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VLD [36.1K]
3 years ago
15

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``
Mathematics
1 answer:
Darina [25.2K]3 years ago
8 0

<u>Choice A: </u>2,2, \sqrt{4}<u />

<u>Choice B: </u>9, 40, 41

<u>Choice C: </u>\sqrt{5}, 10$ and  \sqrt{125}

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.

Hypotenuse^2=Opposite^2+Adjacent^2

Note that the longest side length is always the hypotenuse.

<u>Choice A: </u>2,2, \sqrt{4}<u />

Now, \sqrt{4}=2

Therefore:

2^2\neq 2^2+2^2\\4 \neq 8

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

<u>Choice B: </u>9, 40, 41

The longest side length is 41.

41^2=1681

40^2+9^2=1681

Therefore:

41^2=40^2+9^2

These side lengths form a right triangle.

<u>Choice C: </u>\sqrt{5}, 10$ and  \sqrt{125}

\sqrt{5} \approx 2.24 \\ \sqrt{125}\approx11.18

Therefore, the longest side length is \sqrt{125}

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

These side lengths do not form a right triangle.

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