Answer:
8 cm
Explanation:
Two circles of radii 5 cm and 4 cm intersect at two points D and E.
The distance between their centers (points A and C) is 3 cm.
Consider triangle ACD. In this triangle,
AC = 3 cm (distance between centers)
AD = 5 cm (radius of larger circle)
CD = 4 cm (radius of smaller circle)
So, this is right triangle and therefore, AC is the height of isosceles triangle ADE (triangle ADE is isosceles, because AE = AD).
The height of the isosceles triangle drawn to the base is its median, so CD = CE = 4 cm, hence
DE = 4 + 4 = 8 cm