# 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:

$$\smash{ c = 9+a \\~\\c = 8+b \\~\\c^2 = a^2+b^2 \\~\\ \\~\\ c^2 = (c-9)^2+(c-8)^2 \\~\\-c^2 +34c -145 =0 \\~\\c^2 -34c +145 =0 \\~\\D = 34^2 - 4\cdot 1 \cdot 145 = 576 \\~\\D>0 \\~\\ \\~\\c_{1,2} = \frac{ 34 \pm \sqrt{ 34^2 - 4\cdot 1 \cdot 145} }{ 2 \cdot 1 } = \frac{ 34 \pm \sqrt{ 576 } }{ 2 } \\~\\c_{1,2} = \frac{ 34 \pm 24 }{ 2 } \\~\\c_{1,2} = 17 \pm 12 \\~\\c_{1} = 29 \\~\\c_{2} = 5 \\~\\ \\~\\ (c -29) (c -5) = 0 \\~\\ \\~\\c>9 \\~\\c = c_{2} = 5 \\~\\a = c-9 = 5-9 = -4 \\~\\b = c-8 = 5-8 = -3 \\~\\o = a+b+c = (-4)+(-3)+5 = -2 \ { cm } \\~\\S = a \cdot b/2 = (-4) \cdot (-3)/2 = 6 \ cm^2 }$$

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