# 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:

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

Checkout calculation with our calculator of quadratic equations.

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

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