Answer:
C
Step-by-step explanation:
Since AC is a tangent at C then ∠ACB = 90 and ΔABC is right
The area A of ΔABC = 0.5 × BC × AC ( A = 0.5 bh )
Calculate AC using Pythagoras' identity
The square on the hypotenuse AB is equal to the sum of the squares on the other 2 sides, that is
AC² + BC² = AB², substitute values
AC² + 4² = 8², that is
AC² + 16 = 64 ( subtract 16 from both sides )
AC² = 48 ( take the square root of both sides )
AC = [tex]\sqrt{48}[/tex] = [tex]\sqrt{16}[/tex] × [tex]\sqrt{3}[/tex] = 4[tex]\sqrt{3}[/tex], hence
Area of ΔABC = 0.5 × 4 × 4[tex]\sqrt{3}[/tex] = 8[tex]\sqrt{3}[/tex] → C
