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arlira

  • 2 years ago

Help me PLEASE I'm confused!!! The hypotenuse of a right triangle is 1cm longer than the longer leg. The shorter leg is 7cm shorter than the longer leg. Find the length of the longer leg of the right triangle.Label each side of the triangle. Then substitute into Pythagorean Theorem.

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  1. hartnn
    • 2 years ago
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    at which step are u confused ?

  2. Aditi_Singh
    • 2 years ago
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    Shorter leg = x longer leg = x+7 hypotenuse = x+7+1 = x+8 Do you know how to proceed now ?

  3. arlira
    • 2 years ago
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    no I don't know-how to continue

  4. arlira
    • 2 years ago
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    hartnn i dont know how to figure out what belongs to what leg

  5. hartnn
    • 2 years ago
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    did u get @Aditi_Singh 's explanation ?

  6. arlira
    • 2 years ago
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    nope

  7. elisichi
    • 2 years ago
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    I'm guessing that x is the length of the longer leg, and that this can be written as: \[X^2+ (X-7)^2 = X+1 \] Then, you can solve from there by expanding brackets and you get a quadratic equation: \[x^2 -8x+48=0\]

  8. Aditi_Singh
    • 2 years ago
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    Almost Correct , but x^2 + (x+7)^2 = (x+8)^2 Apply Pythagoras theorem!

  9. elisichi
    • 2 years ago
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    Wait no.. my answer makes sense. Doesnt it? \[x^2+(x−7)^2=(x+1)^2 \] \[= x^2+x^2-7x+49=x^2 + x+1 \] \[x^2-8x+48=0 \]

  10. Aditi_Singh
    • 2 years ago
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    :| , ofcourse it does makes sense. But here, your hypotenuse isn't (x+1) , it is x+8 ..

  11. elisichi
    • 2 years ago
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    Im taking x as the longer leg, not the shorter one

  12. Aditi_Singh
    • 2 years ago
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    Okay, then the longer leg would be x , shorter length would be x-7, and hyp would be x+1 .. Well done buddy :D , you were absolutely correct :D

  13. elisichi
    • 2 years ago
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    Woohoo ;) High five for being right :D

  14. Aditi_Singh
    • 2 years ago
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    High five ;) :D

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