# 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.

Result

o =  -2 cm
S =  6 cm2

#### Solution:

$c=9+a \ \\ c=8+b \ \\ c^2=a^2+b^2 \ \\ \ \\ \ \\ -c^2 +34c -145=0 \ \\ c^2 -34c +145=0 \ \\ \ \\ p=1; q=-34; r=145 \ \\ D=q^2 - 4pr=34^2 - 4\cdot 1 \cdot 145=576 \ \\ D>0 \ \\ \ \\ c_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 34 \pm \sqrt{ 576 } }{ 2 } \ \\ c_{1,2}=\dfrac{ 34 \pm 24 }{ 2 } \ \\ c_{1,2}=17 \pm 12 \ \\ c_{1}=29 \ \\ c_{2}=5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -29) (c -5)=0 \ \\ \ \\ c>9 \ \\ c=c_{2}=5 \ \\ a=c-9=5-9=-4 \ \\ b=c-8=5-8=-3 \ \\ o=a+b+c=(-4)+(-3)+5=-2 \ \text{cm}$

Checkout calculation with our calculator of quadratic equations.

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

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