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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