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question no 5
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Given chords AB=6 cm, CD =12 cm and AB||CD
Draw OP⊥ AB. Let it intersect CD at Q and AB at P
∴ AP = PB = 3 cm and CQ = DQ = 6 cm [ perpendicular draw from the centre of the chord bisects the chord]
Let OD = OB = r
In right ΔOQD, r2 = x2 + 62 [By
r2 = x2 + 36 → (1)
In right ΔOPB, r2 = (x + 3)2 + 32 [By Pythagoras theorem]
Þ r2 = x2 + 6x + 9 + 9 = x2 + 6x + 18 → (2)
(1) and (2) we get
x2 + 36 = x2 + 6x + 18
⇒ 6x = 18
∴ x = 3
Put x = 3 in (1), we get
r2 = 32 + 36 = 9 + 36 = 45
∴ r = √45 = 3√5 cm
Draw OP⊥ AB. Let it intersect CD at Q and AB at P
∴ AP = PB = 3 cm and CQ = DQ = 6 cm [ perpendicular draw from the centre of the chord bisects the chord]
Let OD = OB = r
In right ΔOQD, r2 = x2 + 62 [By
r2 = x2 + 36 → (1)
In right ΔOPB, r2 = (x + 3)2 + 32 [By Pythagoras theorem]
Þ r2 = x2 + 6x + 9 + 9 = x2 + 6x + 18 → (2)
(1) and (2) we get
x2 + 36 = x2 + 6x + 18
⇒ 6x = 18
∴ x = 3
Put x = 3 in (1), we get
r2 = 32 + 36 = 9 + 36 = 45
∴ r = √45 = 3√5 cm
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