For a triangle, 3 sides are given.

What's the radius of its circumcircle?

Are we able to get it without using cosine law or sine law or heron formula?

My guess would be that if you could find the center of the triangle, then everything would be easy from there. THe radius would be the center to any vertex.

It depends on what kind of triangle you are dealing with.....

A triangle with 7.5 cm , 12.5 cm and 10 cm.

My teacher hasn't taught us cosine law or sine law or heron law.

This question appears in a chapter of property of circle,simple properties of circle.

I have learnt the above three laws so I can find a lot of method to solve it but I lack a method without using these untaught things.

Yes.

 

You can find it without using Heron's Formula or Law of Cosine I belive.

how about making an accurate diagram and measuring the radius. it might work?

Umm...isn't that triangle a 3-4-5 triangle?

-Uncool-

is it a right triangle? i think i solved it using the designs of our good buddy pythagorus

 

with those dimensions would it be 6.25 for the radius?

 

circumfrence of 12.5pi?

 

i'm afraid that only works if it's a right triangle though

As uncool already has pointed out, the triangle is a scaled 3-4-5-triangle, and surely it must be a right triangle. You can easily prove it by using Pytagoras: A triangle with sides a, b and c, c >= a, b, is a right triangle if and only if c^2 = a^2 + b^2. (If and only if, as the Law of Cosine would show, or alternatively the simple observation that when you know three sides of a triangle, the triangle is uniquely given.)

Now, to find the radius of the circumcircle, you can use this:

a/sin A = b/sin B = c/sin C = 2R iirc

where a, b, and c are the sides, and A, B, and C are the opposite angles of the corresponding sides, and R is the radius of the circumcircle

-Uncool-

but he said without using sine or cosine or heron laws

If you do not want to use sine and cosine, then you can use similarities instead, and if you do not want to use the Heron formula, you can always work hard with Pythagoras formula. I do not know why that would be better - it is easier just learning the stuff, or possibly even better: how to derive it.

 

 

The point in the specific 7.5-10-12.5- exercise is that right triangles are inscribed in a circle with the hypotenus as a diameter (Thales theorem, a special case of the "periferivinkelsetninga" - I don't know the english name - that the center angle is twice the angle at the circumference. Funny enough, this sentence is not taught at Norwegian schools anymore, although it is very simple and something everyone that know a bit about geometry should know. Last year I was talking to a person who now studies physics and electronics at Princeton, and he had never heard about the theorem, nor did he manage to derive some simple consequences of it. So much for geometry in Norway!)

I assume the triangle is equalateral. Find its height and divide it by two.

The triangle was established to be a scaled 3-4-5 right angled triangle, and thus the hypotheneuse is a diameter of the circle required.

Then you can't possibly get a radius

Isn't the radius simply half of the diameter?

Then you can't possibly get a radius

 

Strange, cos every one else is agreed that the answer is 6.25, half the longest side given.

However there appears to be several possible radii. It is impossible to know the diameter as there are several ratio's to the circumference. Remember the triangle is not equalateral.Not sure I understand this one, but the in- radii of the triangle is 15 cm, according to the formular r=ab/a+b-c. Another site says the hypotenus is the diameter, which means the radius is 6.25.This one's a toughie

Hmm... what about making a list of all possible radii values, then complaining to the Math teacher that the task was too confusing to work out properly.

Give me a break, haven't seen trig since H.School. I was only analizing

The solution:

Since the triangle is a right-angled triangle, the longest side is the diameter.

I think this thread might be closed in a few days.