Rectangle field

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

Final Answer:

x =  102 m
y =  114 m

Step-by-step explanation:

$$\begin{gathered} 119\cdot 19=(119-x)\cdot (19+y) \\ 2(119+19)+24=2\cdot ((119-x)+(19+y)) \\ y=x+12 \\ 119\cdot 19=(119-x)\cdot (19+(x+12)) \\ 119\cdot 19=(119-x)\cdot (19+(x+12)) \\ {x}^2-88x-1428=0 \\ a=1;b=-88;c=-1428 \\ D=b^2-4ac=88^2-4\cdot 1\cdot (-1428)=13456 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{88\pm \sqrt{13456}}{2} \\ {x}_{1,2}=\frac{88\pm 116}{2} \\ {x}_{1,2}=44\pm 58 \\ {x}_1=102 \\ {x}_2=-14 \\ x=x_1=102=102\text{m} \end{gathered}$$

Our quadratic equation calculator calculates it.

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

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