2х + Зу = 5 у= 5 - x also need method
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<u>Given</u>–
AB = 12 Cm
CD = 14 Cm
PO = 10 Cm
AP = 1/2 × 12 = 6 Cm
<u>Construction</u>–
Draw NO such that it is the perpendicular bisector of CB.
Hence,
ND = 1/2 × 14 = 7 Cm
To Find,
Measure of NO
<u>Solution</u>,
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.
