1. **State the problem:** We have an isosceles triangle ABC with sides BA = BC, angle BAC = 70°, and a circle inscribed inside the triangle with radius 6 cm. The length BO = 17.54 cm. We need to find the area of triangle ABC. 2. **Key facts and formulas:** - The circle is inscribed, so it is tangent to all sides. - The radius of the inscribed circle (inradius) is $r = 6$ cm. - The area $A$ of a triangle can be found using the formula $A = r \times s$, where $s$ is the semiperimeter. - Since BA = BC, triangle ABC is isosceles with vertex angle at B. 3. **Analyze the triangle:** - Given angle BAC = 70°, and BA = BC, angles ABC and ACB are equal. - Sum of angles in triangle: $70° + 2x = 180° \Rightarrow 2x = 110° \Rightarrow x = 55°$. - So angles ABC = ACB = 55°. 4. **Use the given length BO = 17.54 cm:** - Point O is the incenter (center of inscribed circle), which lies at the intersection of angle bisectors. - BO is the distance from vertex B to incenter O. 5. **Find side lengths:** - Let the equal sides BA = BC = $a$. - Let base AC = $b$. 6. **Use Law of Sines in triangle ABC:** $$\frac{a}{\sin 70°} = \frac{b}{\sin 55°}$$ 7. **Express $b$ in terms of $a$:** $$b = a \times \frac{\sin 55°}{\sin 70°}$$ 8. **Find coordinates or use trigonometry to find $a$ using BO:** - Since BO is along the angle bisector of angle B (which is 55°), and O is the incenter, we can use the formula for distance from vertex to incenter: $$BO = \frac{r}{\sin(\frac{B}{2})}$$ - Here, angle B = 55°, so $$BO = \frac{6}{\sin(27.5°)}$$ - Calculate: $$\sin(27.5°) \approx 0.4617$$ $$BO = \frac{6}{0.4617} \approx 12.99$$ - But given BO = 17.54 cm, so this suggests angle B is not 55°, so re-examine. 9. **Reconsider angle B:** - Since BA = BC, angles at B and C are equal. - Given angle BAC = 70°, angles at B and C are each 55°. - The incenter distance from vertex B is: $$BO = \frac{r}{\sin(\frac{B}{2})} = \frac{6}{\sin(27.5°)} = 12.99$$ - Given BO = 17.54, so angle B must be smaller. 10. **Calculate angle B from BO:** $$BO = \frac{r}{\sin(\frac{B}{2})} \Rightarrow \sin(\frac{B}{2}) = \frac{r}{BO} = \frac{6}{17.54} = 0.342$$ $$\frac{B}{2} = \arcsin(0.342) = 20°$$ $$B = 40°$$ 11. **Find angle A:** $$A = 180° - 2B = 180° - 80° = 100°$$ 12. **Use Law of Sines to find sides:** Let equal sides be $a$, base $b$. $$\frac{a}{\sin 100°} = \frac{b}{\sin 40°}$$ 13. **Calculate semiperimeter $s$ using inradius formula:** Area $A = r \times s$. 14. **Find area using formula:** Area $A = \frac{1}{2}ab \sin B$. 15. **Express $b$ in terms of $a$:** $$b = a \times \frac{\sin 40°}{\sin 100°}$$ 16. **Calculate area in terms of $a$:** $$A = \frac{1}{2} a \times a \times \frac{\sin 40°}{\sin 100°} \times \sin 40° = \frac{1}{2} a^2 \frac{\sin^2 40°}{\sin 100°}$$ 17. **Calculate semiperimeter $s$:** $$s = \frac{2a + b}{2} = \frac{2a + a \frac{\sin 40°}{\sin 100°}}{2} = a \frac{2 + \frac{\sin 40°}{\sin 100°}}{2}$$ 18. **Use $A = r \times s$ to solve for $a$:** $$A = 6 \times s = 6 \times a \frac{2 + \frac{\sin 40°}{\sin 100°}}{2}$$ 19. **Equate the two expressions for area:** $$\frac{1}{2} a^2 \frac{\sin^2 40°}{\sin 100°} = 6 a \frac{2 + \frac{\sin 40°}{\sin 100°}}{2}$$ 20. **Simplify and solve for $a$:** Multiply both sides by 2: $$a^2 \frac{\sin^2 40°}{\sin 100°} = 12 a \left(1 + \frac{\sin 40°}{2 \sin 100°}\right)$$ Divide both sides by $a$ (assuming $a \neq 0$): $$a \frac{\sin^2 40°}{\sin 100°} = 12 \left(1 + \frac{\sin 40°}{2 \sin 100°}\right)$$ Use \cancel to show cancellation: $$a \cancel{\frac{\sin^2 40°}{\sin 100°}} = 12 \left(1 + \frac{\sin 40°}{2 \sin 100°}\right)$$ 21. **Calculate numerical values:** $$\sin 40° \approx 0.6428$$ $$\sin 100° \approx 0.9848$$ Calculate left side coefficient: $$\frac{\sin^2 40°}{\sin 100°} = \frac{0.6428^2}{0.9848} = \frac{0.4132}{0.9848} = 0.4194$$ Calculate right side term: $$1 + \frac{0.6428}{2 \times 0.9848} = 1 + \frac{0.6428}{1.9696} = 1 + 0.3264 = 1.3264$$ 22. **Solve for $a$:** $$a \times 0.4194 = 12 \times 1.3264 = 15.9168$$ $$a = \frac{15.9168}{0.4194} = 37.95 \text{ cm}$$ 23. **Find base $b$:** $$b = 37.95 \times \frac{0.6428}{0.9848} = 37.95 \times 0.6529 = 24.78 \text{ cm}$$ 24. **Calculate area $A$:** $$A = r \times s$$ Calculate semiperimeter: $$s = \frac{2a + b}{2} = \frac{2 \times 37.95 + 24.78}{2} = \frac{75.9 + 24.78}{2} = \frac{100.68}{2} = 50.34$$ Area: $$A = 6 \times 50.34 = 302.04 \text{ cm}^2$$ **Final answer:** The area of triangle ABC is approximately **302.04 cm²**.