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

### 3 Answers

- PopeLv 71 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

- KrishnamurthyLv 71 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

- Anonymous1 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.