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GEOMETRY

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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 .

Jun 19, 2018

#1
+2298
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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 $$BCDE$$

$$A_{\text{ABCD}}=9+15=24\text{units}^2$$

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Jun 20, 2018

#1
+2298
+1

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 $$BCDE$$

$$A_{\text{ABCD}}=9+15=24\text{units}^2$$

TheXSquaredFactor Jun 20, 2018