Anton used strings to hang two small light
balls on the ceiling as shown in the figure on
the right. The broken line represents the
distance from the point of tangency of the two
light balls to the ceiling.

b. Suppose Anton hangs
40 pairs of light balls on the ceiling of a hall in
preparation for an event. How long is the string that he needs to hang
these light balls if each has a diameter of 12 cm and the point of
tangency of each pair of balls is 30 cm from the ceiling?

[tex] \begin{array}{l} \textsf{Let }x\textsf{ be the length of string used for each light} \\ \textsf{ball.} \\ \\ \textsf{Refer to the diagram drawn.} \\ \\ \textsf{By Tangent - Secant Theorem,} \\ \\ \begin{aligned} \qquad CT^2 &= CA(CB) \\ CT^2 &= CA(CA + AB) \\ 30^2 &= x(x + 12) \\ 900 &= x(x + 12) \\ 900 &= x^2 + 12x \end{aligned} \\ \\ \textsf{By completing the square,} \\ \\ \begin{aligned} x^2 + 12x + 36 &= 900 + 36 \\ (x + 6)^2 &= 936 \\ x + 6 &= \pm \sqrt{936} \\ x + 6 &= \pm \sqrt{36(26)} \\ \implies x &= -6 \pm 6\sqrt{26}\end{aligned} \\ \\ \textsf{We only need the positive value of }x, \textsf{so }\\ x = (-6 + 6\sqrt{26})\textsf{ cm} \\ \\ \textsf{There are 40 pairs of lights balls or a total of }\\ 40(2) = 80\textsf{ light balls.} \\ \\ \textsf{Total length of string to be used :} \\ \implies 80x = 80(-6 + 6\sqrt{26}) \\ \implies 80x \approx \boxed{1967.53\textsf{ cm or }19.675\textsf{ meters}} \end{array} [/tex]

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