It's a geometrical problem from Bangladesh. Can anyone provide the simplest way to solve it?

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  • Pope
    Lv 7
    1 month ago

    Let M be the midpoint of AD, hence the center of the semicircle.

    All linear measures below are in centimeters.

    MA = MQ = 2

    BA = BQ = 4

    MB = 2√(5)

    Quadrilateral ABQM is a kite, so its diagonals are perpendicular.

    (1/2)(MB)(AQ) = (MA)(AB) = area(ABQM)

    (1/2)[2√(5)](AQ) = (2)(4)

    AQ = 8/√(5)

    Let ∠QMB = ∠AMB = ∠QAB = θ.

    θ = tan⁻¹(2)

    sin(θ) = 2/√(5)

    cos(θ) = 1/√(5)

    sin(2θ) = 2sin(θ)cos(θ) = 4/5

    area(sector APQ) = (1/2)(AQ)²θ = (1/2)[8/√(5)]²θ = (32/5)θ

    area(sector MAQ) = (1/2)(MA)²(2θ) = (1/2)(2)²(2θ) = 4θ

    area(∆MAQ) = (1/2)(MA)(MQ)sin(2θ) = (1/2)(2)(2)(4/5) = 8/5

    required area

    = area(sector APQ) - area(sector MAQ) + area(∆MAQ)

    = (32/5)θ - 4θ + 8/5

    = (12/5)θ + 8/5

    = (12/5)tan⁻¹(2) + 8/5

    ≈ 4.257 cm²

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  • 1 month ago

    Start with a square with the side length of 4 cm.

    Draw a quarter circle from B and draw a half circle from the mid point of AD.

    The intersection point is Q.

    Now draw a sement of a circle of the radius AQ from the point A as its center

    which intersects with AB at the point P.

    Find the area of AQP.

    ABCD has an area of 16 cm^2.

    Half circle with diameter 4 cm has an area of 2pi cm^2.

    Quarter circle with radius of 4 cm has an area of 4 pi cm^2

  • Anonymous
    1 month ago

    It’s probably just a cultural thing that has gone on for generations. The best thing to do is to let the Bangladesh people solve their own problems without outside interference.

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