Triangle

Calculate the sides of the triangle if its area S = 630 and the second cathethus is shorter by 17.

Correct result:

a =  45
b =  28
c =  53

Solution:

S=ab2=a(a17)2=630  a217a1260=0  p=1;q=17;r=1260 D=q24pr=17241(1260)=5329 D>0  a1,2=q±D2p=17±53292 a1,2=17±732 a1,2=8.5±36.5 a1=45 a2=28   Factored form of the equation:  (a45)(a+28)=0  a=a1=45S = \dfrac{ab}{2} = \dfrac{a(a-17)}{2} = 630 \ \\ \ \\ a^2 -17a -1260 =0 \ \\ \ \\ p=1; q=-17; r=-1260 \ \\ D = q^2 - 4pr = 17^2 - 4\cdot 1 \cdot (-1260) = 5329 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 17 \pm \sqrt{ 5329 } }{ 2 } \ \\ a_{1,2} = \dfrac{ 17 \pm 73 }{ 2 } \ \\ a_{1,2} = 8.5 \pm 36.5 \ \\ a_{1} = 45 \ \\ a_{2} = -28 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (a -45) (a +28) = 0 \ \\ \ \\ a = a_1 = 45
b=a17=28
c=a2+b2=53

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