ABCD is a parallelogram in which BC is produced to E suh that CE=BC.AE intersects CD at F.if ar(DFB)=3cm2find the area of the parallelogram ABCD.

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To find the area of the parallelogram ABCD, we can use the fact that the area of any triangle is half the product of its base and height.

Since BCF is a triangle, we know that:

Area of triangle BCF = (1/2) * base * height

In this case, the base is BF and the height is the distance between BF and the line parallel to BF passing through A.

Let's solve the problem step-by-step:

1. Draw the given parallelogram ABCD.
2. Extend the line BC to a point E such that CE = BC.
3. Draw a line from A to intersect CD at F.
4. Identify triangle BCF in the diagram.

Now, we need to find the height of triangle BCF in order to calculate its area. To do so, we will use the given information that the area of triangle DFB is 3 cm^2.

Given: Area of triangle DFB = 3 cm^2

We can use the fact that the ratio of the areas of two triangles with a common base is equal to the ratio of their heights.

Therefore, (Area of triangle DFB) / (Area of triangle BCF) = (Height of triangle DFB) / (Height of triangle BCF)

We know that the area of triangle DFB is 3 cm^2. Suppose the height of triangle DFB is h, and the height of triangle BCF is H.

Substituting the given values, we get:

3 cm^2 / (Area of triangle BCF) = h / H

Since the area of triangle BCF is half the area of the parallelogram, let's denote it as A/2, where A is the area of the parallelogram ABCD.

3 cm^2 / (A/2) = h / H
6 cm^2 / A = h / H

Now, we need to find the height of the parallelogram ABCD, which is represented by H. To find H, we need to calculate the area of triangle ABC, which is half the area of the parallelogram.

Let's denote the area of triangle ABC as X, where X = (A/2).

Now, we have:

6 cm^2 / (2X) = h / H
3 cm^2 / X = h / H

We know, triangle ABC and triangle BCF share the same height because they are parallel and have the same base (BC). Therefore, H = X.

Replacing H with X, we have:

3 cm^2 / X = h / X

So, h = 3 cm^2.

Now, we can find the area of triangle BCF:

Area of triangle BCF = (1/2) * base * height
= (1/2) * BF * 3 cm^2
= (3/2) * BF cm^2

Since parallelogram ABCD has the same area as triangle ABC, we have:

Area of parallelogram ABCD = 2 * Area of triangle ABC
= 2 * X
= 2 * (base * height)
= 2 * (BC * BF)

Therefore, the area of parallelogram ABCD is 2 times the product of the length of BC and the length of BF.