Segment AB is divided by points M, N, P, K into 5 equal parts so that AM = MN = NP = PK = KB.

| May 10, 2021 | education

Segment AB is divided by points M, N, P, K into 5 equal parts so that AM = MN = NP = PK = KB. Circles are constructed on segments MB, AP, KB as on diameters. Then the ratio of the area of a circle bounded by a smaller circle to the sum of the areas of circles bounded by two other circles is equal to ..

Without loss of generality, let us assume that AM = MN = NP = PK = KB = 1.
Then MB = 4, AP = 3.
Since KB <MB and KB <AP, then the smaller circle is the circle constructed on the segment KB.
Let’s calculate the area of ​​a circle bounded by a smaller circle. To do this, use the formula: S = (π / 4) * d2. Therefore, the area (SKB) of a circle bounded by a smaller circle is SKB = π / 4.
Now, in a similar way, we calculate the areas of circles bounded by two other circles.
The area (SMB) of a circle bounded by a circle drawn on the segment MB is equal to SMB = (π / 4) * 42 = 4 * π.
The area (SAP) of a circle bounded by a circle drawn on the segment AP is SAP = (π / 4) * 32 = (9 * π) / 4.
Thus, the ratio of the area of ​​a circle bounded by a smaller circle to the sum of the areas of circles bounded by two other circles is (SKB): (SMB + SAP) = (π / 4): (4 * π + (9 * π) / 4 ) = (π / 4): (25 * π / 4) = 1/25.
Answer: 1/25.



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