If the length of AB is 10 cm from the center and has a length of 12, find the approximate length from the center to CD.​

Given–

AB = 12 Cm

CD = 14 Cm

PO = 10 Cm

AP = 1/2 × 12 = 6 Cm

Construction–

Draw NO such that it is the perpendicular bisector of CB.

Hence,

ND = 1/2 × 14 = 7 Cm

To Find,

Measure of NO

Solution,

Here, PO Perpendicular to AB

Hence, APO is a right-angled triangle–

AO² = PO² + AP²

Or, AO² = 10 ² + 6 ²

Or, AO² = 36 + 100

Or, AO = √136 Cm

Or, AO = 11.66 Cm

Also, AO = OD = 11.66 Cm

( Radius of the same circle )

Now, in triangle OND,

OD² = ON² + ND ²

Or, 11.66 ² = ON² + 7²

Or, 136 = ON² + 49

Or, ON ² = 136 – 49

Or, ON ² = 87

Or, ON ² = √87

Or, ON = 9.3 Cm

Therefore, the appropriate length from center to CD is 9.3 Cm.

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