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Area question

Posted February 21st, 2008 by alpha
in

Two parallelograms ABCD and AEFB are drawn on the opposite sides of AB, Prove that
area of parallelogram ABCD+ area of parallelogram AEFB=area of parallelogram EFCD

‹ find the height of the cone Prove that this quadrilateral is a parallelogram ›

Comments

Parallelogram area

On February 27th, 2008 Structure says:
$$Area(FEDC)=|\overrightarrow{FE}\times\overrightarrow{ED}|$$
$$Area(FEAB)=|\overrightarrow{FE}\times\overrightarrow{EA}|$$
$$Area(BADC)=|\overrightarrow{BA}\times\overrightarrow{AD}|=|\overrightarrow{FE}\times\overrightarrow{AD}|$$

But

$$\overrightarrow{ED}=\overrightarrow{EA}+\overrightarrow{AD}$$

Then

$$\overrightarrow{FE}\times\overrightarrow{ED}=\overrightarrow{FE}\times(\overrightarrow{EA}+\overrightarrow{AD})=\overrightarrow{FE}\times\overrightarrow{EA}+\overrightarrow{FE}\times\overrightarrow{AD}$$

As the three vectors are all with the same orientation we have also

$$|\overrightarrow{FE}\times\overrightarrow{ED}|=|\overrightarrow{FE}\times\overrightarrow{EA}|+|\overrightarrow{FE}\times\overrightarrow{AD}|$$

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