(4) In \( \triangle A B C \), if \( a=4 \mathrm{~cm}, b=7 \mathrm{~cm} \), m \( (\angle C)=120^{\circ} \), then the area of the triangle \( =\cdots \cdots \cdots l l \) \( \begin{array}{llll}\text { (a) } 7 \sqrt{3} & \text { (b) } 14 \sqrt{3} & \text { (c) } 7 & \text { (d) } 14\end{array} \)

To find the area of triangle \( ABC \) with sides \( a = 4 \, \text{cm} \), \( b = 7 \, \text{cm} \), and angle \( C = 120^\circ \), we can use the formula for the area of a triangle given two sides and the included angle: \[ \text{Area} = \frac{1}{2}ab \sin(C) \] First, we need to calculate \( \sin(120^\circ) \). We know that: \[ \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \] Now we can substitute the values into the area formula: \[ \text{Area} = \frac{1}{2} \cdot 4 \cdot 7 \cdot \sin(120^\circ) \] Substituting \( \sin(120^\circ) \): \[ \text{Area} = \frac{1}{2} \cdot 4 \cdot 7 \cdot \frac{\sqrt{3}}{2} \] Calculating this step-by-step: 1. Calculate \( 4 \cdot 7 = 28 \). 2. Then, \( \frac{1}{2} \cdot 28 = 14 \). 3. Finally, multiply by \( \frac{\sqrt{3}}{2} \): \[ \text{Area} = 14 \cdot \frac{\sqrt{3}}{2} = 7\sqrt{3} \] Thus, the area of triangle \( ABC \) is: \[ \boxed{7\sqrt{3}} \] The correct answer is (a) \( 7\sqrt{3} \).