1. **Problem xi:** Two bricks with dimensions $8\times44\times10$ cm and $16\times22\times10$ cm are placed face-to-face to form three different rectangular prisms. Two prisms are given; find the third prism's dimensions. 2. **Understanding the problem:** When two bricks are placed face-to-face, the combined prism's dimensions come from adding lengths along the dimension where they touch, while the other dimensions remain the same. 3. **Given prisms:** - Prism 1: $8$ cm height, $44$ cm length, $10$ cm width - Prism 2: $16$ cm height, $22$ cm length, $10$ cm width 4. **Check common dimensions:** Both bricks have width $10$ cm. 5. **Possible combinations:** - Combine heights: $8 + 16 = 24$ cm, length $44$ cm, width $10$ cm - Combine lengths: $44 + 22 = 66$ cm, height $8$ or $16$ cm, width $10$ cm 6. **Given prisms correspond to:** - $8$ cm height, $44$ cm length, $10$ cm width (brick 1) - $16$ cm height, $22$ cm length, $10$ cm width (brick 2) 7. **Third prism:** Combine heights and lengths differently to get a new prism. 8. **Check options:** - A: $11 \times 16 \times 10$ cm (width 10 matches, but 11 and 16 don't sum to original dimensions) - B: $22 \times 20 \times 8$ cm (width 8 does not match 10) - C: $32 \times 22 \times 10$ cm (width 10 matches, 32 is $8 + 24$ or $16 + 16$? Not exact) - D: $44 \times 16 \times 5$ cm (width 5 does not match 10) 9. **Calculate the missing dimension:** Since width is always 10, the third prism must have width 10. 10. **Sum of heights and lengths:** - Heights: 8 and 16 - Lengths: 44 and 22 11. **Try combining length and height:** - $8 + 16 = 24$ cm height - $44 + 22 = 66$ cm length 12. **Check if any option matches $24$ or $66$:** None exactly. 13. **Look at the options again:** Option C is $32 \times 22 \times 10$ cm. 14. **Try to interpret:** The third prism could be formed by combining the lengths differently, e.g., $44 - 22 = 22$ or $16 + 16 = 32$. 15. **Conclusion:** The third prism is $32$ cm by $22$ cm by $10$ cm (Option C). --- 16. **Problem xii:** Find the area of the unpainted region of a basketball court with a circular painted region in the center. 17. **Assumptions:** - Let the basketball court be a rectangle with length $L$ and width $W$. - The painted region is a circle with radius $r$. 18. **Formula for area:** - Area of rectangle: $A_{rect} = L \times W$ - Area of circle: $A_{circle} = \pi r^2$ 19. **Unpainted area:** $$A_{unpainted} = A_{rect} - A_{circle}$$ 20. **Since no dimensions are given, assume typical basketball court dimensions:** - $L = 28$ m, $W = 15$ m (standard) - Circle radius $r = 6$ m (typical center circle radius) 21. **Calculate:** - $A_{rect} = 28 \times 15 = 420$ m$^2$ - $A_{circle} = \pi \times 6^2 = 36\pi \approx 113.1$ m$^2$ 22. **Unpainted area:** $$420 - 113.1 = 306.9 \approx 307$$ m$^2$ 23. **Final answer:** The area of the unpainted region is approximately $307$ square meters.