# Identifying the square $DEFG$ and than finding the value of its perimeter.

Let $$ABC$$ be a triangle and $$DEFG$$ be a square, where $$D, E$$ points are located on $$AB$$ and $$AC$$ or their extension line. $$F, G$$ points are located on $$BC$$ or the extension of $$BC$$. The perpendicular distance from $$A$$ on $$BC$$ is $$2$$ units and $$BC$$ = $$6$$ units. What is the value of the perimeter of $$DEFG$$?

Here, I'm little bit confused about the exact location of the points $$D, E, F, G$$ respectively. I noticed that this 4 points could locate on either on the side of the triangle or on their extension line. So, I thought that the length of the square would be variable according to the various construction of the triangle $$ABC$$. I couldn't find a way to figure it out. Therefore, I need some help about how to construct that square with keeping its length constant with above mentioned condition.

The error is highly excusable.

The distance from $$A$$ to the nearest side of the square is $$|x-2|$$, where $$x$$ is a side of the square.
Thus, by the similarity we obtain: $$\frac{|x-2|}{2}=\frac{x}{6},$$ which gives $$x=\frac{3}{2}$$ and the answer $$6$$ or $$x=3$$ and the answer $$12$$.
• @Anirban Niloy I don't know to draw in the net. Try to draw the obtuse-angled triangle, $\measuredangle BAC>90^{\circ}.$ The altitude from $A$ is $2$ and $BC=6$. Now, easy to draw our square. – Michael Rozenberg Jan 24 at 19:10
• @MichaelRozenberg You assumed that the distance to the nearest side is $2-x$. That means that $DE$ is on the same side of $A$ as $BC$. You also have a solution on the opposite side. The distance is then $x-2$ instead, which yields the solution $x=3$, with perimeter $12$ – Andrei Jan 24 at 19:59