If A, B and C is three angles of a ΔABC, whose area is Δ. Let a, b

1.	If A, B and C is three angles of a ΔABC, whose area is Δ. Let a, b and c be the sides opposite to the angles A, B and C respectively. If \(s=\dfrac{a+b+c}{2}=6\), then the product \(\dfrac{1}{3}s^2 (s-a)(s-b)(s-c)\) is equal to 1. 2Δ 2. 2Δ23. \(\sqrt{2}\Delta\)4. Δ2
Answer» Correct Answer - Option 2 : 2Δ² Concept: The area of any triangle can be defined as: A = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\) where a, b and c are the sides of the triangle and \(\rm s=\dfrac{a+b+c}{2}\) Calculation: Given \(\rm s=\dfrac{a+b+c}{2}=6\) Area of the triangle = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\) = Δ ⇒ s(s - a)(s - b)(s - c) = Δ² Multiplying both side by s ⇒ s × s(s - a)(s - b)(s - c) = s × Δ2 ⇒ s²(s - a)(s - b)(s - c) = 6 Δ2 Multiplying both side by \(\rm 1\over 3\) ⇒ \(\boldsymbol{\rm 1\over 3}\) s2(s - a)(s - b)(s - c) = 2 Δ2

If A, B and C is three angles of a ΔABC, whose area is Δ. Let a, b and c be the sides opposite to the angles A, B and C respectively. If \(s=\dfrac{a+b+c}{2}=6\), then the product \(\dfrac{1}{3}s^2 (s-a)(s-b)(s-c)\) is equal to 1. 2Δ 2. 2Δ23. \(\sqrt{2}\Delta\)4. Δ2

Answer» Correct Answer - Option 2 : 2Δ²

Concept:

The area of any triangle can be defined as:

A = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\)

where a, b and c are the sides of the triangle and \(\rm s=\dfrac{a+b+c}{2}\)

Calculation:

Given \(\rm s=\dfrac{a+b+c}{2}=6\)

Area of the triangle = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\) = Δ

⇒ s(s - a)(s - b)(s - c) = Δ²

Multiplying both side by s

⇒ s × s(s - a)(s - b)(s - c) = s × Δ2

⇒ s²(s - a)(s - b)(s - c) = 6 Δ2

Multiplying both side by \(\rm 1\over 3\)

⇒ \(\boldsymbol{\rm 1\over 3}\) s2(s - a)(s - b)(s - c) = 2 Δ2