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?

Result

x =  100 m

Solution:

a=20+b x=a2+b2 x=2⋅ b−20  (2⋅ b−20)2=(20+b)2+b2 2b2−120b=0  p=2;q=−120;r=0 D=q2−4pr=1202−4⋅2⋅0=14400 D>0  b1,2=−q±D2p=120±144004 b1,2=120±1204 b1,2=30±30 b1=60 b2=0   Factored form of the equation:  2(b−60)b=0b>0 b=b1=60=60 m a=20+b=20+60=80 m x=a2+b2=802+602=100=100  m a = 20 + b \ \\ x = \sqrt{ a^2+b^2 } \ \\ x = 2 \cdot \ b- 20 \ \\ \ \\ (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} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 120 \pm \sqrt{ 14400 } }{ 4 } \ \\ b_{1,2} = \dfrac{ 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 = 0b>0 \ \\ b = b_{ 1 } = 60 = 60 \ m \ \\ a = 20 + b = 20 + 60 = 80 \ m \ \\ x = \sqrt{ a^2+b^2 } = \sqrt{ 80^2+60^2 } = 100 = 100 \ \text{ m }

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