A triangle is given where the equation of one side is \(3x + 5y - 45 = 0\). We need to determine the radius of the circumcircle of this triangle, which is a right-angled triangle.
In a right-angled triangle, the hypotenuse serves as the diameter of the circumcircle. Hence, the radius is:
The equation \(3x + 5y = 45\) gives the intercepts:
So, the triangle has vertices at: \((0, 0), (0, 9), (15, 0)\)
The hypotenuse is between the points \((0, 9)\) and \((15, 0)\):
Rounding to the nearest integer: \(\boxed{9}\)
The required radius (to the nearest integer) is \(\boxed{9}\).
The center of a circle $ C $ is at the center of the ellipse $ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a>b $. Let $ C $ pass through the foci $ F_1 $ and $ F_2 $ of $ E $ such that the circle $ C $ and the ellipse $ E $ intersect at four points. Let $ P $ be one of these four points. If the area of the triangle $ PF_1F_2 $ is 30 and the length of the major axis of $ E $ is 17, then the distance between the foci of $ E $ is:
Let \( ABCD \) be a tetrahedron such that the edges \( AB \), \( AC \), and \( AD \) are mutually perpendicular. Let the areas of the triangles \( ABC \), \( ACD \), and \( ADB \) be 5, 6, and 7 square units respectively. Then the area (in square units) of the \( \triangle BCD \) is equal to: