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# Daily Challenge #281 - Area or Perimeter

You are given the length and width of a 4-sided polygon. The polygon can either be a rectangle or a square.

If it is a square, return its area. If it is a rectangle, return its perimeter.

`area_or_perimeter(6, 10)` --> 32
`area_or_perimeter(4, 4)` --> 16

### Tests

`area_or_perimeter(5, 5)`
`area_or_perimeter(10, 20)`

Good luck!

This challenge comes from `no one` on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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## Discussion (11) Sergiu Pruteanu

learning Go

``````func area_or_perimeter(x int, y int) int {
if x == y {
return x * y
} else {
return 2 * (x + y)
}
}

func main() {
//tests
fmt.Println(area_or_perimeter(5, 5))
fmt.Println(area_or_perimeter(10, 20))
}
`````` Deepak Raj
``````def area_or_perimeter(a, b):
if a == b:
return a * b
else:
return 2 * (a + b)

assert area_or_perimeter(5, 5) == 25
assert area_or_perimeter(10, 20) == 60
`````` peter279k

Here is the simple solution with Python:

``````def area_or_perimeter(l , w):
if l == w:
return l * w
return (l + w) * 2
`````` Hridayesh Sharma • Edited on
``````const area_or_perimeter = (a,b) => a===b ? a*b : 2*(a+b);

`````` Alexander Girke

In Kotlin it would be:

``````fun area_or_perimeter(a: Int, b: Int) =
if (a == b)
a * b
else
2 * (a + b)
`````` Michael Kohl

A solution in Forth:

``````: area-or-perimeter 2dup = if * else + 2 * then ;
4 4 area-or-perimeter . 16  ok
6 10 area-or-perimeter . 32  ok
``````