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Gayatri Rampersaud
Asked: 2022-11-09T20:01:38+05:30 In:

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. 2Δ2
3. \(\sqrt{2}\Delta\)
4. 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. 2Δ2
3. \(\sqrt{2}\Delta\)
4. Δ2
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1 Answer

  1. 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

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