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SNIIX18

  • 2 years ago

Given: AG and CI are common internal tangents of .H and .B, HG =7, ED=18, and ED=EF. What is the measure of EC?@

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  1. SNIIX18
    • 2 years ago
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  2. SNIIX18
    • 2 years ago
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    @RadEn CAN YOU HELP ME PLZ

  3. RadEn
    • 2 years ago
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    EC = EI (let be x) HI = HF = HG = radius = 7 EH = FE + HF = 18+7 = 25 look at the triangle of HIE (it is a right triangle), use the pythagorean theorem to get EI so, EI^2 = EH^2 - HI^2 EI^2 = 25^2 - 7^2 solve for HI

  4. RadEn
    • 2 years ago
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    i mean solve for EI

  5. RadEn
    • 2 years ago
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    i already supposed EI be x so, x^2 = 25^2 - 7^2 solve for x

  6. SNIIX18
    • 2 years ago
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    BUTI DONT KNOW THE MEASUREMENTS ARE THEY THE ONES YOU GAVE ME 7, AND 25

  7. RadEn
    • 2 years ago
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    from information above, HG = 7 nah, HI = HF = HG = radius = 7

  8. SNIIX18
    • 2 years ago
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    I GOT 674

  9. RadEn
    • 2 years ago
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    still wrong

  10. RadEn
    • 2 years ago
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    EI^2 = 25^2 - 7^2 EI^2 = 625 - 49 EI^2 = 576 EI = sqrt(576) = 24

  11. RadEn
    • 2 years ago
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    because EC = EI so, EC = 24

  12. SNIIX18
    • 2 years ago
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    HOLD UP BUT HOW DID YOU GET 625

  13. RadEn
    • 2 years ago
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    25^2 = 25 x 25 = 625 note : ^ sign means square

  14. RadEn
    • 2 years ago
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    7^2 = 7 * 7 = 49

  15. SNIIX18
    • 2 years ago
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    LOL IKNOW AND THANK YOU SO MUCH I LEARNED ALOT TODAY MY HEAD IS ACCTUALLY GOING TO BLOW UP :)

  16. RadEn
    • 2 years ago
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    you're welcome dont eat to more :) wkkkkkkkk

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