You have a right-angled triangle. So you can use Pythagoras' Theorem.

AB² + BC² = AC²

(x + 1)² + (x + 3)² = (2x)²

x² + 2x + 1 + x² + 6x + 9 = 4x²

2x² + 8x + 10 = 4x²

2x² - 8x - 10 = 0

x² - 4x - 5 = 0

(x + 1)(x - 5) = 0

x +1 = 0 or x - 5 = 0

x = -1 (rejected, length cannot be negative)
or
x = 5

Area of triangle

= ½ x AB X BC
= ½ (5 + 1)(5 + 3)
= ½ (6)(8)
= 24

Since it's given that area of rectangle
= area of triangle

And area of rectangle = PQ X QS
= 2y x y
= 2y²

Then 2y² = 24
y² = 12
y = √12 = √(4 x 3) = √4√3 = 2√3

Perimeter of rectangle
= (2y + 2y + y + y) cm
= 6y cm

= 6(2√3) cm
= 12√3 cm

Thank you so much

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