Mathematics
In the adjoining figure, a chord PQ of a circle with center O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of :
(i) PQ
(ii) AP
(iii) BP.
Circles
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Answer
(i) Given,
AB bisects PQ.
∴ OM bisects PQ.
Since, the straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is perpendicular to the chord,
∴ OM ⊥ PQ.
In right △OMP,
⇒ OP2 = OM2 + PM2 (By pythagoras theorem)
⇒ 152 = 92 + PM2
⇒ PM2 = 152 - 92
⇒ PM2 = 225 - 81
⇒ PM2 = 144
⇒ PM = = 12 cm.
PQ = 2PM = 24 cm.
Hence, PQ = 24 cm.
(ii) From figure,
AM = AO + OM = 15 + 9 = 24 cm.
In right △APM,
⇒ AP2 = AM2 + PM2 (By pythagoras theorem)
⇒ AP2 = 242 + 122
⇒ AP2 = 576 + 144
⇒ AP2 = 720
⇒ AP = = cm.
Hence, AP = cm.
(iii) From figure,
MB = OB - OM = 15 - 9 = 6 cm.
In right △MPB,
⇒ BP2 = PM2 + MB2 (By pythagoras theorem)
⇒ BP2 = 122 + 62
⇒ BP2 = 144 + 36
⇒ BP2 = 180
⇒ BP = = cm.
Hence, BP = cm.
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