Question

r1r2+r2r3+r3r1=s²=(a+b+c/2)³​

Answer 1

EXPLANATION.

⇒ r₁r₂ + r₂r₃ + r₃r₁ = s².

As we know that,

Formula of :

⇒ r₁ = Δ/(s - a).

⇒ r₂ = Δ/(s - b).

⇒ r₃ = Δ/(s - c).

Using this formula in the equation, we get.

⇒ [Δ/(s - a) x Δ/(s - b)] + [Δ/(s - b) x Δ/(s - c)] x [Δ/(s - c) x Δ/(s - a)].

⇒ [Δ²/(s - a)(s - b) + Δ²/(s - b)(s - c) + Δ²/(s - c)(s - a)].

⇒ Δ²[1/(s - a)(s - b) + 1/(s - b)(s - c) + 1/(s - c)(s - a)].

⇒ Δ²[(s - c) + (s - a) + (s - b)/(s - a)(s - b)(s - c)].

As we know that,

Formula of :

⇒ a + b + c = 2s.

⇒ s(s - a)(s - b)(s - c) = Δ².

Using this formula in the equation, we get.

⇒ Δ²[3s - (a + b + c)/(s - a)(s - b)(s - c)].

⇒ Δ²[(3s - 2s)/(s - a)(s - b)(s - c)].

⇒ Δ²[s/(s - a)(s - b)(s - c)].

Multiply and divide by s in equation, we get.

⇒ Δ²[s²/s(s - a)(s - b)(s - c)].

⇒ Δ²[s²/Δ²] = s².

Hence Proved.

Answer 2

Step-by-step explanation:

$\color{olive} \pmb{ \begin{aligned} \tt r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1} & \\ & \tt=\frac{\Delta}{(s-a)} \cdot \frac{\Delta}{(s-b)}+\frac{\Delta}{(s-b)} \\ \\ & \tt=\frac{\Delta}{(s-c)}+\frac{\Delta}{(s-c)} \cdot \frac{\Delta}{(s-a)} \\ \\ & \tt=\frac{\Delta^{2}(s-c+s-a+s-b)}{(s-a)(s-b)(s-c)} \\ \\ & \tt=\frac{\Delta^{2}(3 s-(a+b+c))}{(s-a)(s-b)(s-c)} \\ \\ & \tt=\frac{\Delta^{2}(3 s-2 s)}{(s-a)(s-b)(s-c)} \\ \\ & \tt=\frac{\Delta^{2} \times s}{s(s-a)(s-b)(s-c)} \\ \\ & \tt=\frac{\Delta^{2} \times s^{2}}{\Delta^{2}} \\ \\ & \tt=s^{2} \end{aligned} }$