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let $abcd$ be a parallelogram. extend $\overline{bc}$ past $b$ to $f$, and let $e$ be the intersection of $\overline{ab}$ and $\overline{df}$. if the areas of triangles $\triangle bef$ and $\triangle ade$ are $1$ and $9$, respectively, find the area of parallelogram $abcd$.

Sagot :

The area of parallelogram is 27 units

Let's take a look at two triangles, BEF and ADE, which are similar. By three angles. Two angles are vertical, and the rest are equal, just like opposite interior angles between two parallel lines (in this case, BC and AD by parallelogram definition). We know that the ratio of the areas of similar triangles is equal to the square of the coefficient of similarity. k^2=1/9k=1/3. That means the DE/EF ratio is 3:1. On the other side, we can see two more triangles: DFC and BEF. They are also similar in that BE is parallel to DC - the foot of DCF. We recall that DE/EF = 3:1, which means that if EF=x, DE=3x (and DF=4x).

We can say that the similarity coefficient between DFC and BEF is 1 to 4. And the area ratio is 1 to 16. The area of the large triangle (DFC) is 16, and the area of the small triangle (BEF) is 1. The area of BCDE equals the area of DFC minus the area of BEF ⇒16-1=15  So, to find the area of ABCD, we must take the area ADE and add it to the area BCDE

⇒9+15=27

Therefore,The area of parallelogram is 27 units

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