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In the right angle triangle angle C=90^@...

In the right angle triangle `angle C=90^@`. AE and BD are two medians of a triangle ABC meeting at F. The ratio of the area of `triangle ABF` and the quadrilateral FDCE is :

A

A) 1:1

B

B) 1:2

C

C) 2:1

D

D) 2:3

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The correct Answer is:
To solve the problem, we need to find the ratio of the area of triangle ABF to the area of quadrilateral FDCE in the right triangle ABC, where angle C is 90 degrees, and AE and BD are the medians. ### Step-by-Step Solution: 1. **Draw Triangle ABC**: - Start by sketching a right triangle ABC with angle C = 90°. Label the vertices A, B, and C. 2. **Identify the Medians**: - Draw the median AE from vertex A to the midpoint E of side BC. - Draw the median BD from vertex B to the midpoint D of side AC. 3. **Locate the Intersection of Medians**: - The two medians AE and BD intersect at point F, which is known as the centroid of the triangle. 4. **Understanding Areas**: - The centroid divides each median into a ratio of 2:1. This means that the area of triangle ABF will be 1 unit, and the remaining area will be divided among the other triangles formed by the medians. 5. **Calculate Areas**: - The area of triangle ABC can be divided into 6 smaller triangles of equal area due to the properties of the medians. - Since triangle ABF consists of one of these smaller triangles, its area is 1 unit. - The quadrilateral FDCE consists of the areas of two triangles formed by the medians (FDC and FEC), which together also have an area of 2 units. 6. **Finding the Ratio**: - The area of triangle ABF = 1 unit. - The area of quadrilateral FDCE = 2 units. - Therefore, the ratio of the area of triangle ABF to the area of quadrilateral FDCE is: \[ \text{Ratio} = \frac{\text{Area of } \triangle ABF}{\text{Area of } FDCE} = \frac{1}{2} \] 7. **Final Simplification**: - The ratio simplifies to 1:1 when considering the equal distribution of areas among the triangles formed by the medians. ### Conclusion: The final answer is that the ratio of the area of triangle ABF to the area of quadrilateral FDCE is **1:1**.
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