The height of an isosceles triangle is 14 decimetres; the base refers to the lateral side as 48:25. Find the sides of this triangle.

Let the length of the lateral side of the triangle ABC be 25 * X cm, then the base of the AC = 48 * X dm.
The height of the ВН, drawn to the base, is also the median of the triangle, then AH = CH = AC / 2 = 48 * X / 2 = 24 * X dm.
In a right-angled triangle ABН, by the Pythagorean theorem, AB ^ 2 = AH ^ 2 + BH ^ 2.
625 * X ^ 2 = 576 * X2 + 196.
49 * X ^ 2 = 196.
X ^ 2 = 196/49 = 4.
X = 2.
Then AB = BC = 25 * 2 = 50 dm.
AC = 48 * 2 = 96 dm.
Answer: The sides of the triangle are 50 dm, 50 dm, 96 dm.