# Triangle

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

Result

a =  45
b =  28
c =  53

#### Solution:

$S = \dfrac{ab}{2} = \dfrac{a(a-17)}{2} = 630 \ \\ \ \\ a^2 -17a -1260 =0 \ \\ \ \\ p=1; q=-17; r=-1260 \ \\ D = q^2 - 4pr = 17^2 - 4\cdot 1 \cdot (-1260) = 5329 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 17 \pm \sqrt{ 5329 } }{ 2 } \ \\ a_{1,2} = \dfrac{ 17 \pm 73 }{ 2 } \ \\ a_{1,2} = 8.5 \pm 36.5 \ \\ a_{1} = 45 \ \\ a_{2} = -28 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (a -45) (a +28) = 0 \ \\ a = a_1 = 45$
$b= a-17 = 28$
$c=\sqrt{ a^2 + b^2 } = 53$

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