# Thread: Maximum area of rectangle inscribed in a scalene triangle

1. ## Maximum area of rectangle inscribed in a scalene triangle

One side of the rectangle is on the same line as one of the triangles sides. There are not any figures. I know the answer is half of the area of the triangle. If I knew what to make the independent variable I might manage this.

2. ## Re: Maximum area of rectangle inscribed in a scalene triangle

I have solved this my making the independent variable y, the height of the rectangle. I drew a line from the top point of the triangle to the side opposite, in the rectangle somewhere. I found the minimum area of the two inscribed triangles on the left which gives the maximum area of the rectangle on the left. Here y=h/2 and finished with some geometry in writing to explain that the four inscribed triangles are half of the main triangles area.

3. ## Re: Maximum area of rectangle inscribed in a scalene triangle

I would approach this problem with coordinate geometry. Orient the scalene triangle with the side on which the rectangle also lies along the xaxis and the left vertex at the origin. In order for the triangle to be inscribed, its 4 vertices must be tangent to the triangle, so neither of the two lower angles are obtuse.

Without loss of generality let the horizontal side be 1 unit in length. Let the altitude of the triangle be h. The vertices of the triangle are then at the points:

$\displaystyle (0,0),(1,0),(x_0,h)$

The side of the triangle from $\displaystyle (0,0)-(x_0,h)$ lies along the line:

$\displaystyle y=\frac{h}{x_0}x$

The side of the triangle from $\displaystyle (x_0,h)-(1,0)$ lies along the line:

a) If $\displaystyle x_0<1$:

$\displaystyle y=\frac{h}{x_0-1}(x-1)$

b) If $\displaystyle x_0=1$:

$\displaystyle x=1$

First case:

The base of the rectangle is:

$\displaystyle x_2-x_1$

The height of the rectangle is:

$\displaystyle \frac{h}{x_0}x_1=\frac{h}{x_0-1}(x_2-1)$

From this, we get:

$\displaystyle x_2=\frac{x_1}{x_0}(x_0-1)+1$

And so the base of the rectangle is:

$\displaystyle \frac{x_1}{x_0}(x_0-1)+1-x_1=\frac{x_0-x_1}{x_0}$

Thus, the area of the rectangle is:

$\displaystyle \left(\frac{h}{x_0}x_1 \right)\left(\frac{x_0-x_1}{x_0} \right)=\frac{h}{x_0^2}(x_1(x_0-x_1))$

This is a quadratic in $\displaystyle x_1$ having the roots $\displaystyle x_1=0,x_0$ so we know the area is maximized for $\displaystyle x_1=\frac{x_0}{2}$ and so the maximum area is:

$\displaystyle \frac{h}{4}$

Since the area of the triangle is $\displaystyle \frac{h}{2}$, the area of the rectangle is half that of the triangle.

I will let you work out the case of the right triangle.

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### area of scaline triangle in a rectangle

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