Answer:
D
Step-by-step explanation:
Given a tangent and a secant from an external point to the circle, then
The product of the external part and the whole secant is equal to the square of the tangent, that is
8(8 - 6 + 2x) = 12²
8(2 + 2x) = 144 ( divide both sides by 8 )
2 + 2x = 18 ( subtract 2 from both sides )
2x = 16 ( divide both sides by 2 )
x = 8
Then
GH = - 6 + 2x = - 6 + 2(8) = - 6 + 16 = 10 → D
Answer:
[tex]\textsf{GH=10}[/tex]
Step-by-step explanation:
[tex](EF)^{2} =FG(FH)[/tex]
[tex](12)^{2} =8(8-6+2x)[/tex]
[tex]144=8(2+2x)[/tex]
[tex]144=16+16x[/tex]
[tex]16x=144-16[/tex]
[tex]16x=128[/tex]
[tex]x=128/16=8[/tex]
So, [tex]GH=-6+2(8)=[/tex]
[tex]-6+16=[/tex]
[tex]=10[/tex]
[tex]\textsf{OAmalOHopeO}[/tex]