The park problem 2

A city is planning a triangular park. Two of its corners, $B$ and $C$ , are already fixed. The third corner $A$ must lie somewhere on a straight line that runs parallel to the line $(BC)$ .
The city wants the park’s area to be as large as possible. Where should $A$ be placed?

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Solution by Ancient Tree

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The area of this triangle is given by $\frac{1}{2} \;\times$ base $\times$ height. Take $BC$ as the base. Notice that the position of $A$ on the line does not change the corresponding height. So the area of the triangle is always the same, regardless of the choice of position of $A$ . This means that the city can chose $A$ wherever they like!

Solution by goldfinchFR

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L’aire du triangle est donnée par : $\frac{1}{2} \;\times$ base $\times$ hauteur.
Prendre $BC$ comme base
Noter que la position de A sur la droite ne change pas la valeur de la hauteur du triangle. Ainsi la valeur deu triangle reste la même quelle que soit la position de A. La ville a donc toute latitude pour choisir A

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