Diagonal 20

The rectangular town plaza's diagonal pathway is 20 m longer than the width. Suppose the pathway is 20 m shorter than twice the width. How long should the pathway be?

Final Answer:

x =  100 m

Step-by-step explanation:

$$\begin{gathered} a=20+b \\ x=\sqrt{a^2+b^2} \\ x=2\cdot b-20 \\ (2\cdot b-20{)}^2=(20+b{)}^2+b^2 \\ (2\cdot b-20{)}^2=(20+b{)}^2+b^2 \\ 2b^2-120b=0 \\ 2...\text{ prime number} \\ 120=2^3\cdot 3\cdot 5 \\ 0 \\ \text{GCD}(2,120,0)=2 \\ {b}^2-60b=0 \\ p=1;q=-60;r=0 \\ D=q^2-4pr=60^2-4\cdot 1\cdot 0=3600 \\ D>0 \\ {b}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{60\pm \sqrt{3600}}{2} \\ {b}_{1,2}=\frac{60\pm 60}{2} \\ {b}_{1,2}=30\pm 30 \\ {b}_1=60 \\ {b}_2=0 \\ b>0 \\ b=b_1=60\text{m} \\ a=20+b=20+60=80\text{m} \\ x=\sqrt{a^2+b^2}=\sqrt{80^2+60^2}=100\text{m} \end{gathered}$$

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