Diagonal of the rectangle

Calculate the diagonal of the rectangle which area is 54 centimeters square and the circuit is equal to 30 cm.

Result

u =  10.817

Solution:

S=54 o=30 S=a b o=2 (a+b) S=a (o/2a)  54=a (30/2a) a215a+54=0  p=1;q=15;r=54 D=q24pr=1524154=9 D>0  a1,2=q±D2p=15±92 a1,2=15±32 a1,2=7.5±1.5 a1=9 a2=6   Factored form of the equation:  (a9)(a6)=0a=a1=9 b=a2=6 u=a2+b2=92+62=3 1310.8167=10.817S = 54 \ \\ o = 30 \ \\ S = a \cdot \ b \ \\ o = 2 \cdot \ (a+b) \ \\ S = a \cdot \ (o/2-a) \ \\ \ \\ 54 = a \cdot \ (30/2-a) \ \\ a^2 -15a +54 = 0 \ \\ \ \\ p = 1; q = -15; r = 54 \ \\ D = q^2 - 4pr = 15^2 - 4\cdot 1 \cdot 54 = 9 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 15 \pm \sqrt{ 9 } }{ 2 } \ \\ a_{1,2} = \dfrac{ 15 \pm 3 }{ 2 } \ \\ a_{1,2} = 7.5 \pm 1.5 \ \\ a_{1} = 9 \ \\ a_{2} = 6 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (a -9) (a -6) = 0a = a_{ 1 } = 9 \ \\ b = a_{ 2 } = 6 \ \\ u = \sqrt{ a^2+b^2 } = \sqrt{ 9^2+6^2 } = 3 \ \sqrt{ 13 } \doteq 10.8167 = 10.817

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