1. **Problem statement:** We have a circle with points P, Q, R on the circumference and secants PS and RT intersecting outside the circle. Given segment lengths: $PT=2.4$, $PQ=7$, and $RZ=z$ (unknown). We need to find $z$, $PS$, and $RT$. 2. **Relevant theorem:** For two secants intersecting outside a circle, the products of the whole secant segment and its external segment are equal: $$PT \times PS = RT \times RZ$$ where $PS = PT + TQ$ and $RT = RZ + ZT$. 3. **Given:** $PT=2.4$, $PQ=7$, so $PS = PT + PQ = 2.4 + 7 = 9.4$. 4. **Set up the equation:** $$PT \times PS = RT \times RZ$$ Substitute known values: $$2.4 \times 9.4 = RT \times z$$ Calculate left side: $$2.4 \times 9.4 = 22.56$$ 5. **Express $RT$ in terms of $z$:** Since $RT = RZ + ZT = z + 2.4$ (assuming $ZT=2.4$ as the external segment similar to $PT$), then: $$22.56 = (z + 2.4) \times z = z^2 + 2.4z$$ 6. **Solve quadratic equation:** $$z^2 + 2.4z - 22.56 = 0$$ Use quadratic formula: $$z = \frac{-2.4 \pm \sqrt{2.4^2 - 4 \times 1 \times (-22.56)}}{2}$$ Calculate discriminant: $$2.4^2 = 5.76$$ $$4 \times 22.56 = 90.24$$ $$\sqrt{5.76 + 90.24} = \sqrt{96} = 9.79796$$ 7. **Calculate roots:** $$z = \frac{-2.4 \pm 9.79796}{2}$$ Positive root: $$z = \frac{-2.4 + 9.79796}{2} = \frac{7.39796}{2} = 3.69898$$ Negative root is discarded since length cannot be negative. 8. **Find $RT$:** $$RT = z + 2.4 = 3.69898 + 2.4 = 6.09898$$ **Final answers:** $$z \approx 3.7$$ $$PS = 9.4$$ $$RT \approx 6.1$$