Question

in triangle ABC if r=1 R=3 Nd S=5 then the value of a^2+b^2+c^2​

Answer 1

Step-by-step explanation:

You know that

r=Δs

Hence we get

Δ=5

Also

R=abc4Δ

From this we get

abc=60

We also have a formula that

4(s−a)(s−b)(s−c)abc=rR

Hence, on substituting values we get

(5−a)(5−b)(5−c)=5

Which reduces to

125−50s+5∑cycab−abc=5

Further on reducing we get

∑cycab=38

Now

(a+b+c2)2=s2

a2+b2+c2+2∑ab4=25

a2+b2+c2=100−76=24

Hence ∑a2=24