Area of circle inscribed within rhombus?

A circle inscribed in a rhombus touches all four sides of the rhombus. Each side of the rhombus acts as a tangent to the inscribed circle. To find the area of this inscribed circle, we need the lengths of the rhombus diagonals.

Where a and b are the diagonals of the rhombus, r is the radius of the inscribed circle, and O is the center.

Syntax

Area = ? × (a² × b²) / (4 × (a² + b²))

Formula Derivation

Using the property that the area of triangle AOB can be calculated in two ways −

• Area = ½ × OA × OB = ½ × (a/2) × (b/2) = ab/8
• Area = ½ × AB × r, where AB = ?((a/2)² + (b/2)²) = ?(a² + b²)/2

Equating both expressions and solving for r −

r = (a × b) / (2 × ?(a² + b²))

Therefore, Area of circle = ? × r² = ? × (a² × b²) / (4 × (a² + b²))

Example

Let's calculate the area of a circle inscribed in a rhombus with diagonals of length 5 and 10 −

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

int main() {
    int a = 5, b = 10;
    double pi = 3.14159;
    double area = (pi * a * a * b * b) / (4.0 * (a * a + b * b));
    
    printf("Rhombus diagonals: %d and %d
", a, b);
    printf("Area of inscribed circle: %.6f
", area);
    
    return 0;
}

Rhombus diagonals: 5 and 10
Area of inscribed circle: 15.707963

Key Points

• The inscribed circle touches all four sides of the rhombus
• Both diagonals of the rhombus pass through the center of the inscribed circle
• The formula works for any rhombus regardless of its orientation

Conclusion

The area of a circle inscribed in a rhombus depends on both diagonal lengths. Using the derived formula ?(a²b²)/(4(a²+b²)), we can efficiently calculate this area for any rhombus.