Find the side of the squares which are lined in a row, and distance between the centers of first and last square is given
Last Updated :
31 May, 2022
Given here are n squares which touch each other externally, and are lined up in a row. The distance between the centers of the first and last square is given. The squares have equal side length. The task is to find the side of each square.
Examples:
Input: d = 42, n = 4
Output: The side of each square is 14
Input: d = 36, n = 5
Output: The side of each square is 9
Approach:
Suppose there are n squares each having side of length a.
Let, the distance between the first and last squares = d
From the figure, it is clear,
a/2 + a/2 + (n-2)*a = d
a + na – 2a = d
na – a = d
so, a = d/(n-1)
C++
#include <bits/stdc++.h>
using namespace std;
void radius(int n, int d)
{
cout << "The side of each square is "
<< d / (n - 1) << endl;
}
int main()
{
int d = 42, n = 4;
radius(n, d);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void radius(int n, int d)
{
System.out.print( "The side of each square is "
+ d / (n - 1));
}
public static void main (String[] args)
{
int d = 42, n = 4;
radius(n, d);
}
}
|
Python3
def radius(n, d):
print("The side of each square is ",
d / (n - 1));
d = 42; n = 4;
radius(n, d);
|
C#
using System;
class GFG
{
static void radius(int n, int d)
{
Console.Write( "The side of each square is "
+ d / (n - 1));
}
public static void Main ()
{
int d = 42, n = 4;
radius(n, d);
}
}
|
Javascript
<script>
function radius(n , d)
{
document.write( "The side of each square is "
+ d / (n - 1));
}
var d = 42, n = 4;
radius(n, d);
</script>
|
Output: The side of each square is 14
Time Complexity: O(1)
Auxiliary Space: O(1)
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