1. **State the problem:** We need to find the length of side $SR$ in triangle $PSR$, which is isosceles with $PS = SR$. Given $PR = 16$ cm and the area of triangle $PSQ$ is 40 cm², and $PQR$ is tangent to the circle at $Q$. 2. **Identify known information and formulas:** - Triangle $PSR$ is isosceles with $PS = SR$. - $PR = 16$ cm. - Area of triangle $PSQ = 40$ cm². - $PQR$ is tangent to the circle at $Q$. 3. **Use the area formula for triangle $PSQ$:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ Here, base $PQ$ and height $SQ$ (perpendicular from $S$ to $PQ$). 4. **Express area in terms of $PQ$ and $SQ$:** $$40 = \frac{1}{2} \times PQ \times SQ \implies PQ \times SQ = 80$$ 5. **Use the tangent property:** Since $PQR$ is tangent at $Q$, $SQ$ is perpendicular to $PQ$ (tangent radius property), so $SQ$ is height. 6. **Express $PR$ in terms of $PQ$ and $QR$:** Since $PQR$ is a straight line, $PR = PQ + QR = 16$ cm. 7. **Use isosceles triangle property:** $PS = SR$, so triangle $PSR$ has two equal sides. 8. **Use Pythagoras theorem in triangles $PSQ$ and $QSR$:** Since $SQ$ is perpendicular to $PQ$, triangles $PSQ$ and $QSR$ are right-angled at $Q$. 9. **Let $PQ = x$, then $QR = 16 - x$.** 10. **Express $PS$ and $SR$ using Pythagoras:** $$PS = \sqrt{SQ^2 + x^2}$$ $$SR = \sqrt{SQ^2 + (16 - x)^2}$$ 11. **Since $PS = SR$, set equal:** $$\sqrt{SQ^2 + x^2} = \sqrt{SQ^2 + (16 - x)^2}$$ Square both sides: $$SQ^2 + x^2 = SQ^2 + (16 - x)^2$$ Simplify: $$x^2 = (16 - x)^2$$ $$x^2 = 256 - 32x + x^2$$ Cancel $x^2$: $$0 = 256 - 32x$$ $$32x = 256$$ $$x = 8$$ 12. **Find $SQ$ using area relation:** $$PQ \times SQ = 80 \implies 8 \times SQ = 80 \implies SQ = 10$$ 13. **Calculate $SR$:** $$SR = \sqrt{SQ^2 + (16 - x)^2} = \sqrt{10^2 + 8^2} = \sqrt{100 + 64} = \sqrt{164} = 2\sqrt{41} \approx 12.81$$ **Final answer:** $$SR = 2\sqrt{41} \text{ cm} \approx 12.81 \text{ cm}$$