# 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:

$S=24 \ \\ o=20 \ \\ S=ab \ \\ o=2(a+b) \ \\ b=10-a \ \\ \ \\ 24=a \cdot \ (10-a) \ \\ a^2 -10a +24=0 \ \\ \ \\ p=1; q=-10; r=24 \ \\ D=q^2 - 4pr=10^2 - 4\cdot 1 \cdot 24=4 \ \\ D>0 \ \\ \ \\ a_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 10 \pm \sqrt{ 4 } }{ 2 } \ \\ a_{1,2}=\dfrac{ 10 \pm 2 }{ 2 } \ \\ a_{1,2}=5 \pm 1 \ \\ a_{1}=6 \ \\ a_{2}=4 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (a -6) (a -4)=0 \ \\ a=a_{1}=6 \ \\ b=10-a=10-6=4 \ \\ u=\sqrt{ a^2+b^2 }=\sqrt{ 6^2+4^2 } \doteq 2 \ \sqrt{ 13 } \doteq 7.2111 \doteq 7.211 \ \text{cm}$

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