Question

If Δ is the area and 2s the sum of three sides of a triangle, then

विकल्प

• `Δ ≤ s^2/(3sqrt(3)`

• `Δ ≤ s/2`

• `Δ > s^2/sqrt(3)`

• None of these

Answer 1

`Δ ≤ s^2/(3sqrt(3)`

Explanation:

We have 2s = a + b + c, Δ² = s(s – a)(s – b)(s – c)

∵ A.M. ≥ G.M.

∴ `((s - a) + (s - b) + (s - c))/3 ≥ (s - a)(s - b)(s - c)^(1/3)`

⇒ `(3s - (a + b + c))/3 ≥ (s - a)(s - b)(s - c)^(1/3)`

Since a + b + c = 2s we have

⇒ `(3s - 2s)/3 ≥ (s - a)(s - b)(s - c)^(1/3)`

⇒ `s/3 ≥ (s - a)(s - b)(s - c)^(1/3)`

Cubing both sides, we get

⇒ `(s/3)^3 > (s - a)(s - b)(s - c)`

⇒ `s/27 > (s - a)(s - b)(s - c)`

Rearranging, we get

⇒ `s^3/27 > A^2/s`

⇒ `s^4 ≥ 27A^2`

∴ `A^2 ≤ s^4/27`

or `A ≤ s^2/(3sqrt(3)`