Question : In a circle of radius 8 units, a chord of length 10 units is drawn. What is the perpendicular distance of the chord from the centre of the circle in units?
Option 1: $\sqrt{39}$ Units
Option 2: $\sqrt{35}$ Units
Option 3: $\sqrt{30}$ Units
Option 4: $\sqrt{33}$ Units
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Correct Answer: $\sqrt{39}$ Units
Solution : Let O be the centre of the circle of radius 8 units. Let AB be the chord of length 10 units. Let OD be perpendicular from the centre to the chord AB. We know the perpendicular from the centre to a chord bisects the chord. So, AD = BD = 5 units Applying Pythagoras theorem in $\triangle$OBD, OB 2 = OD 2 + DB 2 ⇒ 8 2 = OD 2 + 5 2 ⇒ OD 2 = 64 – 25 $\therefore$ OD = $\sqrt{39}$ units Hence, the correct answer is $\sqrt{39}$ units.
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