The radius of the inscribed circle in triangle RST is 5√17/2. Point X, determined as the incenter, is the intersection of angle bisectors, serving as the center for the inscribed circle touching all sides.
A. To find the radius of the inscribed circle of triangle RST, we need to know the lengths of the sides of the triangle and the area of the triangle. Then we can use the formula:
r = A / s
where r is the radius of the inscribed circle, A is the area of the triangle, and s is the semiperimeter of the triangle, given by:
s = (a + b + c) / 2
where a, b, and c are the lengths of the sides of the triangle.
From the image, we can see that the equations of the lines XU and XW are given by:
XU = -4y + 20
XW = 6y + 10
We can solve these equations simultaneously to find the coordinates of point X, which is the center of the inscribed circle. We get:
x = 5, y = 5/2
Then we can use the distance formula to find the length of XU, which is the same as the radius of the inscribed circle. We get:
r = sqrt((5-0)^2 + (5/2-10)^2) = 5sqrt(17)/2
Therefore, the radius of the inscribed circle of triangle RST is 5sqrt(17)/2.
B. Point X represents the point of concurrency of the angle bisectors of triangle RST. This point is also called the incenter of the triangle. It is the center of the inscribed circle that touches all three sides of the triangle.