Answer:
x = 2[tex]\sqrt{13}[/tex]
Step-by-step explanation:
Given a tangent and a secant from an external point to the circle
Then the square of the measure of the tangent is equal to the product of the measures of the secant's external part and the entire secant , that is
x² = 4(4 + 9) = 4 × 13 = 52 ( take square root of both sides )
[tex]\sqrt{x^2}[/tex] = [tex]\sqrt{52}[/tex] , then
x = [tex]\sqrt{52}[/tex] = [tex]\sqrt{4(13)}[/tex] = [tex]\sqrt{4}[/tex] × [tex]\sqrt{13}[/tex] = 2[tex]\sqrt{13}[/tex]
