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What is the area of the shaded regions in the following figures? (1) A rectangle with a triangle removed. (2) A rectangle with a semicircle removed.

Mathematics

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Here are the areas for each shaded region:

1)

  1. Identify the shapes.

    The figure is a rectangle with a triangle removed from its left side.
    The rectangle has a length of 15 ft15 \text{ ft} and a width of 8 ft8 \text{ ft}.
    The triangle has a base of 8 ft8 \text{ ft} (same as the rectangle's width) and a height of 7 ft7 \text{ ft}.

  2. Calculate the area of the rectangle.

    Arectangle=length×width=15 ft×8 ft=120 ft2A_{\text{rectangle}} = \text{length} \times \text{width} = 15 \text{ ft} \times 8 \text{ ft} = 120 \text{ ft}^2
  3. Calculate the area of the unshaded triangle.

    Atriangle=12×base×height=12×8 ft×7 ft=28 ft2A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \text{ ft} \times 7 \text{ ft} = 28 \text{ ft}^2
  4. Subtract the area of the triangle from the area of the rectangle to find the shaded area.

    Ashaded=Arectangle−Atriangle=120 ft2−28 ft2=92 ft2A_{\text{shaded}} = A_{\text{rectangle}} - A_{\text{triangle}} = 120 \text{ ft}^2 - 28 \text{ ft}^2 = 92 \text{ ft}^2

    The shaded area is 92.00 ft2\boxed{\text{92.00 ft}^2}.

    2)

  5. Identify the shapes.

    The figure is a rectangle with a semicircle removed from its left side.
    The rectangle has a length of 10 m10 \text{ m} and a width of 8 m8 \text{ m}.
    The semicircle has a diameter of 8 m8 \text{ m}, so its radius is r=82=4 mr = \frac{8}{2} = 4 \text{ m}.

  6. Calculate the area of the rectangle.

    Arectangle=length×width=10 m×8 m=80 m2A_{\text{rectangle}} = \text{length} \times \text{width} = 10 \text{ m} \times 8 \text{ m} = 80 \text{ m}^2
  7. Calculate the area of the unshaded semicircle.

    The area of a full circle is πr2\pi r^2. The area of a semicircle is 12πr2\frac{1}{2} \pi r^2.

    Asemicircle=12π(4 m)2=12π(16 m2)=8π m2≈8×3.14159 m2≈25.1327 m2A_{\text{semicircle}} = \frac{1}{2} \pi (4 \text{ m})^2 = \frac{1}{2} \pi (16 \text{ m}^2) = 8\pi \text{ m}^2 \approx 8 \times 3.14159 \text{ m}^2 \approx 25.1327 \text{ m}^2
  8. Subtract the area of the semicircle from the area of the rectangle to find the shaded area.

    Ashaded=Arectangle−Asemicircle=80 m2−25.1327 m2=54.8673 m2A_{\text{shaded}} = A_{\text{rectangle}} - A_{\text{semicircle}} = 80 \text{ m}^2 - 25.1327 \text{ m}^2 = 54.8673 \text{ m}^2
  9. Round to two decimal places.

    Ashaded≈54.87 m2A_{\text{shaded}} \approx 54.87 \text{ m}^2

    The shaded area is 54.87 m2\boxed{\text{54.87 m}^2}.

    3)

  10. Identify the shapes.

    The figure is a circle with a square removed from its center.
    The circle has a diameter of 14 cm14 \text{ cm}, so its radius is r=142=7 cmr = \frac{14}{2} = 7 \text{ cm}.
    The square has a side length of 10 cm10 \text{ cm}.

  11. Calculate the area of the circle.

    Acircle=πr2=π(7 cm)2=49π cm2≈49×3.14159 cm2≈153.938 cm2A_{\text{circle}} = \pi r^2 = \pi (7 \text{ cm})^2 = 49\pi \text{ cm}^2 \approx 49 \times 3.14159 \text{ cm}^2 \approx 153.938 \text{ cm}^2
  12. Calculate the area of the unshaded square.

    Asquare=side×side=10 cm×10 cm=100 cm2A_{\text{square}} = \text{side} \times \text{side} = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2
  13. Subtract the area of the square from the area of the circle to find the shaded area.

    Ashaded=Acircle−Asquare=153.938 cm2−100 cm2=53.938 cm2A_{\text{shaded}} = A_{\text{circle}} - A_{\text{square}} = 153.938 \text{ cm}^2 - 100 \text{ cm}^2 = 53.938 \text{ cm}^2
  14. Round to two decimal places.

    Ashaded≈53.94 cm2A_{\text{shaded}} \approx 53.94 \text{ cm}^2

    The shaded area is 53.94 cm2\boxed{\text{53.94 cm}^2}.

    4)

  15. Identify the shapes.

    The figure is a rectangle with a circle removed from its center.
    The rectangle has a length of 12 yd12 \text{ yd} and a width of 7 yd7 \text{ yd}.
    The circle has a diameter of 4 yd4 \text{ yd}, so its radius is r=42=2 ydr = \frac{4}{2} = 2 \text{ yd}.

  16. Calculate the area of the rectangle.

    Arectangle=length×width=12 yd×7 yd=84 yd2A_{\text{rectangle}} = \text{length} \times \text{width} = 12 \text{ yd} \times 7 \text{ yd} = 84 \text{ yd}^2
  17. Calculate the area of the unshaded circle.

    Acircle=πr2=π(2 yd)2=4π yd2≈4×3.14159 yd2≈12.5664 yd2A_{\text{circle}} = \pi r^2 = \pi (2 \text{ yd})^2 = 4\pi \text{ yd}^2 \approx 4 \times 3.14159 \text{ yd}^2 \approx 12.5664 \text{ yd}^2
  18. Subtract the area of the circle from the area of the rectangle to find the shaded area.

    Ashaded=Arectangle−Acircle=84 yd2−12.5664 yd2=71.4336 yd2A_{\text{shaded}} = A_{\text{rectangle}} - A_{\text{circle}} = 84 \text{ yd}^2 - 12.5664 \text{ yd}^2 = 71.4336 \text{ yd}^2
  19. Round to two decimal places.

    Ashaded≈71.43 yd2A_{\text{shaded}} \approx 71.43 \text{ yd}^2

    The shaded area is 71.43 yd2\boxed{\text{71.43 yd}^2}.

    5)

  20. Identify the shapes.

    The figure is a large triangle with an unshaded trapezoid removed from its base.
    The large triangle has a base of 10 in10 \text{ in} and a height of 13 in13 \text{ in}.
    The unshaded trapezoid has parallel bases of 8 in8 \text{ in} and 10 in10 \text{ in}, and a height of 5 in5 \text{ in}.

  21. Calculate the area of the large triangle.

    Alarge triangle=12×base×height=12×10 in×13 in=65 in2A_{\text{large triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \text{ in} \times 13 \text{ in} = 65 \text{ in}^2
  22. Calculate the area of the unshaded trapezoid.

    Atrapezoid=12×(base1+base2)×height=12×(8 in+10 in)×5 in=12×18 in×5 in=45 in2A_{\text{trapezoid}} = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} = \frac{1}{2} \times (8 \text{ in} + 10 \text{ in}) \times 5 \text{ in} = \frac{1}{2} \times 18 \text{ in} \times 5 \text{ in} = 45 \text{ in}^2
  23. Subtract the area of the trapezoid from the area of the large triangle to find the shaded area.

    Ashaded=Alarge triangle−Atrapezoid=65 in2−45 in2=20 in2A_{\text{shaded}} = A_{\text{large triangle}} - A_{\text{trapezoid}} = 65 \text{ in}^2 - 45 \text{ in}^2 = 20 \text{ in}^2

    The shaded area is 20.00 in2\boxed{\text{20.00 in}^2}.

    6)

  24. Identify the shapes.

    The figure is a large square with a smaller square removed from its center.
    The large square has a side length of 14 m14 \text{ m}.
    The small square has a side length of 8 m8 \text{ m}.

  25. Calculate the area of the large square.

    Alarge square=side×side=14 m×14 m=196 m2A_{\text{large square}} = \text{side} \times \text{side} = 14 \text{ m} \times 14 \text{ m} = 196 \text{ m}^2
  26. Calculate the area of the unshaded small square.

    Asmall square=side×side=8 m×8 m=64 m2A_{\text{small square}} = \text{side} \times \text{side} = 8 \text{ m} \times 8 \text{ m} = 64 \text{ m}^2
  27. Subtract the area of the small square from the area of the large square to find the shaded area.

    Ashaded=Alarge square−Asmall square=196 m2−64 m2=132 m2A_{\text{shaded}} = A_{\text{large square}} - A_{\text{small square}} = 196 \text{ m}^2 - 64 \text{ m}^2 = 132 \text{ m}^2

    The shaded area is 132.00 m2\boxed{\text{132.00 m}^2}.

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