Rectangle diagonals

It is given rectangle with area 24 cm2 a circumference 20 cm. The length of one side is 2 cm larger than length of second side. Calculate the length of the diagonal. Length and width are yet expressed in natural numbers.

Result

u =  7.211 cm

Solution:

S=24 o=20 S=ab o=2(a+b) b=10−a  24=a⋅ (10−a) a2−10a+24=0  p=1;q=−10;r=24 D=q2−4pr=102−4⋅1⋅24=4 D>0  a1,2=−q±D2p=10±42 a1,2=10±22 a1,2=5±1 a1=6 a2=4   Factored form of the equation:  (a−6)(a−4)=0 a=a1=6 b=10−a=10−6=4 u=a2+b2=62+42≐2 13≐7.2111≐7.211 cmS=24 \ \\ o=20 \ \\ S=ab \ \\ o=2(a+b) \ \\ b=10-a \ \\ \ \\ 24=a \cdot \ (10-a) \ \\ a^2 -10a +24=0 \ \\ \ \\ p=1; q=-10; r=24 \ \\ D=q^2 - 4pr=10^2 - 4\cdot 1 \cdot 24=4 \ \\ D>0 \ \\ \ \\ a_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 10 \pm \sqrt{ 4 } }{ 2 } \ \\ a_{1,2}=\dfrac{ 10 \pm 2 }{ 2 } \ \\ a_{1,2}=5 \pm 1 \ \\ a_{1}=6 \ \\ a_{2}=4 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (a -6) (a -4)=0 \ \\ a=a_{1}=6 \ \\ b=10-a=10-6=4 \ \\ u=\sqrt{ a^2+b^2 }=\sqrt{ 6^2+4^2 } \doteq 2 \ \sqrt{ 13 } \doteq 7.2111 \doteq 7.211 \ \text{cm}

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