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The altitude to the hypotenuse of a right triangle bisects the hypotenuse. How does the length of the altitude compare with the length of the segments of the hypotenuse. Explain completely. Use Geogebra to help you set up a diagram and show evidence that supports your conclusion.

*Search up Geogebra and take a pic of your triangle*

And explain and show evidence that supports your conclusion.

Any questions please feel free, I will answer your questions.

Sagot :

[tex] \frac{1}{2}[/tex]will be the answer.

Step-by-step explanation:-

If the altitude to the hypotenuse of a right-angled triangle bisects the hypotenuse, then we can know that the triangle is Isosceles right triangle. (AB=AC)

Make AB=AC=2, so,

BC=[tex] \sqrt{ {2}^{2} + {2}^{2} } [/tex]= [tex] \sqrt{4 + 4} [/tex]= [tex] \sqrt{8} [/tex]

Prime factorizing, we get,

BC = [tex] \sqrt{2 \times 2 \times 2} [/tex]

Take two 2's outside by taking common because of "square" root,

⇢BC = [tex]2 \sqrt{2} [/tex]

BD = DC = [tex] \sqrt{2} [/tex],

AD = [tex] \sqrt{AB^{2} - BD^{2} } [/tex]= [tex] \sqrt{4 - 2} [/tex]= [tex] \sqrt{2} [/tex]

So, AD=BD=CD= [tex] \frac{1}{2}[/tex]BC

∴⇢[tex] \frac{AD}{BC} = \frac{1}{2} [/tex]

Read more about hypotenuse altitude question;

https://brainly.com/question/3160811

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