1. **Problem 6: Find the lateral surface area of a triangular prism with an equilateral base of side 1.5 in and length 10 in.** 2. The lateral surface area (LSA) of a prism is the perimeter of the base times the height (length) of the prism: $$\text{LSA} = \text{Perimeter of base} \times \text{length}$$ 3. Since the base is an equilateral triangle with side $1.5$ in, its perimeter is: $$P = 3 \times 1.5 = 4.5 \text{ in}$$ 4. The length of the prism is $10$ in, so: $$\text{LSA} = 4.5 \times 10 = 45 \text{ in}^2$$ 5. **Answer for problem 6:** The lateral surface area is $45$ square inches. 6. **Problem 7a: Find the volume of the cube with edge length 3 ft.** 7. The volume $V$ of a cube with edge length $s$ is: $$V = s^3$$ 8. Substitute $s=3$ ft: $$V = 3^3 = 27 \text{ ft}^3$$ 9. **Answer for 7a:** The volume of the cube is $27$ cubic feet. 10. **Problem 7b: Find the volume of the pyramid with the same base and height as the cube.** 11. The volume $V$ of a pyramid is: $$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$ 12. The base of the pyramid is the same as the cube's base, which is a square of side 3 ft, so base area: $$\text{Base Area} = 3 \times 3 = 9 \text{ ft}^2$$ 13. The height of the pyramid is the same as the cube's height, which is 3 ft. 14. Substitute values: $$V = \frac{1}{3} \times 9 \times 3 = \frac{1}{3} \times 27 = 9 \text{ ft}^3$$ 15. **Answer for 7b:** The volume of the pyramid is $9$ cubic feet. 16. **Problem 8a: Classify and name two obtuse central angles in the circle with diameters AE and BD.** 17. Central angles are angles with vertex at the center $C$. 18. Since AE and BD are diameters, they divide the circle into four arcs. 19. The obtuse central angles are those greater than $90^\circ$ but less than $180^\circ$. 20. Two obtuse central angles are $\angle ACB$ and $\angle ECD$. 21. **Problem 8b: Name two acute central angles.** 22. Acute central angles are less than $90^\circ$. 23. Two acute central angles are $\angle BCE$ and $\angle DCA$. 24. **Problem 9a: Find the area of each triangle formed by points on the circle.** 25. Consider triangle $ABC$. 26. Since $AE$ and $BD$ are diameters intersecting at $C$, and $DE=8$ cm, radius $=3$ cm. 27. The triangle $ABC$ is right-angled because $BD$ is a diameter and angle $BAC$ subtends a semicircle. 28. Area of triangle $ABC$ is: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 29. Base $= 8$ cm (segment $DE$), height $= 3$ cm (radius from $C$). 30. Substitute: $$\text{Area} = \frac{1}{2} \times 8 \times 3 = 12 \text{ cm}^2$$ 31. **Answer for 9a:** The area of each triangle is $12$ square centimeters. 32. **Problem 9b: Classify triangle $ABC$ by its sides.** 33. Since $ABC$ is inscribed in a circle with diameters and has a right angle, and sides include diameter and radius segments, it is an isosceles right triangle if two sides are equal. 34. Given the radius is 3 cm and base 8 cm, sides are not equal. 35. Therefore, triangle $ABC$ is a scalene right triangle. 36. **Answer for 9b:** Triangle $ABC$ is scalene and right-angled. **Summary:** - Problem 6 lateral surface area: $45$ in$^2$ - Problem 7a cube volume: $27$ ft$^3$ - Problem 7b pyramid volume: $9$ ft$^3$ - Problem 8a obtuse central angles: $\angle ACB$, $\angle ECD$ - Problem 8b acute central angles: $\angle BCE$, $\angle DCA$ - Problem 9a area of triangle: $12$ cm$^2$ - Problem 9b triangle $ABC$ is scalene right triangle.