# Rectangle field

The field has a shape of a rectangle having a length of 119 m and a width of 19 m. , How many meters have to shorten its length and increase its width to maintain its area and circumference increased by 24 m?

Result

x =  102 m
y =  114 m

#### Solution:

$119 \cdot \ 19 = (119-x) \cdot \ (19+y) \ \\ 2(119+19) + 24 = 2 \cdot \ ((119-x)+(19+y)) \ \\ \ \\ y = x+12 \ \\ \ \\ \ \\ 119 \cdot \ 19 = (119-x) \cdot \ (19+(x+12)) \ \\ x^2 -88x -1428 = 0 \ \\ \ \\ a = 1; b = -88; c = -1428 \ \\ D = b^2 - 4ac = 88^2 - 4\cdot 1 \cdot (-1428) = 13456 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 88 \pm \sqrt{ 13456 } }{ 2 } \ \\ x_{1,2} = \dfrac{ 88 \pm 116 }{ 2 } \ \\ x_{1,2} = 44 \pm 58 \ \\ x_{1} = 102 \ \\ x_{2} = -14 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -102) (x +14) = 0 \ \\ x = x_{ 1 } = 102 = 102 \ \text { m }$

Checkout calculation with our calculator of quadratic equations.

$y = x+12 = 102+12 = 114 \ \\ a = 119-x = 119-102 = 17 \ m \ \\ b = 19+y = 19+114 = 133 \ m \ \\ \ \\ y = 114 = 114 \ \text { m }$

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