1. **Problem 1: Calculate the area of the shape made from triangle ABC and a sector of a circle with center C, radius CA = 7 cm, CB = 16 cm, and angle ACB = 150°**. 2. The shape consists of a sector of a circle with radius $r=7$ cm and central angle $\theta=150^\circ$, plus triangle ABC with sides CA = 7 cm and CB = 16 cm. 3. **Formula for area of sector:** $$\text{Area}_{\text{sector}} = \frac{\theta}{360^\circ} \pi r^2$$ 4. Calculate the area of the sector: $$\text{Area}_{\text{sector}} = \frac{150}{360} \pi (7)^2 = \frac{5}{12} \pi \times 49 = \frac{245}{12} \pi \approx 64.17 \text{ cm}^2$$ 5. **Formula for area of triangle using two sides and included angle:** $$\text{Area}_{\triangle} = \frac{1}{2} ab \sin C$$ 6. Here, sides are CA = 7 cm, CB = 16 cm, and included angle $\angle ACB = 150^\circ$. 7. Calculate the area of triangle ABC: $$\text{Area}_{\triangle} = \frac{1}{2} \times 7 \times 16 \times \sin 150^\circ = 56 \times 0.5 = 28 \text{ cm}^2$$ 8. **Total area of the shape:** $$\text{Area}_{\text{total}} = \text{Area}_{\text{sector}} + \text{Area}_{\triangle} = 64.17 + 28 = 92.17 \text{ cm}^2$$ --- 1. **Problem 2: Calculate the area of the shaded part in a circle with diameter AOD, radius 9 cm, and angle ADC = 35°**. 2. The shaded area is the segment of the circle cut off by chord AC corresponding to angle ADC = 35°. 3. **Formula for area of segment:** $$\text{Area}_{\text{segment}} = \text{Area}_{\text{sector}} - \text{Area}_{\triangle}$$ 4. Radius $r=9$ cm, central angle $\theta = 35^\circ$. 5. Calculate area of sector AOC: $$\text{Area}_{\text{sector}} = \frac{35}{360} \pi (9)^2 = \frac{35}{360} \pi \times 81 = \frac{2835}{360} \pi = 7.875 \pi \approx 24.73 \text{ cm}^2$$ 6. Calculate area of triangle AOC (isosceles with sides 9 cm and angle 35°): $$\text{Area}_{\triangle} = \frac{1}{2} r^2 \sin \theta = \frac{1}{2} \times 81 \times \sin 35^\circ = 40.5 \times 0.574 = 23.25 \text{ cm}^2$$ 7. Calculate area of shaded segment: $$\text{Area}_{\text{segment}} = 24.73 - 23.25 = 1.48 \text{ cm}^2$$ 8. Rounded to 3 significant figures: $1.48$ cm$^2$. --- 1. **Problem 3: Find the total surface area of the remaining solid after removing a smaller cone of height 4 cm from the top of a cone of height 12 cm and base radius 6 cm**. 2. The original cone has height $H=12$ cm and base radius $R=6$ cm. 3. The smaller cone removed has height $h=4$ cm. Since the cut is parallel to the base, the smaller cone is similar to the original. 4. Find radius $r$ of the smaller cone: $$\frac{r}{R} = \frac{h}{H} \Rightarrow r = R \times \frac{h}{H} = 6 \times \frac{4}{12} = 2 \text{ cm}$$ 5. Calculate slant heights: - Original cone slant height $l = \sqrt{H^2 + R^2} = \sqrt{12^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5} \approx 13.42$ cm - Smaller cone slant height $l_s = \sqrt{h^2 + r^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \approx 4.47$ cm 6. Total surface area of remaining solid = lateral surface area of original cone minus lateral surface area of smaller cone plus area of the new base (the cut surface). 7. Lateral surface area formula: $$A_{lat} = \pi r l$$ 8. Calculate lateral areas: - Original cone: $A_{lat1} = \pi \times 6 \times 13.42 = 80.52\pi$ cm$^2$ - Smaller cone: $A_{lat2} = \pi \times 2 \times 4.47 = 8.94\pi$ cm$^2$ 9. Area of the cut surface (circle with radius 2 cm): $$A_{cut} = \pi r^2 = \pi \times 2^2 = 4\pi$$ 10. Total surface area: $$A_{total} = A_{lat1} - A_{lat2} + A_{cut} = 80.52\pi - 8.94\pi + 4\pi = (80.52 - 8.94 + 4)\pi = 75.58\pi \approx 237.3 \text{ cm}^2$$ **Final answers:** - Problem 1 area: $92.17$ cm$^2$ - Problem 2 shaded area: $1.48$ cm$^2$ - Problem 3 total surface area: $237.3$ cm$^2$