in Any triangle ABC,r1r2+r2r3+r3r1=
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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.
Answered by
1
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.
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