1. **State the problem:** We have triangle PQR with an inscribed circle touching sides at points A, B, and C. Given: - RA = 0.9 cm (segment on side PR) - PA = 2.6 cm (segment on side PQ) - Perimeter of triangle PQR = 11.8 cm We need to find the length of QB. 2. **Recall the tangent segment property:** Tangents drawn from the same external point to a circle are equal in length. So, - From vertex P: PA = PC = 2.6 cm - From vertex R: RA = RB = 0.9 cm - From vertex Q: QB = QC (unknown, call it x) 3. **Express the sides of the triangle in terms of these segments:** - Side PQ = PA + PC = 2.6 + 2.6 = 5.2 cm - Side PR = RA + AP = 0.9 + 2.6 = 3.5 cm - Side QR = QB + BR = x + 0.9 cm 4. **Use the perimeter to find x:** Perimeter = PQ + PR + QR $$11.8 = 5.2 + 3.5 + (x + 0.9)$$ Simplify: $$11.8 = 8.7 + x + 0.9$$ $$11.8 = 9.6 + x$$ Subtract 9.6 from both sides: $$x = 11.8 - 9.6 = 2.2$$ 5. **Conclusion:** The length of QB is $\boxed{2.2}$ cm.