Q4 of 15 Page 87

In a circle, two parallel chords of lengths 4 and 6 centimetres are 5 centimetres apart. What is the radius of the circle?


O is the centre of the circle.


EF and GH are two parallel chords 5 cm apart.


EF = 6 cm and GH = 4 cm.


OJ is perpendicular on EF and OI is perpendicular drawn on GH.


IJ = 5 cm


Let, OJ = x cm and OI = (5 – x) cm


Let, radius of the circle = r cm


OE = OG = r cm


In ΔOIG we have,


∠OIG = 90° [∵OI is perpendicular drawn on GH]


IG = 4/2 = 2 cm [∵perpendicular drawn from centre bisects chord]


OI = (5 – x) cm


OG = r cm


…… (1)


In ΔOJE we have,


∠OJE = 90° [∵ OJ is perpendicular on EF]


JE = 6/2 = 3 cm [∵perpendicular drawn from centre bisects chord]


OJ = x cm


OE = r cm


…… (2)


From (1) and (2) we have,



⇒ 4 + 25 – 10x + x2 = 9 + x2


⇒ 29 – 10x = 9


⇒ 10x = 20


∴ x = 2


Putting the value x = 2 in (2) we get,



⇒ r = √13


∴ radius of the circle = √13 cm


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