Answer:
At the midpoint of the hypotenuse
Step-by-step explanation:
The circumcentre is the point of intersection of the 3 perpendicular bisectors.
It is positioned.
Inside all Acute triangles.
Outside all Obtuse triangles.
At the midpoint of the hypotenuse on all Right triangles.
Let x be a <span>positive number
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![x^2+(x+2)^2=130 \ \ \ \ \ \ [x\ \textgreater \ 0] \\ x^2+x^2+4x+4-130=0 \\ 2x^2+4x-126=0 \\ x^2+2x-63=0 \\ D=b^2-4ac=2^2-4*1*(-63)=4+252=256 \\ x_{1,2}= \frac{-bб \sqrt{D} }{2a}= \frac{-2б \sqrt{256} }{2*1}= \frac{-2б16 }{2}\\x_1=-9 \ \ \ \ \O\\x_2=7](https://tex.z-dn.net/?f=x%5E2%2B%28x%2B2%29%5E2%3D130%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5Bx%5C%20%5Ctextgreater%20%5C%200%5D%20%5C%5C%20x%5E2%2Bx%5E2%2B4x%2B4-130%3D0%20%5C%5C%202x%5E2%2B4x-126%3D0%20%5C%5C%20x%5E2%2B2x-63%3D0%20%5C%5C%20D%3Db%5E2-4ac%3D2%5E2-4%2A1%2A%28-63%29%3D4%2B252%3D256%20%5C%5C%20x_%7B1%2C2%7D%3D%20%5Cfrac%7B-b%D0%B1%20%5Csqrt%7BD%7D%20%7D%7B2a%7D%3D%20%5Cfrac%7B-2%D0%B1%20%20%5Csqrt%7B256%7D%20%20%7D%7B2%2A1%7D%3D%20%20%5Cfrac%7B-2%D0%B116%20%7D%7B2%7D%5C%5Cx_1%3D-9%20%5C%20%5C%20%5C%20%5C%20%5CO%5C%5Cx_2%3D7)
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Answer: the </span><span>number</span> is 7.
Check:
7² + (7+2)² = 49 + 9² = 49 + 81 = 130
It would be 1444
hope this helps
is there an included angle or lengths for now it's unsolvable
The Perimeter of Triangle ABC would be 60.
You can solve this by setting up proportions.