In a circle , centre at o and radius 5cm , AB and CD are two parallel chords ,AB=8cm , CD=6cm . Then the length of the chord AC is
Answers
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Given : radius 5 cm AB and CD are two parallel chords AB=8 cm, CD= 6 cm.
To find : Length of Chord AC
Solution:
Perpendicular bisector of chord meets at center of Circle
AB ║ CD
Let say MN ⊥ AB & CD passing through center
M mid point of AB & N mid point of CD
AM = BM = AB/2 = 8/2 = 4 cm
CN = DN = CD/2 = 6/2 = 3 cm
OA = OB = OC = OD = Radius = 5 cm
∠CON = Sin⁻¹ (3/5)
∠AOM = Sin⁻¹ (4/5)
∠CON + ∠AOM = Sin⁻¹ (3/5) + Sin⁻¹ (4/5) = π/2
∠CON + ∠AOM + ∠AOC = π
=> ∠AOC = π - π/2 = π/2
Now join AC
Draw OP ⊥ AC
∠COP = ∠AOP = ∠AOC/2 = π/4
Sinπ/4 = CP/OC
=> 1/√2 = CP/5
=> CP = 5/√2
AC = 2 * CP = 2 * 5/√2 = 5√2
length of the chord AC is 5√2
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