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How can I find the Area of this figure?

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It is quite curious because it is a particular case of this sequence:

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Anyone know how to find the area of this sequence as a function of the number of segments?

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2 Answers 2

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Suppose you are looking at the shape given by the lines $L_{x,y}$ linking $(x,0)$ to $(0,y)$ when $x+y=n$ (your examples are $n=3$ and $n=8$). For $0\le k < n$, call $(x_k,y_k)$ the intersection of $L_{n-k,k}$ with $L_{n-k-1,k+1}$. A few calculations show that $(x_k,y_k) = (\frac{(n-k)(n-k-1)}n,\frac {k(k+1)}n)$

Then, decompose your shape in $n-1$ triangles of base $1$ and height $y_k$ : The total area is then $\frac 1 2 \sum_{k=1}^{n-1} \frac {k(k+1)}n = \frac {(n+1)n(n-1)}{6n} = \frac{n^2-1}6$

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Write the points as Cartesian coordinates. So $O=(0,0)$, $A=(0,2k)$, $B=(2k,0)$, and $D$, the point where the the two hypotenuses meet (solving two equations $2y+x = 2k$ and $y+2x=2k$) is $(\frac{2k}{3},\frac{2k}{3})$.

The two triangles, $ODA$ and $ODB$ are congruent, so the total area is twice the area of $ODA$. But that's just the area of the parallelogram, $O,D,A,D+A$, which, if you remember your linear algebra, is just the determinant, $\frac{2k}{3}2k - \frac{2k}{3}0 = \frac{4k^2}{3}$

The more general solution is $n(n+2)/6$ where $n=2$ in your first example, and $n=7$ in your second.

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