Biggest Reuleaux Triangle inscirbed within a square inscribed in a semicircle?

Here we will see how to calculate the area of the biggest Reuleaux triangle inscribed within a square which is inscribed in a semicircle. Given the radius of the semicircle as R, the side of the square as 'a', and the height of the Reuleaux triangle as h.

Syntax

float areaReuleaux(float radius);

We know that the side of the square inscribed in a semicircle is −

a = (2R)/?5

The height of the Reuleaux triangle equals the side of the square, so h = a. The area of the Reuleaux triangle is −

Area = (? - ?3) × h²/2

Substituting h = (2R)/?5, we get −

Area = (? - ?3) × (2R/?5)²/2

Example

This program calculates the area of the biggest Reuleaux triangle inscribed in a square within a semicircle −

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

float areaReuleaux(float r) {
    if (r < 0) // Invalid radius
        return -1;
    
    float side = (2 * r) / sqrt(5); // Side of inscribed square
    float area = ((M_PI - sqrt(3)) * side * side) / 2;
    return area;
}

int main() {
    float radius = 8.0;
    float area = areaReuleaux(radius);
    
    printf("Semicircle radius: %.2f<br>", radius);
    printf("Area of Reuleaux Triangle: %.4f<br>", area);
    
    return 0;
}

Semicircle radius: 8.00
Area of Reuleaux Triangle: 36.0819

Key Points

• The square inscribed in a semicircle has side length (2R)/?5
• The Reuleaux triangle's height equals the square's side length
• The area formula is (? - ?3) × h²/2 where h is the height

Conclusion

The area of the biggest Reuleaux triangle inscribed in a square within a semicircle can be calculated using the formula (? - ?3) × (2R/?5)²/2, where R is the semicircle's radius.