In the given diagram 'O' is the centre of the circle. Chord AB is parallel to chord
CD. AB = 64 cm, CD = 48 cm and radius of the circle is 40 cm. Find the distance
between the two chords.
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Answer:
The distance between the two chords is 56 cm.
Step-by-step explanation:
consider AB and CD are the chords parallel to each other
AB = 64 cm
CD = 48 cm
radius of the circle = 40 cm
construct OL ⊥ AB and OM⊥ CD
join the diagonals OA and OC
we know that OA = OC= 40 cm
perpendicular from the center of the circle to a chord bisects the chord
therfore
AL =
AL=
AL = 32 cm
similarly,
CM = 24 cm
consider
using pythagoras theorem
OA²= AL²+ OL²
(40)²= (32)²+OL²
OL²= (40)²-(32)² = 1600 - 1024 = 576
OL²= 576
OL = 24 cm
similarly
consider OMC
using again pythagoras theorem
OC²= CM² + OM²
(40)² = (24)² + OM²
OM² = (40)²- (24)²= 1600 - 576 = 1024
OM²= 1024
OM= 32 cm
distance between the chords = OM+OL
= 24 + 32 cm
= 56 cm
hence,
The distance between the two chords is 56 cm.
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