Side AB of rectangle ABCD is a chord of the circle that touches side CD. The extension of side Cb

| August 15, 2021 | education

Side AB of rectangle ABCD is a chord of the circle that touches side CD. The extension of side Cb meets the circle at point K. Chord AP meets BK at point M, with AM = 15, MK = 7, PM = 4.2. Find the side of the rectangle CB.

AR and BK are intersecting chords.
According to the sv-wu of intersecting chords: AM · PM = BM · KM ⇒ BM = AM · PM: KM = 15 4.2: 7 = 9.
В ΔАВМ AB² = АМ² – ВМ² = 15² – 9² = 144,
AB = 12.
В ΔАВК BК = BM + КМ = 9 + 7 = 16.
AK = √ (AB² + BK²) = √ (12² + 16²) = 20.
The center of the circle, point O divides the diagonal in half: OK = AK: 2 = 10.
OE is the radius of the circle, OE = OK = 10.
Let’s draw a perpendicular OH to the BK side. BH = BK: 2 = 16: 2 = 8.
OK = OE = 10.
In a rectangle OHCE HC = OE.
ВС = ВН + НС = 8 + 10 = 18.
Answer: 18.



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