When three circles are drawn with the sides of a right triangle as diameters, the region contained within all three circles forms a shape known as a Reuleaux triangle. In this case, since the triangle is a right triangle with side lengths 3, 4, and 5, the circles' diameters are 3, 4, and 5 units.
The Reuleaux triangle is a curved shape formed by taking three circles of equal radius and arranging them such that each circle's center lies on the circumference of the other two circles. The resulting shape has rounded corners that touch the midpoint of each side of the original triangle.
To calculate the area of the Reuleaux triangle, we can subtract the areas of three sectors from the area of the equilateral triangle formed by connecting the midpoints of the original right triangle's sides.
1. Calculate the area of the equilateral triangle:
The original right triangle's sides have lengths 3 and 4, making the hypotenuse 5. Half of the hypotenuse is 2.5, and this is also the radius of the circles.
The equilateral triangle's side length is equal to the diameter of the larger circle (5 units). The height of the equilateral triangle can be found using the Pythagorean theorem: \(h = \sqrt{5^2 - 2.5^2} = \sqrt{25 - 6.25} = \sqrt{18.75} \approx 4.33\).
The area of an equilateral triangle is \(A_{\text{equilateral}} = \frac{\sqrt{3}}{4} \times \text{side length}^2\). Plugging in the side length:
\[A_{\text{equilateral}} = \frac{\sqrt{3}}{4} \times 5^2 \approx 10.82.\]
2. Calculate the area of the sectors:
The central angle of each sector is \(120^\circ\) because the circles are tangent at the midpoint of each side of the original triangle.
The area of a sector of a circle is \(A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2\), where \(\theta\) is the central angle and \(r\) is the radius.
For the circle with a diameter of 3 (radius 1.5):
\[A_{\text{sector}} = \frac{120^\circ}{360^\circ} \times \pi \times 1.5^2 \approx 1.178.\]
Calculate the area for the other two circles in the same way, each with a central angle of \(120^\circ\).
3. Subtract the total area of the sectors from the area of the equilateral triangle:
\[A_{\text{Reuleaux}} = A_{\text{equilateral}} - 3 \times A_{\text{sector}} \approx 10.82 - 3 \times 1.178 \approx 6.286.\]
So, the area of the region contained within all three circles (the Reuleaux triangle) is approximately \(6.286\) square units.
