Question

Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm, then, the area of the triangle EOD, in sq. cm, is

Answer 1

Step 1: Area Ratio of Triangles △ABD and △BDC

Given: \[ \text{Area of } △ABD : \text{Area of } △BDC = 1 : 1 \] Therefore, \[ \text{Area of } △ABD = \frac{1}{2} \text{Area of } △ABC \] Let’s assume area of △ABC = \( 108 \, \text{sq cm} \), so: \[ \text{Area of } △ABD = 54 \, \text{sq cm} \]

Step 2: Area Ratio of Triangles △EDB and △ADE

Given: \[ \text{Area of } △EDB : \text{Area of } △ADE = 1 : 1 \] And since △ABD = △EDB + △ADE = 54: \[ \text{Area of } △ADE = \frac{54}{2} = 27 \, \text{sq cm} \]

Step 3: Using the Centroid Property

The centroid \( O \) divides each median in the ratio \( 2:1 \). So, the area of a triangle is divided accordingly. In triangle △ADE, point \( O \) lies on median, so: \[ \text{Area of } △BEO : \text{Area of } △EOD = 2 : 1 \] So, \[ \text{Area of } △ADE = \text{Area of } △BEO + \text{Area of } △EOD = 27 \] Let the area of △EOD = \( x \), then area of △BEO = \( 2x \) \[ x + 2x = 27 \Rightarrow 3x = 27 \Rightarrow x = 9 \]

Final Answer:

\[ \boxed{9 \, \text{sq cm}} \]