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Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Question

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Solution

1. Break Down the Problem

To find the area of the shaded region, we'll need to:

  1. Calculate the area of square ABCD.
  2. Calculate the area of semicircles APD and BPC.
  3. Subtract the combined area of the semicircles from the area of the square to obtain the area of the shaded region.

2. Relevant Concepts

  • The area of a square is given by the formula: Areasquare=side2 \text{Area}_{\text{square}} = \text{side}^2
  • The area of a semicircle is given by the formula: Areasemicircle=12πr2 \text{Area}_{\text{semicircle}} = \frac{1}{2} \pi r^2

3. Analysis and Detail

  1. Calculate the area of square ABCD:

    • Side length = 14 cm Areasquare=142=196cm2 \text{Area}_{\text{square}} = 14^2 = 196 \, \text{cm}^2
  2. Calculate the area of semicircles APD and BPC:

    • The radius r r of each semicircle equals half of the side of the square: r=142=7cm r = \frac{14}{2} = 7 \, \text{cm}
    • Area of one semicircle: Areasemicircle=12π(72)=12π(49)=49π2cm2 \text{Area}_{\text{semicircle}} = \frac{1}{2} \pi (7^2) = \frac{1}{2} \pi (49) = \frac{49\pi}{2} \, \text{cm}^2
    • Since there are two semicircles, we multiply the area of one by 2: Combined Areasemicircles=2×49π2=49πcm2 \text{Combined Area}_{\text{semicircles}} = 2 \times \frac{49\pi}{2} = 49\pi \, \text{cm}^2
  3. Calculate the area of the shaded region: Areashaded=AreasquareCombined Areasemicircles \text{Area}_{\text{shaded}} = \text{Area}_{\text{square}} - \text{Combined Area}_{\text{semicircles}} Areashaded=19649πcm2 \text{Area}_{\text{shaded}} = 196 - 49\pi \, \text{cm}^2

4. Verify and Summarize

To summarize, after calculating the area of the square and the semicircles, we found the area of the shaded region to be: Areashaded=19649πcm2 \text{Area}_{\text{shaded}} = 196 - 49\pi \, \text{cm}^2

Final Answer

The area of the shaded region is 19649πcm2 196 - 49\pi \, \text{cm}^2 (approx. 196 - 153.94 = 42.06 cm² for numerical approximation).

This problem has been solved

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