Answer on Question #54831 - Math - Geometry
Find the perimeter of the figure.
Solution
The figure consists of two squares and two quarters of circle. The perimeter equals P=AB+BM⃗+MN+NK+KD⃗+DAP = AB + \vec{BM} + MN + NK + \vec{KD} + DAP=AB+BM+MN+NK+KD+DA .
It is given that AB=MN=NK=AD=4AB = MN = NK = AD = 4AB=MN=NK=AD=4
Arcs BM⃗,KD⃗\vec{BM},\vec{KD}BM,KD are equal, because they have the same radius R=4R = 4R=4 ( R=MC=CKR = MC = CKR=MC=CK ) and subtend equal angles 90∘90{}^{\circ}90∘ .
So BM⃗=KD⃗=14∗2πR=πR2=2π\vec{BM} = \vec{KD} = \frac{1}{4} * 2\pi R = \frac{\pi R}{2} = 2\piBM=KD=41∗2πR=2πR=2π .
Finally P=4+2π+4+4+2π+4=16+4πP = 4 + 2\pi + 4 + 4 + 2\pi + 4 = 16 + 4\piP=4+2π+4+4+2π+4=16+4π
Answer: P=16+4πP = 16 + 4\piP=16+4π
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