Question

In a triangle with one of the angles \( 120^\circ \), the lengths of the sides form an A.P. If the length of the greatest side is 7 m, then the area of the triangle is

• A. \( \frac{15\sqrt{3}}{4} \text{ m}^2 \)
• B. \( \frac{15\sqrt{3}}{2} \text{ m}^2 \)
• C. \( \frac{15}{2} \text{ m}^2 \)
• D. \( \frac{15}{4} \text{ m}^2 \)

Hint:

For sides in A.P. with a $120^\circ$ angle, the common difference ratio is fixed; 3-5-7 is the standard integer triangle with a $120^\circ$ angle.

Answer 1

The Correct Option is A

Step 1: Define Sides
Sides in A.P.: $a, b, 7$. Let them be $7-2d, 7-d, 7$ (where 7 is greatest).
In a triangle, the greatest side is opposite the greatest angle. So $c = 7, C = 120^\circ$.
Step 2: Cosine Rule
$c^2 = a^2 + b^2 - 2ab \cos 120^\circ \implies 49 = a^2 + b^2 + ab$.
Substitute $a=7-2d, b=7-d$. Solving gives $d=2$.
Sides are 3, 5, 7.
Step 3: Calculate Area
Area $= \frac{1}{2} ab \sin 120^\circ = \frac{1}{2} (3)(5) \frac{\sqrt{3}}{2}$.
Step 4: Conclusion
Area $= \frac{15\sqrt{3}}{4}$ m$^2$.
Final Answer:(A)