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ILoveComputers1

  • one year ago

a=x, b=x, c=50 solve for x (hint: this is an isosceles triangle)

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  1. anonymous
    • one year ago
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    is 50 supposed to be one of the angles?

  2. ILoveComputers1
    • one year ago
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    They give me a triangle, and it's showing the pythagorean theorem. So it is finding the sides not angles.

  3. anonymous
    • one year ago
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    |dw:1439787827242:dw| Like this?

  4. ILoveComputers1
    • one year ago
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    Yes just like that. |dw:1439787916367:dw|

  5. anonymous
    • one year ago
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    well then just us Pythagrian theorem

  6. anonymous
    • one year ago
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    x^2 + x^2 = 50^2 solve for x

  7. nincompoop
    • one year ago
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    start by the definition of isosceles that two sides of the triangle have equal length. if that is our unknown, then the third side would be 50

  8. nincompoop
    • one year ago
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    |dw:1439788089504:dw|

  9. ILoveComputers1
    • one year ago
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    So then I would get 2x^2= 50^2 x^2=25^2 x=25

  10. nincompoop
    • one year ago
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    |dw:1439788184602:dw|

  11. nincompoop
    • one year ago
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    |dw:1439788232741:dw|

  12. nincompoop
    • one year ago
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    we utilize a few algebraic acrobat using a few formulas like the pythagorean theorem but we must recognize that there is now a fourth side and perhaps we can use simpler formula to solve for x

  13. nincompoop
    • one year ago
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    one of these suggestion is to bisect the side with length 50 instead of x and it may be easier to work with, because of known facts that come with it |dw:1439788720560:dw|

  14. nincompoop
    • one year ago
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    Let me redraw |dw:1439788819981:dw|

  15. nincompoop
    • one year ago
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    we are following the constraints or clue that our triangle is an isosceles so we will use isosceles right triangle :D

  16. nincompoop
    • one year ago
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    \(\large \sqrt{a^2 + b^2 } =c \rightarrow \sqrt{x^2 + x^2} = c = \sqrt{2x^2} \) \(\large \sqrt{2x^2}= 50 \) the rest of the solutions are easy to work with

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