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