1. Given Information: - The semi-perimeter \( s \) of the triangle is: \[ s = \frac{a + b + c}{2} = 6 \] - The area \( \Delta \) of the triangle is given by Heron's formula: \[ \Delta = \sqrt{s(s - a)(s - b)(s - c)} \] 2. Square Both Sides: To eliminate the square root, square both sides of the area formula: \[ \Delta^2 = s(s - a)(s - b)(s - c) \] 3. Multiply Both Sides by \( s \): Multiply both sides of the equation by \( s \): \[ s \times \Delta^2 = s \times s(s - a)(s - b)(s - c) \] \[ s \Delta^2 = s^2(s - a)(s - b)(s - c) \] 4. Multiply Both Sides by \( \frac{1}{3} \): Multiply both sides of the equation by \( \frac{1}{3} \): \[ \frac{1}{3} s \Delta^2 = \frac{1}{3} s^2(s - a)(s - b)(s - c) \] Given \( s = 6 \), substitute: \[ \frac{1}{3} \times 6 \Delta^2 = \frac{1}{3} \times 6^2 (s - a)(s - b)(s - c) \] \[ 2 \Delta^2 = \frac{1}{3} \times 36 (s - a)(s - b)(s - c) \] \[ 2 \Delta^2 = 12 (s - a)(s - b)(s - c) \] Therefore, the derived expression is: \[ \boxed{\frac{1}{3} s^2(s - a)(s - b)(s - c) = 2 \Delta^2} \]