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.

Result

a =  4 m
b =  3 m
c =  5 m

Solution:

a=c1 b=c2  a2+b2=c2   c26c+5=0  p=1;q=6;r=5 D=q24pr=62415=16 D>0  c1,2=q±D2p=6±162 c1,2=6±42 c1,2=3±2 c1=5 c2=1   Factored form of the equation:  (c5)(c1)=0 c>2 c=c1=5  a=c1=51=4=4  m a = c-1 \ \\ b = c-2 \ \\ \ \\ a^2+b^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 = 4 \ \text { m }

Checkout calculation with our calculator of quadratic equations.

b=c2=52=3=3  m b = c-2 = 5-2 = 3 = 3 \ \text { m }
c=5=5  m c = 5 = 5 \ \text { m }



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Looking for help with calculating roots of a quadratic equation? Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.

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