Hello! I'm stuck on Pythagorean theorem in geometry. There are some numbers that suppose to have a line on top but I don't know how to put it on top of the number, I just put this `. So, here's the problem: Which side lengths form a right triangle? A. 2,2, v```4``

Respuesta :

Choice A: [tex]2,2, \sqrt{4}[/tex]

Choice B: 9, 40, 41

Choice C: [tex]\sqrt{5}, 10$ and \sqrt{125}[/tex]

Answer:

(B)9, 40, 41

Step-by-step explanation:

To check if the sides form a right triangle, you check to see if they satisfy the Pythagorean theorem.

[tex]Hypotenuse^2=Opposite^2+Adjacent^2[/tex]

Note that the longest side length is always the hypotenuse.

Choice A: [tex]2,2, \sqrt{4}[/tex]

Now, [tex]\sqrt{4}=2[/tex]

Therefore:

[tex]2^2\neq 2^2+2^2\\4 \neq 8[/tex]

These side lengths form an equilateral triangle. They do not satisfy the theorem.

Choice B: 9, 40, 41

The longest side length is 41.

[tex]41^2=1681[/tex]

[tex]40^2+9^2=1681[/tex]

Therefore:

[tex]41^2=40^2+9^2[/tex]

These side lengths form a right triangle.

Choice C: [tex]\sqrt{5}, 10$ and \sqrt{125}[/tex]

[tex]\sqrt{5} \approx 2.24 \\ \sqrt{125}\approx11.18[/tex]

Therefore, the longest side length is [tex]\sqrt{125}[/tex]

[tex](\sqrt{125})^2=125\\(\sqrt{5})^2+10^2=5+100=105\\\\(\sqrt{125})^2 \neq (\sqrt{5})^2+10^2[/tex]

These side lengths do not form a right triangle.

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