1

I am interested to know if, there exist any other set of geometrical figures/shapes which have same perimeter and same area (considering only integer values). For example:

(1) Rectangle with measures of sides as: length = 6, breadth = 2 and

(2) A parallelogram with sides as 4, 5 and such that the length of vertical dropped from a vertex to base/ height would be 3.

Here, in this case, we can see that area of rectangle = $2 × 6 = 12$ and perimeter of the rectangle = $2 × (2 + 6) = 16$

Also, for the parallelogram,the area = $4 × 3 = 12$ and it's perimeter = $2×(3 + 5)= 16$.

That is - two different geometrical figures - having their perimeters same and areas too!

Is there a formal way of approach to arrive at the possible solution sets, rather than trial and error?

Karri Chandrasekhar
  • 233
  • 1
    Related: "Can two figures (of any kind, bounded by curves or not) have the same area and perimeter, but different shapes?". This doesn't address strategies for finding such figures, so it's not really a duplicate. – Blue Nov 17 '21 at 18:05
  • Take any two squares with sides $a > b$, glue the small square to the the "interior" of an edge of the larger square. The resulting figure has perimeter $4a + 2b$ and area $a^2+b^2$. – achille hui Nov 17 '21 at 18:35
  • 1
    The perimeter of your parallelogram is $2\times(4+5)=18$. The smaller side is $4$, not $3$. – Andrei Nov 17 '21 at 18:56
  • If you have a figure with (integer) area and smaller numerical perimeter, you can shear it to keep the area fixed while increasing the perimeter slowly, hence eventually arise at the same area. So, there are infinitely many such shapes, and you're unlikely to get a simply way to describe all of them. EG If you start with any square of length $l \geq 4$, this approach yields a parallelogram of side length $ l , l^2/2-l $ $\quad$ Note: We can always scale up a shape with factor $k$ which makes the area $k^2$ and perimeter $k$ times, so eventually the perimeter will be numerically smaller. – Calvin Lin Nov 17 '21 at 20:20
  • Your might be interested in this previous Question which concerns geometric figures (quadrilaterals and triangles) with integer sides whose (integer) perimeter equals its area. So a special case of what you ask about is a $5:12:13$ right triangle and a $5:5:6:14$ cyclic quadrilateral (both having perimeter = area = $30$). Sides of a cyclic quadrilateral can be permuted freely without changing area (or perimeter), so the latter quadrilateral is realized both as a trapezoid and non-trapezoid. – hardmath Nov 17 '21 at 20:22
  • @Andrei - yeah, you are correct, my bad. The height and base values got interchanged. Please read them as : 5 and 3 with an altitude of 4.Thanks. – Karri Chandrasekhar Nov 18 '21 at 02:49

0 Answers0