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 c2=a2+b2   −c2+34c−145=0 c2−34c+145=0  p=1;q=−34;r=145 D=q2−4pr=342−4⋅1⋅145=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:  (c−29)(c−5)=0  c>9 c=c2=5 a=c−9=5−9=−4 b=c−8=5−8=−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}

Checkout calculation with our calculator of quadratic equations.

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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