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two parallel chords of a circle which are on the same side of the centre subtend angles of 72 degrees and 144 degrees respectively at the centre prove that the perpendicular distance between the chords is half the radius of the circle

two parallel chords of a circle which are on the same side of the centre subtend angles of 72 degrees and 144 degrees respectively at the centre
prove that the perpendicular distance between the chords is half the radius of the circle
 

Grade:10

1 Answers

Himanshu
103 Points
5 years ago
Please draw a figure while reading the following explanation.
Let the two chords be AB and CD of the circle with centre O, where AB>CD.
AOB + OAB + OBA = 180`
144` + OAB + OAB = 180`
OBA=OAB=18`
 
COD + OCD + ODC = 180`
72` + OCD + ODC = 180`
OCD = ODC = 54`
Let OX be the perpendicular to chords which cut AB at M and CD at N.
In triangle OAM, sin18` = OM / radius
                           radius × sin18`= OM               ….i)
In triangle OBN, sin54` = ON / radius
                           radius × sin54` = ON               ….ii)
 
Subtract i) and ii) ,
ON – OM = radius (sin18` – sin54`)
NM = radius / 2 (sin18` – sin54`= 2)
Hence, proved.

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