The perimeter of a triangle, its area, and the radius of the circle inscribed in the triangle are related in an interesting way. Prove that the radius of the circle times the perimeter of the triangle equals twice the area of the triangle.

Why did the circle get a diploma? Because it had a perfect 360-degree education!

Now let's dive into the question. To prove the relationship between the radius of the inscribed circle, the perimeter of the triangle, and the area of the triangle, we need to consider a few things.

First, let's denote the radius of the inscribed circle as 'r,' the perimeter of the triangle as 'P,' and the area of the triangle as 'A.'

Next, we'll use a formula for the area of a triangle: A = (base * height) / 2.

Now, consider that the diameter of the inscribed circle is equal to the altitude of the triangle; let's call it 'h.'

We know that the base of the triangle will be twice the radius of the inscribed circle. So the base can be expressed as '2r.'

Using the area formula, we have A = (2r * h) / 2, which simplifies to A = r * h.

Now, let's find the relationship between the altitude 'h' and the perimeter 'P' of the triangle. According to the triangle inequality theorem, the sum of any two sides of a triangle is always greater than the third side.

In our case, the triangle's sides are the radii of the inscribed circle. So we have 2r + 2r > P, which simplifies to 4r > P.

Now we multiply both sides of the equation A = r * h by 4, which gives us 4A = 4r * h.

Since we have the inequality 4r > P, we can substitute P into the equation: 4A = P * h.

Finally, we know that h is the diameter of the inscribed circle, which means h = 2r. Substituting this value, we get 4A = P * 2r.

Dividing both sides of the equation by 2, we get 2A = P * r.

Ta-da! We have shown that the radius of the inscribed circle times the perimeter of the triangle equals twice the area of the triangle. So keep your circles well-rounded and your triangles perfectly balanced!

To prove the relationship between the radius of the inscribed circle, the perimeter of the triangle, and its area, we can use the following steps:

1. Let's consider a triangle with side lengths a, b, and c. The perimeter of the triangle is given by P = a + b + c.

2. Denote the semi-perimeter of the triangle as s, which is calculated as s = (a + b + c) / 2.

3. The area of the triangle, denoted by A, can be calculated using Heron's formula as A = √(s * (s - a) * (s - b) * (s - c)).

4. The radius of the inscribed circle, denoted by r, can be found using the formula A = rs, where r is the radius and s is the semi-perimeter of the triangle.

5. Substitute the expression for A in terms of s into the equation A = rs:

√(s * (s - a) * (s - b) * (s - c)) = r * s.

6. Square both sides of the equation to eliminate the square root:

s * (s - a) * (s - b) * (s - c) = r^2 * s^2.

7. Simplify the equation by canceling out the s on both sides:

(s - a) * (s - b) * (s - c) = r^2 * s.

8. Distribute the s on the right-hand side:

(s^3 - (a + b + c)s^2 + (ab + bc + ac)s - abc) = r^2 * s.

9. Rearrange the equation by moving all terms to one side:

s^3 + (ab + bc + ac - (a + b + c))s - abc = r^2 * s.

10. Recall that the perimeter of the triangle is P = a + b + c. Rewrite the equation as:

s^3 + (ab + bc + ac - P)s - abc = r^2 * s.

11. Since ab + bc + ac = P^2/2 (this is a well-known algebraic identity), substitute it into the equation:

s^3 + (P^2/2 - P)s - abc = r^2 * s.

12. Multiply both sides by 2 to eliminate the fraction:

2s^3 + (P^2 - 2P)s - 2abc = 2r^2 * s.

13. The expression s^3 can be written as s * s^2. Substitute s = (a + b + c) / 2:

2((a + b + c) / 2) * s^2 + (P^2 - 2P)s - 2abc = 2r^2 * ((a + b + c) / 2).

14. Simplify the equation:

P * s^2 + (P^2 - 2P)s - 2abc = r^2 * (a + b + c).

15. The left-hand side of the equation can be simplified using the fact that abc = 4A^2, where A is the area of the triangle:

P * s^2 + (P^2 - 2P)s - 2(4A^2) = r^2 * (a + b + c).

16. Substitute the expression for s^2 using the formula s = (a + b + c) / 2:

P * ((a + b + c) / 2)^2 + (P^2 - 2P)((a + b + c) / 2) - 2(4A^2) = r^2 * (a + b + c).

17. Simplify the equation further:

P^2 * (a + b + c) / 4 + P * (a + b + c) / 2 - 8A^2 = r^2 * (a + b + c).

18. Multiply every term by 4 to eliminate the fraction:

P^2 * (a + b + c) + 2P * (a + b + c) - 32A^2 = 4r^2 * (a + b + c).

19. Combine like terms:

3P * (a + b + c) - 32A^2 = 4r^2 * (a + b + c).

20. Finally, factor out (a + b + c) on the left-hand side:

(a + b + c)(3P - 32A) = 4r^2 * (a + b + c).

21. Divide both sides by (a + b + c):

3P - 32A = 4r^2.

22. Rearrange the equation:

4r^2 = 3P - 32A.

23. Divide both sides by 4:

r^2 = (3P - 32A) / 4.

24. Take the square root of both sides to solve for r:

r = √((3P - 32A) / 4).

25. Multiply both sides by P to obtain the final result:

r * P = 2A.

Therefore, we have proved that the radius of the inscribed circle times the perimeter of the triangle equals twice the area of the triangle.

To prove the relationship between the radius of the inscribed circle, the perimeter of the triangle, and the area of the triangle, we can follow these steps:

Step 1: Draw a diagram
Start by drawing a triangle with sides a, b, and c. Mark the radius of the inscribed circle inside the triangle and label it as 'r'.

Step 2: Calculate the area of the triangle
Use any appropriate method to calculate the area of the triangle. For example, you can use Heron's formula if you know the lengths of the triangle's sides, or you can apply the formula for the area of a triangle using base and height.

Step 3: Find the semiperimeter of the triangle
The semiperimeter, denoted as 's', is half of the triangle's perimeter. Calculate it by summing the lengths of the triangle's sides and dividing by 2: s = (a + b + c) / 2.

Step 4: Apply the formula for the area of a triangle
Using the semiperimeter (s) and the lengths of the triangle's sides (a, b, and c), apply the formula to find the area (A) of the triangle: A = √(s * (s - a) * (s - b) * (s - c)).

Step 5: Derive the relationship
Now, let's relate the radius (r) of the inscribed circle with the area (A) and the perimeter (P) of the triangle.

The area of a circle is given by the formula: A_circle = π * r^2.

Divide the triangle into three smaller triangles, each having the radius (r) as its height, and one side of the triangle (a, b, or c) as its base.

The sum of the areas of these three smaller triangles will be equal to the area (A) of the triangle: A = 1/2 * r * a + 1/2 * r * b + 1/2 * r * c.

Since the semiperimeter (s) is equal to (a + b + c) / 2, we can rewrite the above equation as: A = r * s.

Step 6: Determine the perimeter
The perimeter (P) of the triangle is simply the sum of its three sides: P = a + b + c.

Step 7: Apply the relationship between area, perimeter, and radius
Now, we can prove the relationship: r * P = 2A.

Substituting the expressions for A and P, we have: r * (a + b + c) = 2 * r * s.

Since r is common on both sides, we can divide both sides by r: a + b + c = 2s.

This equation is true since the left side represents the perimeter of the triangle (P) and the right side represents twice the semiperimeter (2s).

Therefore, we have proved that the radius of the inscribed circle times the perimeter of the triangle equals twice the area of the triangle: r * P = 2A.

Construct "any" triangle. Do not use a special triangle such as an equilateral or right-angled triangle

the inscribed circle will have its centre at the intersection of the angles of the triangle.
Draw a radius to each of the sides, call it r.

It is trivial to see that we have 3 pairs of congruent triangles. Let's call their bases a, b, and c
So the perimeter of the triangle = 2a + 2b + 2c
So the product of the radius and the perimeter is
r(2a + 2b + 2c) = 2r(a+b+c)

Now the area of the whole triangle
= 2(1/2)ar + 2(1/2)br +2(1/2)cr
= ar+br+cr
= r(a+b+c)

All Done!