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
2Δ2
Concept:
The area of any triangle can be defined as follows:
A = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\)
where a, b, and c are the sides of the triangle and s=(a+b+c)/2
Calculation:
It is given that \(\rm s=\dfrac{a+b+c}{2}=6\)
The area of the triangle = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\) = Δ
⇒ s(s - a)(s - b)(s - c) = Δ2
Multiplying both sides by s
⇒ s × s(s - a)(s - b)(s - c) = s × Δ2
⇒ s2(s - a)(s - b)(s - c) = 6 Δ2
Multiplying both sides by ⅓
⇒ (⅓) s2(s - a)(s - b)(s - c) = 2 Δ2
Important Questions from Properties of Triangles
What is the perimeter of the triangle ?
Consider the following statements :
1. ABC is right angled triangle
2. The angles of the triangle are in AP
Which of the statements given above is/are correct ?
What is the nature of the triangle ?
If c = 8, what is the area of the triangle ?
What is the value of n ?
