SNIIX18
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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SNIIX18
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SNIIX18
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@RadEn CAN YOU HELP ME PLZ
RadEn
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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
RadEn
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i mean solve for EI
RadEn
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i already supposed EI be x
so,
x^2 = 25^2 - 7^2
solve for x
SNIIX18
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BUTI DONT KNOW THE MEASUREMENTS ARE THEY THE ONES YOU GAVE ME 7, AND 25
RadEn
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from information above, HG = 7
nah, HI = HF = HG = radius = 7
SNIIX18
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I GOT 674
RadEn
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still wrong
RadEn
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EI^2 = 25^2 - 7^2
EI^2 = 625 - 49
EI^2 = 576
EI = sqrt(576) = 24
RadEn
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because EC = EI
so, EC = 24
SNIIX18
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HOLD UP BUT HOW DID YOU GET 625
RadEn
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25^2 = 25 x 25 = 625
note : ^ sign means square
RadEn
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7^2 = 7 * 7 = 49
SNIIX18
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LOL IKNOW AND THANK YOU SO MUCH I LEARNED ALOT TODAY MY HEAD IS ACCTUALLY GOING TO BLOW UP :)
RadEn
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you're welcome
dont eat to more :) wkkkkkkkk