The diagonals of a rectangle divide it into 4 triangles, the perimeters of which are 9/14 and 4/7 of the rectangle’s

The diagonals of a rectangle divide it into 4 triangles, the perimeters of which are 9/14 and 4/7 of the rectangle’s perimeter. Find the ratio of the smallest side of the rectangle to the largest.

The diagonals of a rectangle have the same length, they intersect and are halved at the intersection.
Let a and b be the sides and d the diagonals of the rectangle.
Then, the perimeter of the rectangle is 2 * (a + b); 2 triangles have perimeter along a + d, and the other 2 – along b + d.
We have, (a + d) / (2 * (a + b)) = 9/14 and (b + d) / (2 * (a + b)) = 4/7.
Or 7 * a + 7 * d = 9 * a + 9 * b and 7 * b + 7 * d = 8 * a + 8 * b.
Let’s define: 7 * d = 2 * a + 9 * b = 8 * a + b, whence b / a = ¾ = 0.75.
Answer: 0.75.