Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I find the Area of this figure?

enter image description here

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

enter image description here

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

share|cite|improve this question
up vote 3 down vote accepted

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 $\frac 1 2 \sum_{k=1}^{n-1} \frac {k(k+1)}n = \frac {(n+1)n(n-1)}{6n} = \frac{n^2-1}6$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.