Question

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?@

Answer 1

@RadEn CAN YOU HELP ME PLZ

Answer 2

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

Answer 3

i mean solve for EI

Answer 4

i already supposed EI be x
so,
x^2 = 25^2 - 7^2
solve for x

Answer 5

BUTI DONT KNOW THE MEASUREMENTS ARE THEY THE ONES YOU GAVE ME 7, AND 25

Answer 6

from information above, HG = 7
nah, HI = HF = HG = radius = 7

Answer 7

I GOT 674

Answer 8

still wrong

Answer 9

EI^2 = 25^2 - 7^2
EI^2 = 625 - 49
EI^2 = 576
EI = sqrt(576) = 24

Answer 10

because EC = EI
so, EC = 24

Answer 11

HOLD UP BUT HOW DID YOU GET 625

Answer 12

25^2 = 25 x 25 = 625
note : ^ sign means square

Answer 13

7^2 = 7 * 7 = 49

Answer 14

LOL IKNOW AND THANK YOU SO MUCH I LEARNED ALOT TODAY MY HEAD IS ACCTUALLY GOING TO BLOW UP :)

Answer 15

you're welcome

dont eat to more :) wkkkkkkkk