Hypotenuse and height

In a right triangle is length of the hypotenuse c = 56 cm and height hc = 4 cm. Determine the length of both trangle legs.

Correct result:

a =  55.9 cm
b =  4 cm

Solution:

56=x+y 42=xy  x(56x)=42  x256x+16=0  a=1;b=56;c=16 D=b24ac=5624116=3072 D>0  x1,2=b±D2a=56±30722=56±3232 x1,2=28±27.712812921102 x1=55.712812921102 x2=0.28718707889796   Factored form of the equation:  (x55.712812921102)(x0.28718707889796)=0   a=x156=55.9 cm56=x+y \ \\ 4^2=xy \ \\ \ \\ x(56-x) = 4^2 \ \\ \ \\ x^2 -56x +16 =0 \ \\ \ \\ a=1; b=-56; c=16 \ \\ D = b^2 - 4ac = 56^2 - 4\cdot 1 \cdot 16 = 3072 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 56 \pm \sqrt{ 3072 } }{ 2 } = \dfrac{ 56 \pm 32 \sqrt{ 3 } }{ 2 } \ \\ x_{1,2} = 28 \pm 27.712812921102 \ \\ x_{1} = 55.712812921102 \ \\ x_{2} = 0.28718707889796 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -55.712812921102) (x -0.28718707889796) = 0 \ \\ \ \\ \ \\ a = \sqrt{ x_1 \cdot 56 } = 55.9 \ \text{cm}
b=x256=4 cmb = \sqrt{ x_2 \cdot 56 } = 4 \ \text{cm}



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