Question

in Any triangle ABC,r1r2+r2r3+r3r1=​

Answer 1

Given: A triangle.

To find: The value of r1r2 + r2r3 + r1r3

Solution:

• Now we know that:

                   r1 = Δ/(s - a)

                   r2 = Δ/(s - b)

                   r3 = Δ/(s - c)

• Here s is semi-perimeter, a, b, c are sides of the triangle.

• So putting these values in given term, we get:

                   (Δ/(s-a) x Δ/(s-b) ) + (Δ/(s-b) x (s-c) ) + ( Δ/(s-c) x (s-a)  )

                   Δ² ( 1/(s-a)(s-b) + 1/(s-b)(s-c) + 1/(s-c)(s-a) )

• After solving this, we get:

                   Δ² ( 3s- (a+b+c)/(s-a)(s-b)(s-c) )

• Now we know:

                   s = a + b + c / 2, applying this, we get:

                   Δ² ( 3s- 2s /(s-a)(s-b)(s-c) )

                   s x ( √(s(s-a)(s-b)(s-c)) )² /(s-a)(s-b)(s-c) )

                   s x (s(s-a)(s-b)(s-c)) /(s-a)(s-b)(s-c) )

• Cancelling common terms, we get:

                   s x s

                   s²

Answer:

           So the value of r1r2 + r2r3 + r1r3 is s².

Answer 2

Answer:

Given: A triangle.

To find: The value of r1r2 + r2r3 + r1r3

Solution:

Now we know that:

                  r1 = Δ/(s - a)

                  r2 = Δ/(s - b)

                  r3 = Δ/(s - c)

Here s is semi-perimeter, a, b, c are sides of the triangle.

So putting these values in given term, we get:

                  (Δ/(s-a) x Δ/(s-b) ) + (Δ/(s-b) x (s-c) ) + ( Δ/(s-c) x (s-a)  )

                  Δ² ( 1/(s-a)(s-b) + 1/(s-b)(s-c) + 1/(s-c)(s-a) )

After solving this, we get:

                  Δ² ( 3s- (a+b+c)/(s-a)(s-b)(s-c) )

Now we know:

                  s = a + b + c / 2, applying this, we get:

                  Δ² ( 3s- 2s /(s-a)(s-b)(s-c) )

                  s x ( √(s(s-a)(s-b)(s-c)) )² /(s-a)(s-b)(s-c) )

                  s x (s(s-a)(s-b)(s-c)) /(s-a)(s-b)(s-c) )

Cancelling common terms, we get:

                  s x s

                  s²

Answer:

          So the value of r1r2 + r2r3 + r1r3 is s².