Let ABCD be a parallelogram. Extend BC past B to F, and let E be the intersection of AB and FD. If the areas of triangles BEF and ADE are 1 and 9, respectively, find the area of parallelogram .

Answer 1

We know that the area of triangle \( BEF \) is given as \( 1 \) and the area of triangle \( ADE \) is given as \( 9 \).

To find the area of parallelogram \( ABCD \), we can first consider the relationship between the areas of the triangles and the parallelogram they form.

Let \( [X] \) denote the area of figure \( X \).

The area of parallelogram \( ABCD \) can be represented in terms of the areas of triangles \( ADE \) and \( BCE \).

Since \( ABCD \) is a parallelogram, triangles \( ADE \) and \( BEC \) are equal in area, as they share the same base \( DE \) and lie between the same two parallels \( AB \) and \( CD \). Thus, we can express:

\[ [BEC] = [ADE] = 9 \]

Now, considering triangle \( BEF \), we can find the area of triangle \( BEC \):

\[ [BEC] = [BEF] + [BCE] = 1 + [BCE] \]

This gives us:

\[ 9 = 1 + [BCE] \]

By simplifying, we find:

\[ [BCE] = 9 - 1 = 8 \]

Now, we've established the area of triangle \( BCE \) and know that the area of triangle \( ADE = 9 \). Therefore, the total area of the parallelogram \( ABCD \) is the sum of the areas of triangles \( ADE \) and \( BCE \):

\[ [ABCD] = [ADE] + [BCE] = 9 + 8 = 17 \]

Thus, the area of parallelogram \( ABCD \) is:

\[ \boxed{17} \]