In $\Delta ABC$, if $\frac{cos~A}{a}=\frac{cos~B}{b}=\frac{cos~C}{c}$ and $a=2,$ then area of $\Delta ABC=$

Question

Accepted Answer

To solve this problem, we need to find the area of triangle $\Delta ABC$ given the condition $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ and $a = 2$.

### Explanation:

1. **Given Condition:**
   The condition $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ implies that the triangle is an equilateral triangle. This is because, in an equilateral triangle, all angles are equal, and hence their cosines are equal. Therefore, the sides must be proportional to each other in the same way, which is consistent with the given condition.

2. **Equilateral Triangle Properties:**
   In an equilateral triangle, all sides are equal, and all angles are $60^\circ$. Since $a = 2$, it follows that $b = 2$ and $c = 2$ as well.

3. **Area of an Equilateral Triangle:**
   The formula for the area of an equilateral triangle with side length $s$ is given by:
   $$
   \text{Area} = \frac{\sqrt{3}}{4} s^2
   $$
   Substituting $s = 2$, we get:
   $$
   \text{Area} = \frac{\sqrt{3}}{4} \times 2^2 = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3}
   $$

4. **Conclusion:**
   Therefore, the area of $\Delta ABC$ is $\sqrt{3}$, which corresponds to Option 4.

### Why Other Options are Incorrect:

- **Option 1 (1):** This option suggests that the area is 1, which is incorrect because, as calculated, the area of an equilateral triangle with side length 2 is $\sqrt{3}$.

- **Option 2 ($\frac{1}{2}$):** This option is incorrect because it significantly underestimates the area. The correct calculation shows that the area is $\sqrt{3}$.

- **Option 3 ($\frac{\sqrt{3}}{2}$):** This option is incorrect because it is half of the correct area. The correct area, as derived, is $\sqrt{3}$.

In conclusion, the correct answer is Option 4, $\sqrt{3}$, based on the properties of an equilateral triangle and the given conditions.