Right triangle eq2

Hypotenuse of a right triangle is 9 cm longer than one leg and 8 cm longer than the second leg. Determine the circumference and area of a triangle.


o =  -2 cm
S =  6 cm2


c=9+a c=8+b c2=a2+b2   c2+34c145=0 c234c+145=0  p=1;q=34;r=145 D=q24pr=34241145=576 D>0  c1,2=q±D2p=34±5762 c1,2=34±242 c1,2=17±12 c1=29 c2=5   Factored form of the equation:  (c29)(c5)=0  c>9 c=c2=5 a=c9=59=4 b=c8=58=3 o=a+b+c=(4)+(3)+5=2 cmc=9+a \ \\ c=8+b \ \\ c^2=a^2+b^2 \ \\ \ \\ \ \\ -c^2 +34c -145=0 \ \\ c^2 -34c +145=0 \ \\ \ \\ p=1; q=-34; r=145 \ \\ D=q^2 - 4pr=34^2 - 4\cdot 1 \cdot 145=576 \ \\ D>0 \ \\ \ \\ c_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 34 \pm \sqrt{ 576 } }{ 2 } \ \\ c_{1,2}=\dfrac{ 34 \pm 24 }{ 2 } \ \\ c_{1,2}=17 \pm 12 \ \\ c_{1}=29 \ \\ c_{2}=5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -29) (c -5)=0 \ \\ \ \\ c>9 \ \\ c=c_{2}=5 \ \\ a=c-9=5-9=-4 \ \\ b=c-8=5-8=-3 \ \\ o=a+b+c=(-4)+(-3)+5=-2 \ \text{cm}

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S=a b/2=(4) (3)/2=6 cm2S=a \cdot \ b/2=(-4) \cdot \ (-3)/2=6 \ \text{cm}^2

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