find the length of the chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm
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Answered by
9
P be the mid point of AB, So
AP = PB
OP = 12 cm = distance of chord from center
And
AO = 13 cm = Radius
Now, OPA forms a right angle triangle.
Now, by pythagorous theorem,
AO2 = OP2 + AP2
AP2 = AO2 - OP2
AP2 = 132 - 122
AP2 = 169 - 144
AP2 = 25
AP = 5 cm.
Now,
AB = AP + PB = AP + AP = 5 +5 = 10 cm.
Hope it helps! ^^
Answered by
358
AB is chord of a circle with center O and OA is its radius OM ⊥ AB
Therefore,
OA = 13 cm, OM = 12 cm
Now from right angled triangle OAM,
OA2 = OM2 + AM2 by using Pythagoras theorem,
132 = 122 + AM2
AM2 = 132 – 122
AM2 = 169 – 144
AM2 = 25
AM = 52
We know that OM perpendicular to AB
Therefore, M is the midpoint of AB
AB = 2 AM
AB = 2 (5)
AB = 10 cm
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