Sides of right angled triangle

One leg is 1 m shorter than the hypotenuse, and the second leg is 2 m shorter than the hypotenuse. Find the lengths of all sides of the right-angled triangle.

Correct result:

a =  4 m
b =  3 m
c =  5 m

Solution:

$$\begin{gathered} a=c-1 \\ b=c-2 \\ {a}^2+b^2=c^2 \\ (c-1{)}^2+(c-2{)}^2=c^2 \\ {c}^2-6c+5=0 \\ p=1;q=-6;r=5 \\ D=q^2-4pr=6^2-4\cdot 1\cdot 5=16 \\ D>0 \\ {c}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{6\pm \sqrt{16}}{2}} \\ {c}_{1,2}={\displaystyle \frac{6\pm 4}{2}} \\ {c}_{1,2}=3\pm 2 \\ {c}_1=5 \\ {c}_2=1 \\ \text{ Factored form of the equation: } \\ (c-5)(c-1)=0 \\ c>2 \\ c=c_1=5 \\ a=c-1=5-1=4\text{ m} \end{gathered}$$

Our quadratic equation calculator calculates it.

$b=c-2=5-2=3\text{ m}$

$c=5=5\text{ m}$

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