# 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:

$a=c-1 \ \\ b=c-2 \ \\ \ \\ a^2+b^2=c^2 \ \\ \ \\ (c-1)^2 + (c-2)^2=c^2 \ \\ \ \\ c^2 -6c +5=0 \ \\ \ \\ p=1; q=-6; r=5 \ \\ D=q^2 - 4pr=6^2 - 4\cdot 1 \cdot 5=16 \ \\ D>0 \ \\ \ \\ c_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 6 \pm \sqrt{ 16 } }{ 2 } \ \\ c_{1,2}=\dfrac{ 6 \pm 4 }{ 2 } \ \\ c_{1,2}=3 \pm 2 \ \\ c_{1}=5 \ \\ c_{2}=1 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -5) (c -1)=0 \ \\ \ \\ c>2 \ \\ c=c_{1}=5 \ \\ \ \\ a=c-1=5-1=4 \ \text{m}$

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