+194 Let ABCD be a parallelogram. Extend BC past B to F, and let E be the intersection of AB and DF. If the areas of triangles BEF and ADE are 1 and 9, respectively, find the area of parallelogram .
+2441 This question requires some observation combined with knowledge of properties of parallelograms and triangle similarity.
Parallelograms, by definition, are quadrilaterals with opposite sides parallel. In this example, \(\overline{AD}\parallel \overline {FC}\) because of the previous statement. Because \(\overline{DF}\) intersects multiple segments, namely \(\overline{AD}\text{ and } \overline {FC}\), \(\overline{DF}\) is classified as a transversal. \(\angle ADF\text{ and }\angle F\) are a pair of angles known as alternate interior angles because they exist on opposite sides of the transversal while remaining on the "inner" side of the intersecting lines. Because \(\overline{AD}\parallel \overline {FC}\), the Alternate Interior Angles Theorem states that \(\angle ADF \cong \angle F\). \(\overline{AB}\) is another transversal bounded by parallel lines. Therefore, \(\angle A\cong \angle FBE\). Because there are two pairs of angles that are congruent in corresponding triangles, namely \(\angle ADF \cong \angle F\text{ and }\angle A\cong \angle FBE\), the Angle-Angle Similarity Theorem states that the triangles are similar. In this case, \(\angle AED\sim \triangle BEF\)
Even after all the logic above, I still do not any side lengths yet. There exists a constant of proportionality between similar triangles. Since the area of \(\triangle BEF\) requires an enlargement by a factor of 9 to obtain the area of \(\angle AED\), the constant of proportionality would be the square root of this. \(\sqrt{9}=3\). Therefore, each side length on the larger triangle is enlarged by a factor of 3. Because \(\overline{BF}\) and \(\overline{AD}\) are corresponding sides according to the triangle similarity statement and \(\overline{AD}\) is enlarged by 3, \(AD=3BF\).
One property of parallelograms is that opposite side lengths are also congruent. Because \(BC=AD\text{ and }AD=3BF \), \(BC=3FC\).
Using the principle of the Segment Addition Postulate \(FC=BF+BC=BF+3BF=4BF\).
Since the constant of proportionality for \(\triangle BEF\text{ and }\triangle CDF\) is 4, the area of \(\triangle CDF\) is 16.
16-1=15, which is the area of the quadrilateral \(BCDE\).
\(A_{\text{ABCD}}=9+15=24\text{units}^2\)
+2441 This question requires some observation combined with knowledge of properties of parallelograms and triangle similarity.
Parallelograms, by definition, are quadrilaterals with opposite sides parallel. In this example, \(\overline{AD}\parallel \overline {FC}\) because of the previous statement. Because \(\overline{DF}\) intersects multiple segments, namely \(\overline{AD}\text{ and } \overline {FC}\), \(\overline{DF}\) is classified as a transversal. \(\angle ADF\text{ and }\angle F\) are a pair of angles known as alternate interior angles because they exist on opposite sides of the transversal while remaining on the "inner" side of the intersecting lines. Because \(\overline{AD}\parallel \overline {FC}\), the Alternate Interior Angles Theorem states that \(\angle ADF \cong \angle F\). \(\overline{AB}\) is another transversal bounded by parallel lines. Therefore, \(\angle A\cong \angle FBE\). Because there are two pairs of angles that are congruent in corresponding triangles, namely \(\angle ADF \cong \angle F\text{ and }\angle A\cong \angle FBE\), the Angle-Angle Similarity Theorem states that the triangles are similar. In this case, \(\angle AED\sim \triangle BEF\)
Even after all the logic above, I still do not any side lengths yet. There exists a constant of proportionality between similar triangles. Since the area of \(\triangle BEF\) requires an enlargement by a factor of 9 to obtain the area of \(\angle AED\), the constant of proportionality would be the square root of this. \(\sqrt{9}=3\). Therefore, each side length on the larger triangle is enlarged by a factor of 3. Because \(\overline{BF}\) and \(\overline{AD}\) are corresponding sides according to the triangle similarity statement and \(\overline{AD}\) is enlarged by 3, \(AD=3BF\).
One property of parallelograms is that opposite side lengths are also congruent. Because \(BC=AD\text{ and }AD=3BF \), \(BC=3FC\).
Using the principle of the Segment Addition Postulate \(FC=BF+BC=BF+3BF=4BF\).
Since the constant of proportionality for \(\triangle BEF\text{ and }\triangle CDF\) is 4, the area of \(\triangle CDF\) is 16.
16-1=15, which is the area of the quadrilateral \(BCDE\).
\(A_{\text{ABCD}}=9+15=24\text{units}^2\)
