The sum of the areas of a rectangle and a square is 361 cm². The breadth of the rectangle is 5 cm more than each side of the square. If the length of the rectangle is 20 cm, then find the difference between the perimeters of the rectangle and the square.

ATQ, Let the length of each side of the square be ‘x’ cm Therefore, breadth of the rectangle = (x + 5) cm Area of the rectangle = 20(x + 5) cm² Area of the square = x² cm² According to the question, 20(x + 5) + x² = 361 Or, x² + 20x + 100 − 361 = 0 Or, x² + 20x − 261 = 0 Or, x² − 9x + 29x − 261 = 0 Or, x(x − 9) + 29(x − 9) = 0 Or, (x − 9)(x + 29) = 0 Or, x = 9, −29 Since length cannot be negative, x = 9 Breadth of rectangle = x + 5 = 14 cm Perimeter of rectangle = 2(20 + 14) = 2 × 34 = 68 cm Each side of the square = x = 9 cm Perimeter of the square = 4 × 9 = 36 cm Required difference = 68 − 36 = 32 cm