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.

Correct result:

o =  -2 cm
S =  6 cm²

Solution:

$$\begin{gathered} c=9+a \\ c=8+b \\ {c}^2=a^2+b^2 \\ {c}^2=(c-9{)}^2+(c-8{)}^2 \\ -c^2+34c-145=0 \\ {c}^2-34c+145=0 \\ p=1;q=-34;r=145 \\ D=q^2-4pr=34^2-4\cdot 1\cdot 145=576 \\ D>0 \\ {c}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{34\pm \sqrt{576}}{2}} \\ {c}_{1,2}={\displaystyle \frac{34\pm 24}{2}} \\ {c}_{1,2}=17\pm 12 \\ {c}_1=29 \\ {c}_2=5 \\ \text{ Factored form of the equation: } \\ (c-29)(c-5)=0 \\ c>9 \\ c=c_2=5 \\ a=c-9=5-9=-4 \\ b=c-8=5-8=-3 \\ o=a+b+c=(-4)+(-3)+5=-2\text{ cm} \end{gathered}$$

Our quadratic equation calculator calculates it.

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

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