The hypotenuse of a right triangle is 41 and the sum of legs is 49. Calculate the length of its legs.


a =  40
b =  9


c=41 a+b=49 a2+b2=c2=1681 (a+b)2=492 a2+2ab+b2=2401  2ab=24011681 2a(49a)=720 249a2a2=720  2a298a+720=0  p=2;q=98;r=720 D=q24pr=98242720=3844 D>0  a1,2=q±D2p=98±38444 a1,2=98±624 a1,2=24.5±15.5 a1=40 a2=9   Factored form of the equation:  2(a40)(a9)=0 c= 41 \ \\ a+b = 49 \ \\ a^2 + b^2 = c^2 = 1681 \ \\ (a+b)^2 = 49^2 \ \\ a^2+2ab+b^2 = 2401 \ \\ \ \\ 2ab = 2401-1681 \ \\ 2a(49-a) = 720 \ \\ 2\cdot 49 a - 2a^2 = 720 \ \\ \ \\ 2a^2 -98a +720 =0 \ \\ \ \\ p=2; q=-98; r=720 \ \\ D = q^2 - 4pr = 98^2 - 4\cdot 2 \cdot 720 = 3844 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 98 \pm \sqrt{ 3844 } }{ 4 } \ \\ a_{1,2} = \dfrac{ 98 \pm 62 }{ 4 } \ \\ a_{1,2} = 24.5 \pm 15.5 \ \\ a_{1} = 40 \ \\ a_{2} = 9 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (a -40) (a -9) = 0 \ \\
b=a2=9b= a_2 = 9

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