two parallel chords 10 cm and 24 cm long are drawn on same side of centre of a circle of radius 13 cm find the distance between chords
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4
Answer:
Drop a perpendicular from the centre and mar the intersecting point E
We know that the perpendicular from the centre bisects the chord.
=> CE = 1/2 CD
=> CE = 1/2 × 24
=> CE = 12 cm
Now, ∆ODC is a right triangle
=> OD² + DC² = OC²
=> OD² + 12² = 13²
=> OD² = 13² - 12²
=> OD² = (13 - 12)(13 + 12)
=> OD² = 25
=> OD = √25
=> OD = 5 cm
Now again, extend the perpendicular such that it bisects AB at F
=> AF = 1/2 AB
=> AF = 5 cm
=> OF² + AF² = OA²
=> OF² + 5² = 13²
=> OF² = 13² - 5²
=> OF² = (13 - 5)(13 + 5)
=> OF² = (8)(18)
=> OF² = 144
=> OF = 12 cm
So distance between them
= OF - OE
= 12 - 5
= 7 cm
Answered by
7
Distance between chords =12−5=7
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