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Find the area of the following shape. Use Use pi=3.142 and give your answer correct to 1 decimal place.

Mathematics

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Q1: Find the area of the following shape. Use π=3.142\pi=3.142 and give your answer correct to 1 decimal place.

  1. Calculate the area of the rectangular part.

    The rectangle has a length of 16 cm and a width of 7 cm.

    Arearectangle=length×width=16 cm×7 cm=112 cm2\text{Area}_{\text{rectangle}} = \text{length} \times \text{width} = 16\, \text{cm} \times 7\, \text{cm} = 112\, \text{cm}^2
  2. Calculate the area of the semicircular part.

    The diameter of the semicircle is 7 cm, so the radius is r=72=3.5r = \frac{7}{2} = 3.5 cm.

    Areasemicircle=12πr2=12×3.142×(3.5 cm)2\text{Area}_{\text{semicircle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times (3.5\, \text{cm})^2 Areasemicircle=12×3.142×12.25 cm2=19.24925 cm2\text{Area}_{\text{semicircle}} = \frac{1}{2} \times 3.142 \times 12.25\, \text{cm}^2 = 19.24925\, \text{cm}^2
  3. Add the areas to find the total area and round to one decimal place.

    Total Area=Arearectangle+Areasemicircle=112 cm2+19.24925 cm2=131.24925 cm2\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{semicircle}} = 112\, \text{cm}^2 + 19.24925\, \text{cm}^2 = 131.24925\, \text{cm}^2

    Rounding to one decimal place:

    Total Area≈131.2 cm2\text{Total Area} \approx 131.2\, \text{cm}^2

    The final answer is 131.2 cm2\boxed{\text{131.2 cm}^2}.

    Q2: Find the area of the shaded region. Give your answer to the nearest cm².

  4. Calculate the area of the large rectangle.

    The rectangle has a length of 44 cm and a width of 28 cm.

    Arearectangle=length×width=44 cm×28 cm=1232 cm2\text{Area}_{\text{rectangle}} = \text{length} \times \text{width} = 44\, \text{cm} \times 28\, \text{cm} = 1232\, \text{cm}^2
  5. Calculate the area of the unshaded triangle.

    The triangle has a base of 34 cm and a height of 18 cm.

    Areatriangle=12×base×height=12×34 cm×18 cm=306 cm2\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 34\, \text{cm} \times 18\, \text{cm} = 306\, \text{cm}^2
  6. Subtract the area of the triangle from the area of the rectangle to find the shaded area and round to the nearest cm².

    Areashaded=Arearectangle−Areatriangle=1232 cm2−306 cm2=926 cm2\text{Area}_{\text{shaded}} = \text{Area}_{\text{rectangle}} - \text{Area}_{\text{triangle}} = 1232\, \text{cm}^2 - 306\, \text{cm}^2 = 926\, \text{cm}^2

    The final answer is 926 cm2\boxed{\text{926 cm}^2}.

    Q3: Find the area of the following trapezium (correct to 2 decimal places).

  7. Identify the lengths of the parallel sides and the height.

    The parallel sides are a=5.54a = 5.54 cm and b=7.64b = 7.64 cm.
    The height is h=3.9h = 3.9 cm.

  8. Use the formula for the area of a trapezium.

    The area of a trapezium is given by Area=12(a+b)h\text{Area} = \frac{1}{2} (a+b) h.

    Area=12(5.54 cm+7.64 cm)×3.9 cm\text{Area} = \frac{1}{2} (5.54\, \text{cm} + 7.64\, \text{cm}) \times 3.9\, \text{cm} Area=12(13.18 cm)×3.9 cm\text{Area} = \frac{1}{2} (13.18\, \text{cm}) \times 3.9\, \text{cm} Area=6.59 cm×3.9 cm=25.701 cm2\text{Area} = 6.59\, \text{cm} \times 3.9\, \text{cm} = 25.701\, \text{cm}^2
  9. Round the answer to two decimal places.

    Area≈25.70 cm2\text{Area} \approx 25.70\, \text{cm}^2

    The final answer is 25.70 cm2\boxed{\text{25.70 cm}^2}.

    Q4: Find the surface area of the Triangular prism.

  10. Calculate the area of the two triangular bases.

    The base of the triangle is B=35B = 35 and the height is H=28H = 28.

    Areabase triangle=12×B×H=12×35×28=490\text{Area}_{\text{base triangle}} = \frac{1}{2} \times B \times H = \frac{1}{2} \times 35 \times 28 = 490

    Since there are two such bases, their combined area is 2×490=9802 \times 490 = 980.

  11. Calculate the areas of the three rectangular faces.

    The length of the prism is L=25L = 25. The sides of the triangular base are 35, 33, and 33 (from the diagram, assuming the S=33S=33 refers to the equal sides of the isosceles triangle, which is consistent with 282+(35/2)2≈33\sqrt{28^2 + (35/2)^2} \approx 33).

    • Area of the bottom rectangular face: 35×25=87535 \times 25 = 875
    • Area of the first slanted rectangular face: 33×25=82533 \times 25 = 825
    • Area of the second slanted rectangular face: 33×25=82533 \times 25 = 825
  12. Sum all the areas to find the total surface area.

    Total Surface Area=(2×Areabase triangle)+Areaface 1+Areaface 2+Areaface 3\text{Total Surface Area} = (2 \times \text{Area}_{\text{base triangle}}) + \text{Area}_{\text{face 1}} + \text{Area}_{\text{face 2}} + \text{Area}_{\text{face 3}} Total Surface Area=980+875+825+825=3505\text{Total Surface Area} = 980 + 875 + 825 + 825 = 3505

    The units are not specified in the question or diagram for this problem, so we will assume generic units.
    The final answer is 3505 square units\boxed{\text{3505 square units}}.

    Q5: Find the Total surface area of the following closed cone with the following dimensions: r=9r=9, H=18H=18, L=20L=20. Use π=3.14\pi=3.14 and give your answer to the nearest square units.

  13. Identify the given dimensions.

    Radius r=9r = 9.
    Slant height L=20L = 20.
    Height H=18H = 18.
    Use π=3.14\pi = 3.14.

  14. Calculate the area of the circular base.

    Areabase=πr2=3.14×(9)2=3.14×81=254.34\text{Area}_{\text{base}} = \pi r^2 = 3.14 \times (9)^2 = 3.14 \times 81 = 254.34
  15. Calculate the lateral surface area of the cone.

    Arealateral=πrL=3.14×9×20=3.14×180=565.2\text{Area}_{\text{lateral}} = \pi r L = 3.14 \times 9 \times 20 = 3.14 \times 180 = 565.2
  16. Add the base area and lateral surface area to find the total surface area and round to the nearest square unit.

    Total Surface Area=Areabase+Arealateral=254.34+565.2=819.54\text{Total Surface Area} = \text{Area}_{\text{base}} + \text{Area}_{\text{lateral}} = 254.34 + 565.2 = 819.54

    Rounding to the nearest square unit:

    Total Surface Area≈820\text{Total Surface Area} \approx 820

    The final answer is 820 square units\boxed{\text{820 square units}}.

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