two concentric circles are of radii 26cm and 10cm. find the length of the chord of the larger circle which touches the smaller circle
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Answer:
let r be 26 cm
and R be 10 cm
A is centre of both circles
AB perpendicular to DC
as radius from centre is perpendicular to the tangent at the point of contact
Thus In ∆ ABC
R^2 + BC ^2 = r^2
BC^2 = 26^2 - 10^2
BC^2 = 676-100
BC^2 = 576
BC = √576
BC = 24 cm
Similarly, DB = BC
Thus DC = 2BC
DC = 24×2
DC = 48 cm = Chord length
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Answer:
length of half chord = √26^2 -10^2
=√576 =24
therefore length of full chord = 2 × 24=48cm
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