Biggest Reuleaux Triangle inscribed within a square which is inscribed within a hexagon?

Here we will see the area of biggest Reuleaux triangle inscribed within a square which is inscribed in a regular hexagon. Suppose 'a' is the side of the hexagon. The side of the square is x and the height of the Reuleaux triangle is h.

Syntax

float areaReuleaux(float hexagonSide);

From the formula of each side of inscribed square inside one hexagon is −

x = 1.268a

The height of the Reuleaux triangle is the same as x. So x = h. So the area of the Reuleaux triangle is −

Area = ((? - ?3) * h²)/2
     = ((? - ?3) * (1.268a)²)/2

Example

This program calculates the area of the biggest Reuleaux triangle inscribed within a square which is inscribed within a regular hexagon −

#include <stdio.h>
#include <math.h>

float areaReuleaux(float a) { 
    if (a < 0) 
        return -1;
    
    float x = 1.268 * a;  // side of inscribed square
    float area = ((3.1415 - sqrt(3)) * x * x) / 2;
    return area;
}

int main() {
    float side = 5;
    printf("Hexagon side: %.2f<br>", side);
    printf("Area of Reuleaux Triangle: %.4f<br>", areaReuleaux(side));
    return 0;
}

Hexagon side: 5.00
Area of Reuleaux Triangle: 28.3268

Key Points

• The relationship between hexagon side 'a' and inscribed square side is x = 1.268a
• The Reuleaux triangle height equals the square's side length
• The formula uses (? - ?3) as the key geometric constant

Conclusion

The area of the biggest Reuleaux triangle inscribed in a square within a hexagon depends on the hexagon's side length. The formula combines geometric relationships between these three shapes to calculate the final area efficiently.