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: Understanding the Question:
The sides of a triangle are in Arithmetic Progression and the largest angle is \( 120^\circ \). We need to find the side lengths and the area.
Step 2: Key Formula or Approach:
1. Let sides be \( a-d, a, a+d \). The largest side is \( a+d = 7 \).
2. Use Cosine Rule: \( c^2 = a^2 + b^2 - 2ab \cos C \).
3. Area \( = \frac{1}{2} ab \sin C \).
Step 3: Detailed Explanation:
The greatest side \( c = 7 \) is opposite to \( 120^\circ \).
Let the sides be \( 7-2d, 7-d, 7 \)? No, let them be \( a, a+d, a+2d \).
Better: let sides be \( x-d, x, x+d \). Greatest side \( x+d = 7 \implies d = 7 - x \).
Cosine rule:
\[ 7^2 = (x-d)^2 + x^2 - 2x(x-d) \cos 120^\circ \] \[ 49 = x^2 - 2xd + d^2 + x^2 + x(x-d) = 3x^2 - 3xd + d^2 \] Substitute \( d = 7-x \):
\[ 49 = 3x^2 - 3x(7-x) + (7-x)^2 \] \[ 49 = 3x^2 - 21x + 3x^2 + 49 - 14x + x^2 \] \[ 49 = 7x^2 - 35x + 49 \implies 7x^2 - 35x = 0 \] Since \( x \ne 0 \), \( x = 5 \).
Then \( d = 7 - 5 = 2 \).
Sides are \( 5-2 = 3 \), \( 5 \), and \( 7 \).
Area \( = \frac{1}{2} \times 3 \times 5 \times \sin 120^\circ = \frac{15}{2} \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4} \text{ m}^2 \).
Step 4: Final Answer:
The area is \( \frac{15\sqrt{3}}{4} \text{ m}^2 \).