Diagonal 20

Diagonal pathway for the rectangular town plaza whose length is 20 m longer than the width. if the pathway is 20 m shorter than twice the width. How long should the pathway be?

Correct result:

x =  100 m

Solution:

$$\begin{gathered} a=20+b \\ x=\sqrt{a^2+b^2} \\ x=2\cdot b-20 \\ (2\ast b-20{)}^2=(20+b{)}^2+b^2 \\ (2\cdot b-20{)}^2=(20+b{)}^2+b^2 \\ 2b^2-120b=0 \\ p=2;q=-120;r=0 \\ D=q^2-4pr=120^2-4\cdot 2\cdot 0=14400 \\ D>0 \\ {b}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{120\pm \sqrt{14400}}{4}} \\ {b}_{1,2}={\displaystyle \frac{120\pm 120}{4}} \\ {b}_{1,2}=30\pm 30 \\ {b}_1=60 \\ {b}_2=0 \\ \text{ Factored form of the equation: } \\ 2(b-60)b=0 \\ b>0 \\ b=b_1=60=60\text{ m } \\ a=20+b=20+60=80\text{ m } \\ x=\sqrt{a^2+b^2}=\sqrt{80^2+60^2}=100\text{ m} \end{gathered}$$

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