Answer:
x = 4[tex]\sqrt{3}[/tex]
Step-by-step explanation:
Given a tangent and a secant from an external point to a circle.
Then the square of the tangent is equal to the product of the secant's external part and the entire secant , that is
x² = 4(4 + 8) = 4 × 12 = 48 ( take square root of both sides )
x = [tex]\sqrt{48}[/tex] = [tex]\sqrt{16(3)}[/tex] = 4[tex]\sqrt{3}[/tex]
