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.

Correct result:

u =  7.2111 cm

Solution:

$$\begin{gathered} S=24 \\ o=20 \\ S=ab \\ o=2(a+b) \\ b=10-a \\ 24=a\ast (10-a) \\ 24=a\cdot (10-a) \\ {a}^2-10a+24=0 \\ p=1;q=-10;r=24 \\ D=q^2-4pr=10^2-4\cdot 1\cdot 24=4 \\ D>0 \\ {a}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{10\pm \sqrt{4}}{2}} \\ {a}_{1,2}={\displaystyle \frac{10\pm 2}{2}} \\ {a}_{1,2}=5\pm 1 \\ {a}_1=6 \\ {a}_2=4 \\ \text{ Factored form of the equation: } \\ (a-6)(a-4)=0 \\ a=a_1=6 \\ b=10-a=10-6=4 \\ u=\sqrt{a^2+b^2}=\sqrt{6^2+4^2}=2\sqrt{13}=7.2111\text{ cm} \end{gathered}$$

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