yes; in this case, pythag would be 5.195^2 + x^2 = 10.39^2. So then we solve for x.

OpenStudy (vishweshshrimali5):

@ontour is this triangle equilateral or any general triangle ?

OpenStudy (lncognlto):

It's equilateral...

OpenStudy (vishweshshrimali5):

Ohh

OpenStudy (vishweshshrimali5):

You could also have used solutions of triangle's formula and would have found out area and perimeter using trigonometry. But @incognIto 's mthod is also correct. I didn't know that it was equilateral.

OpenStudy (lncognlto):

I know there is a way to calculate the area by use of trig, but I'd have to look it up again. But yeah, that would be an equally valid method.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!