The three medians AD, BE and CF of `Delta`ABC intersect at point G. If the area of `DeltaABC` is 60 sq.cm. then the area of the quadrilateral BDGF is :

Options

A. 10 sq.cm
B. 15 sq.cm
C. 20 sq.cm
D. 30 sq.cm

Answer

To solve the problem, we need to find the area of the quadrilateral BDGF given that the area of triangle ABC is 60 sq. cm. The medians of the triangle intersect at point G, which is the centroid of the triangle.

### Step-by-Step Solution:

1. **Understanding the Medians**:
The medians of a triangle are the line segments that connect each vertex to the midpoint of the opposite side. In triangle ABC, the medians are AD, BE, and CF.

2. **Identifying the Centroid**:
The point G where all three medians intersect is called the centroid. The centroid divides each median into two segments, with the segment connecting the vertex to the centroid being twice the length of the segment connecting the centroid to the midpoint of the side.

3. **Area Division by Medians**:
The three medians divide triangle ABC into six smaller triangles of equal area. Since the area of triangle ABC is 60 sq. cm, each of these smaller triangles has an area of:
\[ \text{Area of each smaller triangle} = \frac{\text{Area of } \Delta ABC}{6} = \frac{60}{6} = 10 \text{ sq. cm} \]

4. **Identifying Quadrilateral BDGF**:
The quadrilateral BDGF is formed by two of these smaller triangles: BGD and DGF. Since each of these triangles has an area of 10 sq. cm, the area of quadrilateral BDGF is:
\[ \text{Area of } BDGF = \text{Area of } \triangle BGD + \text{Area of } \triangle DGF = 10 + 10 = 20 \text{ sq. cm} \]

5. **Conclusion**:
Therefore, the area of quadrilateral BDGF is 20 sq. cm.

### Final Answer:
The area of quadrilateral BDGF is **20 sq. cm**.