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OnePieceFTW

  • one year ago

Zoe is using the figure shown below to prove Pythagorean Theorem using triangle similarity: In the given triangle ABC, angle A is 90 degrees and segment AD is perpendicular to segment BC.

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  1. OnePieceFTW
    • one year ago
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  2. OnePieceFTW
    • one year ago
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    Which of these could be a step to prove that BC^2 = AB^2 + AC^2? By distribution, AC^2 plus AB^2 = BC multiplied by the quantity DC plus AB. By distribution, AC^2 plus AB^2 = BC multiplied by the quantity DC plus BD. By distribution, AC^2 plus AD^2 = BC multiplied by the quantity DC plus AB. By distribution, AC^2 plus AD^2 = BC multiplied by the quantity DC plus BD.

  3. anonymous
    • one year ago
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    @jim_thompson5910

  4. anonymous
    • one year ago
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    Could you help me with this? Its the same question I have

  5. jolo_yolo
    • one year ago
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    @ParthKohli HI, can you please help me with this?

  6. jolo_yolo
    • one year ago
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    @phi help me with this, please?

  7. phi
    • one year ago
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    the question is pretty nebulous. But it looks like they are using the idea that segment BC is the same as BD+DC so BC^2 = BC * BC = BC* (BD+DC)

  8. jolo_yolo
    • one year ago
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    so that narrows it down to answers D and B, right?

  9. phi
    • one year ago
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    also, it seems they are using AB^2 + AC^2 = BC^2

  10. phi
    • one year ago
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    so we have AB^2 + AC^2 = BC* (BD+DC) I'm not sure how this is a proof, but it's true.

  11. phi
    • one year ago
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    and the other choices don't look true.

  12. jolo_yolo
    • one year ago
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    That makes much more sense, you're a life saver thank you so much!

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