1. **State the problem:** We have triangle PQR with point S on segment QR. Given: $QP = QR = 9$ cm, $PR = PS = 6$ cm. We need to find the length of $SR$. 2. **Analyze the triangle:** Since $QP = QR$, triangle PQR is isosceles with base $PQ = QR = 9$ cm. 3. **Use the given equal lengths:** $PR = PS = 6$ cm means point S lies on QR such that $PS = PR$. 4. **Apply the Law of Cosines in triangles PQR and PSR:** - In triangle PQR, by Law of Cosines for side $PR$: $$PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\angle PQR)$$ - Since $PQ = QR = 9$ and $PR = 6$: $$6^2 = 9^2 + 9^2 - 2 \cdot 9 \cdot 9 \cdot \cos(\angle PQR)$$ $$36 = 81 + 81 - 162 \cos(\angle PQR)$$ $$36 = 162 - 162 \cos(\angle PQR)$$ $$162 \cos(\angle PQR) = 162 - 36 = 126$$ $$\cos(\angle PQR) = \frac{126}{162} = \frac{7}{9}$$ 5. **Find length SR:** Since S lies on QR, let $QS = x$ and $SR = 9 - x$. 6. **Use Law of Cosines in triangle PSR:** - $PS = 6$, $PR = 6$, and $SR = 9 - x$. - Triangle PSR shares side $PR$ and $PS$ equal, so triangle PSR is isosceles with $PS = PR$. 7. **Use the fact that $PS = PR$ and point S lies on QR:** - Using the distance formula or coordinate geometry, place points: - Let $Q$ at $(0,0)$, - $R$ at $(9,0)$, - $P$ somewhere above the x-axis. 8. **Find coordinates of P:** - Since $PQ = 9$, $Q=(0,0)$, $P=(x_p,y_p)$ satisfies: $$\sqrt{(x_p - 0)^2 + (y_p - 0)^2} = 9$$ $$x_p^2 + y_p^2 = 81$$ - Since $PR = 6$, $R=(9,0)$: $$\sqrt{(x_p - 9)^2 + y_p^2} = 6$$ $$(x_p - 9)^2 + y_p^2 = 36$$ 9. **Subtract equations:** $$x_p^2 + y_p^2 = 81$$ $$(x_p - 9)^2 + y_p^2 = 36$$ Subtract second from first: $$x_p^2 + y_p^2 - (x_p - 9)^2 - y_p^2 = 81 - 36$$ $$x_p^2 - (x_p^2 - 18x_p + 81) = 45$$ $$x_p^2 - x_p^2 + 18x_p - 81 = 45$$ $$18x_p = 126$$ $$x_p = 7$$ 10. **Find $y_p$:** $$7^2 + y_p^2 = 81$$ $$49 + y_p^2 = 81$$ $$y_p^2 = 32$$ $$y_p = \sqrt{32} = 4\sqrt{2}$$ So, $P = (7, 4\sqrt{2})$. 11. **Find coordinates of S:** - $S$ lies on $QR$ between $Q(0,0)$ and $R(9,0)$, so $S = (s,0)$ for some $s$ between 0 and 9. 12. **Use $PS = 6$:** $$PS = \sqrt{(7 - s)^2 + (4\sqrt{2} - 0)^2} = 6$$ $$(7 - s)^2 + (4\sqrt{2})^2 = 36$$ $$(7 - s)^2 + 32 = 36$$ $$(7 - s)^2 = 4$$ $$7 - s = \pm 2$$ 13. **Solve for $s$:** - If $7 - s = 2$, then $s = 5$. - If $7 - s = -2$, then $s = 9$. Since $S$ lies between $Q(0)$ and $R(9)$, $s=5$ is valid. 14. **Find $SR$:** $$SR = 9 - s = 9 - 5 = 4$$ **Final answer:** $$\boxed{4 \text{ cm}}$$