1. **Problem statement:** We have a circle with radius 5 cm and a chord PQ of length 8 cm. Tangents at points P and Q intersect at point T. We need to find the length of segment TP. 2. **Key concepts:** - The radius of the circle is $RO = 5$ cm. - The chord $PQ = 8$ cm. - The tangents at P and Q meet at T. - The line $OT$ (from center O to T) is perpendicular to chord PQ and bisects it, so $PR = RQ = 4$ cm. - Triangles $TRP$ and $PRO$ are right triangles and similar by AA similarity. 3. **Using similarity:** Since $\triangle TRP \sim \triangle PRO$, corresponding sides are proportional: $$\frac{TP}{PO} = \frac{RP}{RO}$$ Given $RO = 5$ cm and $RP = 4$ cm, and $PO = 3$ cm (since $O$ is center and $PR = 4$, $PQ=8$, $RO=5$, $PO$ is perpendicular segment from center to chord, so $PO = \sqrt{5^2 - 4^2} = 3$ cm), we get: $$\frac{TP}{3} = \frac{4}{5} \implies TP = \frac{4}{5} \times 3 = \frac{12}{5} = 2.4 \text{ cm}$$ 4. **Using Pythagoras theorem (alternative method):** Let $TP = x$ and $TR = y$. - In right triangle $PRT$: $$x^2 = y^2 + 4^2 = y^2 + 16$$ - In right triangle $OPT$: $$x^2 + 5^2 = (y + 3)^2$$ Subtracting the first from the second: $$x^2 + 25 = y^2 + 6y + 9$$ $$x^2 = y^2 + 6y - 16$$ Equate with the first equation: $$y^2 + 16 = y^2 + 6y - 16$$ Simplify: $$16 = 6y - 16 \implies 6y = 32 \implies y = \frac{16}{3}$$ Substitute back to find $x$: $$x^2 = \left(\frac{16}{3}\right)^2 + 16 = \frac{256}{9} + 16 = \frac{256}{9} + \frac{144}{9} = \frac{400}{9}$$ $$x = \frac{20}{3} \approx 6.67 \text{ cm}$$ 5. **Final answer:** $$\boxed{TP = \frac{20}{3} \text{ cm} \approx 6.67 \text{ cm}}$$ This length is the distance from the tangent intersection point T to point P on the circle.