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Find the maximum area of a rectangle placed in a right angle triangle $\triangle ABC$.

My Work:

  • Pythagoras: $$c^2=a^2+b^2$$
  • Area of a triangle: $$\text{area triangle}=\frac{1}{2}\times\text{height triangle}\times\text{width triangle}=\frac{1}{2}\times a \times b$$
  • Area of a rectangle: $$\text{area rectangle}=\text{height rectangle}\times\text{width rectangle}= h \times w$$
  • And logically we know that: $$\text{area triangle}>\text{area rectangle}$$

Now, how to continue?

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  • $\begingroup$ Try to generalise the result given here. $\endgroup$ – Rohan Feb 1 '17 at 17:17
  • $\begingroup$ There are many other questions on this site that are essentially the same as this, but with specific numbers instead of $a,$ $b,$ and $c.$ The solutions easily generalize. Did you try looking for any of them? (Search for "rectangle right triangle", for example.) $\endgroup$ – David K Feb 1 '17 at 17:18
  • $\begingroup$ In particular, look at math.stackexchange.com/questions/1911117/… and math.stackexchange.com/questions/1895231/… $\endgroup$ – David K Feb 1 '17 at 17:19
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The area of the Rectangle is half the area of the triangle, and this is even universal and not dependent on the triangle be right angled.

Pick the base of the triangle that is longest that has length $b$ and associated height $h$. Now you can place on there a triangle of width $b/2$ and height $h/2$.

There must be a position where it fits since at half height the triangle has still half its width by the interception theorem.

So the area of the rectangle is given by $\frac{b}{2} \frac{h}{2} = b h / 4$ and as such has half the area of the triangle.

For illustration look at http://jwilson.coe.uga.edu/emt725/Class/Pearman/rectangle.triangle/rect.tri.html

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    $\begingroup$ In a right triangle or acute triangle, you don't even need to use the largest side to place the base of your triangle. That is required only for obtuse triangles. $\endgroup$ – David K Feb 1 '17 at 17:25
  • $\begingroup$ The top of the rectangle is the line joining the midpoints of the other two sides. $\endgroup$ – Joffan Feb 1 '17 at 17:27
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For $\triangle ABC$, choose the longest side, $AB$, as your base, length $b$. Then you can choose the width of the rectangle as a proportion $p$ of that length, and the height will be the reverse proportion, $(1-p)$ of the altitude of $C$, $h$, by choosing the rectangle corners to be $p$ of the side length from $C$ to $A$ and $B$.

So the area of the rectangle is $pb(1-p)h = p(1-p)2T$, where $T$ is area of $\triangle ABC$. $p(1-p)$ has zeros at $0$ and $1$ and is maximum at $p=0.5$. Thus the maximum rectangle with area of $T/2$ is produced by joining the midpoints of the shorter sides and dropping perpendicular to the long side.

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enter image description here

We have to consider two cases: an inscribed rectangle with a side on the hypotenuse; an inscribed rectangle with a vertex at the right angle. In the first case the maximum area is achieved when a side of the rectangle crosses the midpoint of the depicted height; in the second case the maximum area is achieved when a vertex of the rectangle is the midpoint of the hypotenuse. In both cases the maximum area is just half the area of the original triangle, hence $\frac{ab}{4}$. On the other hand, the area of an inscribed rectangle cannot exceed half the area of the original triangle, since by "folding" the white regions on the blue inscribed rectangle, we entirely cover the rectangle: enter image description here

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