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What are the distances between the following locations: a) Dream School to Excellence, b) Tassah to Olympic Glory, c) Excellence to Gayap, d) Dream School to Gayap?

Mathematics

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here are the solutions to the remaining problems:

3a)

  1. Identify the path from Dream School to Excellence.

    The path is Dream School →\to Tassan →\to Excellence.

  2. Find the distances for each segment from the legend.

    Dream School to Tassan = 8 km8 \text{ km}
    Tassan to Excellence = 10 km10 \text{ km}

  3. Add the distances.

    Total distance = 8 km+10 km=18 km8 \text{ km} + 10 \text{ km} = 18 \text{ km}
    It is 18 km\boxed{18 \text{ km}} from Dream School to Excellence.

    3b)

  4. Identify the path from Tassah to Olympic Glory.

    The path is Tassah →\to Gayap →\to Olympic Glory.

  5. Find the distances for each segment from the legend.

    Tassah to Gayap = 16 km16 \text{ km}
    Gayap to Olympic Glory = 4 km4 \text{ km}

  6. Add the distances.

    Total distance = 16 km+4 km=20 km16 \text{ km} + 4 \text{ km} = 20 \text{ km}
    It is 20 km\boxed{20 \text{ km}} from Tassah to Olympic Glory.

    3c)

  7. Identify the path from Excellence to Gayap.

    The path is Excellence →\to Tassan →\to Gayap.

  8. Find the distances for each segment from the legend.

    Excellence to Tassan = 10 km10 \text{ km}
    Tassan to Gayap = 16 km16 \text{ km}

  9. Add the distances.

    Total distance = 10 km+16 km=26 km10 \text{ km} + 16 \text{ km} = 26 \text{ km}
    It is 26 km\boxed{26 \text{ km}} from Excellence to Gayap.

    3d)

  10. Calculate the distance from Dream School to Gayap.

    Path: Dream School →\to Tassan →\to Gayap
    Distance = 8 km+16 km=24 km8 \text{ km} + 16 \text{ km} = 24 \text{ km}

  11. Calculate the total distance for the round trip.

    Total distance = 2×24 km=48 km2 \times 24 \text{ km} = 48 \text{ km}

  12. Calculate the time taken for the journey.

    Speed = 12 km/h12 \text{ km/h}
    Time = DistanceSpeed=48 km12 km/h=4 hours\frac{\text{Distance}}{\text{Speed}} = \frac{48 \text{ km}}{12 \text{ km/h}} = 4 \text{ hours}

  13. Determine the return time.

    Starting time = 11:50 am11:50 \text{ am}
    Add 4 hours to the starting time:
    11:50 am+4 hours=15:5011:50 \text{ am} + 4 \text{ hours} = 15:50
    In 12-hour format, 15:5015:50 is 3:50 pm3:50 \text{ pm}.
    Mr. Anselm returns to Dream School at 3:50 pm\boxed{3:50 \text{ pm}}.

    4a)

  14. Identify the relationship between the pool's radius and the yard's length CD.

    The length CD is the diameter of the semi-circular swimming pool.

  15. Use the given radius to find the diameter.

    Radius (rr) = 3.5 m3.5 \text{ m}
    Diameter (dd) = 2×r=2×3.5 m=7 m2 \times r = 2 \times 3.5 \text{ m} = 7 \text{ m}
    The length of the yard CD is 7 m\boxed{7 \text{ m}}.

    4b)

  16. Identify the dimensions of the rectangular yard ABCD.

    Length (CD) = 7 m7 \text{ m} (from 4a)
    Width (AD) = 5 m5 \text{ m} (given in the diagram)

  17. Calculate the area of the rectangle ABCD.

    Area of rectangle = Length ×\times Width

    =7 m×5 m= 7 \text{ m} \times 5 \text{ m} =35 m2= 35 \text{ m}^2

    The area of the yard ABCD is 35 m2\boxed{35 \text{ m}^2}.

    4c)

  18. Calculate the area of the semi-circular swimming pool.

    Radius (rr) = 3.5 m3.5 \text{ m}
    Area of a full circle = πr2\pi r^2
    Area of semi-circle = 12πr2\frac{1}{2} \pi r^2
    Using π≈227\pi \approx \frac{22}{7}:

    Area of semi-circle=12×227×(3.5)2\text{Area of semi-circle} = \frac{1}{2} \times \frac{22}{7} \times (3.5)^2 =12×227×12.25= \frac{1}{2} \times \frac{22}{7} \times 12.25 =11×12.257= 11 \times \frac{12.25}{7} =11×1.75= 11 \times 1.75 =19.25 m2= 19.25 \text{ m}^2
  19. Calculate the area to be concreted (shaded area).

    Area to concrete = Area of rectangle ABCD - Area of semi-circular pool

    =35 m2−19.25 m2= 35 \text{ m}^2 - 19.25 \text{ m}^2 =15.75 m2= 15.75 \text{ m}^2
  20. Calculate the cost of concreting.

    Cost per square metre = 2000 frs2000 \text{ frs}
    Total cost = Area to concrete ×\times Cost per square metre

    =15.75 m2×2000 frs/m2= 15.75 \text{ m}^2 \times 2000 \text{ frs/m}^2 =31500 frs= 31500 \text{ frs}

    The cost of concreting the yard is 31500 frs\boxed{31500 \text{ frs}}.

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