ABC is an equilateral triangle. P and Q are two points on `bar(AB)` and `bar(AC)` respectively such that `bar(PQ) || bar(BC)`. If `bar(PQ) = 5cm`, then area of `Delta APQ` is :

Options

A. `(25)/(4)` sq. cm
B. `(25)/(sqrt(3))` sq. cm
C. `(25sqrt(3))/(4)` sq. cm
D. `25sqrt(3)` sq. cm

Answer

To solve the problem, we need to find the area of triangle APQ given that PQ is parallel to BC and PQ = 5 cm.

### Step-by-Step Solution:

1. **Understanding the Triangle**:
- We have an equilateral triangle ABC where all sides are equal and all angles are 60 degrees.
- Points P and Q are on sides AB and AC respectively such that line segment PQ is parallel to side BC.

2. **Properties of Parallel Lines**:
- Since PQ is parallel to BC, the angles formed are equal due to the properties of corresponding angles.
- Therefore, angle APQ = angle ABC = 60 degrees and angle AQP = angle ACB = 60 degrees.

3. **Identifying Triangle APQ**:
- Since angles APQ and AQP are both 60 degrees, triangle APQ is also an equilateral triangle.

4. **Finding the Length of Sides of Triangle APQ**:
- Given that PQ = 5 cm and since triangle APQ is equilateral, all sides of triangle APQ are equal. Thus, PA = AQ = PQ = 5 cm.

5. **Calculating the Area of Triangle APQ**:
- The formula for the area of an equilateral triangle with side length 'a' is given by:
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]
- Here, a = 5 cm. Therefore, substituting the value:
\[ \text{Area} = \frac{\sqrt{3}}{4} (5)^2 = \frac{\sqrt{3}}{4} \times 25 = \frac{25\sqrt{3}}{4} \text{ cm}^2 \]

### Final Answer:
The area of triangle APQ is \(\frac{25\sqrt{3}}{4} \text{ cm}^2\).

---