Rectangles are drawn on line segments of fixed lengths. When the breadths are 6m and 5m respectively the sum of the areas of the rectangles is 83 m². But if the breadths are 5m and 4m respectively the sum of the areas is 68 m². Find the sum of the areas of squares drawn on the line segments.

Let length of first fixed line segment be x and second line segment be y.

In first case, when the breadths are 6m and 5m the sum of the areas = 83 m².

6x + 5y = 83 ……(i)

In second case, when the breadths are 5m and 4m the sum of the areas = 68 m².

5x + 4y = 68 ……(ii)

Multiplying (i) by 4 and (ii) by 5 we get,

24x + 20y = 332 ……(iii)

25x + 20y = 340 ……(iv)

Subtracting (iii) from (iv) we get,

25x + 20y - (24x + 20y) = 340 - 332

⇒ x = 8.

On substituting value of x in (i) we get,

⇒ 6(8) + 5y = 83

⇒ 48 + 5y = 83

⇒ 5y = 83 - 48

⇒ 5y = 35

⇒ y = 7.

Sum of areas of squares on these two line segments = x² + y² = 8² + 7² = 64 + 49 = 113 m².

Hence, sum of areas of squares on these two line segments = 113 m².