In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to

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Solution

We know that, the perpendicular from the centre of a circle to a chord bisects the chord.
$A C = C B = \frac{1}{2} A B = \frac{1}{2} \times 8 = 4 c m g i v e n, O A = 5 c m$

[ 1 mark ]

$A O^{2} = A C^{2} + O C^{2} [u s i n g p y t h a g o r a s t h e o r e m] (5)^{2} = (4)^{2} + O C^{2} 25 = 16 + O C^{2} O C^{2} = 25 - 16 = 9 O C = 3 c m$

[taking positive square root, because length is always positive]
OA = OD [same radius of a circle]
OD = 5 cm
CD = OD - OC = 5 - 3 = 2 cm

[ 1 mark ]

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In Figure, if $ OA=5 cm$, $ AB=8 cm$ and $ OD$ is perpendicular to $ AB$, then $ CD$ is equal to:

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