1. ## Area of a Triangle Formula: inradius & semiperimeter

How would you show a triangle's area is equal to the product of its inradius and its semiperimeter?

3. I learned something there. I had never heard of an inradius.

4. ## Re: Area of a Triangle Formula: inradius & semiperimeter

Originally Posted by Trenters4325
How would you show a triangle's area is equal to the product of its inradius and its semiperimeter?

The internal circle tangent to the three sides and the incenter as center.
The radius of the inscribed circle is r = A/s where A = the area of the triangle and s
= the semi-perimeter = (a + b + c)/2, a, b, and c being the three sides.
The radius of the inscribed circle may also be derived from r = ab/(a + b + c).
The radius of the inscribed circle may also be derived from the particular m and n
used in deriving a Pythagoraen Triple triangle by r = n(m - n).
If x, y, and z are the points of contact of the incircle with the sides of the triangle
A, B, C, then Cx = Cy = s - c, Bx = Bz = s - b, and Ay = Az = s - a.

5. ## Re: Area of a Triangle Formula: inradius & semiperimeter

Originally Posted by TchrWill
The radius of the inscribed circle is r = A/s where A = the area of the triangle and s...
Where did A/s come from?

Also, on the Wolfram website, where did they right side of the first line come from?

6. Well, O.K.
You have been given two rather clear answers.
Now, we have no way of knowing your background: what theorems you know; how well you understand triangles; how far along in the Euclidean Geometry you are.

You have got to do somethings for yourself. The questions you have asked seem to indicate a real lack of understanding of the particulars of this problem on your part.

7. I'm actually in calculus, so I studied geometry a while ago, and although I understand it pretty well, I'm kind of rusty.

8. ## Re: Area of a Triangle Formula: inradius & semiperimeter

Originally Posted by Trenters4325
How would you show a triangle's area is equal to the product of its inradius and its semiperimeter?
1--Consider an acute triangle ABC, A lower left, B lower right and C at the top.
2--Bisect each angle
3--The intersection of the bisectors is the center of the inscribed circle of the triangle with radius r..
4--Let the center of this incircle be called O.and the three sides a, b and c.
5--Consider the three triangles AOB, BOC and COA
6--The areas of these triangles are cr/2, ar/2 and br/2
7--Therefore the total area of the triangle ABC is A = cr/2 + ar/2 + br/2
8--This simplifies to A = r(a + b + c)/2
9--(a + b + c) is the semi-perimeter of the triangle, s.
10--Therefore, A = rs.

9. ## Re: Area of a Triangle Formula: inradius & semiperimeter

Originally Posted by TchrWill
3--The intersection of the bisectors is the center of the inscribed circle of the triangle with radius r..
How do you know that the point of intersection is equidistant from a, b, and c?

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•