If two medians BE and CF of a triangle ABC , intersect each other at G and if `BG=CG, angle BGC=60^(@), BC=8 cm` then area of the triangle ABC is :

Answer

To find the area of triangle ABC given the conditions in the problem, we can follow these steps:

### Step 1: Understand the properties of medians
In triangle ABC, BE and CF are medians, which means they divide the opposite sides into two equal parts. The point G is the centroid of the triangle, which divides each median in the ratio 2:1.

### Step 2: Set up the known values
We know:
- \( BG = CG \) (since G is the centroid)
- \( \angle BGC = 60^\circ \)
- \( BC = 8 \, \text{cm} \)

Since \( BG = CG \), we can denote \( BG = CG = x \).

### Step 3: Use the Law of Cosines in triangle BGC
In triangle BGC, we can apply the Law of Cosines to find the length of BC:
\[ BC^2 = BG^2 + CG^2 - 2 \cdot BG \cdot CG \cdot \cos(\angle BGC) \]
Substituting the known values:
\[ 8^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we have:
\[ 64 = 2x^2 - 2x^2 \cdot \frac{1}{2} \]
This simplifies to:
\[ 64 = 2x^2 - x^2 \]
\[ 64 = x^2 \]
Thus, \( x = 8 \).

### Step 4: Calculate the area of triangle BGC
The area of triangle BGC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \cdot BG \cdot CG \cdot \sin(\angle BGC) \]
Substituting the values:
\[ \text{Area} = \frac{1}{2} \cdot 8 \cdot 8 \cdot \sin(60^\circ) \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ \text{Area} = \frac{1}{2} \cdot 8 \cdot 8 \cdot \frac{\sqrt{3}}{2} = \frac{64\sqrt{3}}{4} = 16\sqrt{3} \, \text{cm}^2 \]

### Step 5: Calculate the area of triangle ABC
Since G is the centroid, the area of triangle ABC is three times the area of triangle BGC:
\[ \text{Area of } \triangle ABC = 3 \cdot \text{Area of } \triangle BGC = 3 \cdot 16\sqrt{3} = 48\sqrt{3} \, \text{cm}^2 \]

Thus, the area of triangle ABC is \( 48\sqrt{3} \, \text{cm}^2 \).

---