Question

In a circle of radius $10$ cm, a chord is drawn $6$ cm from the centre. If a chord half the length of the original chord were drawn, its distance in centimeters from the centre would be

• A. $\sqrt{84}$
• B. $9$
• C. $8$
• D. $3\pi$

Answer 1

Answer: (A)

$\sqrt{84}$

Let the circle be with center O and radius $10$ cm. Let there be a chord AB
Draw a perpendicular from O on AB to meet at P. The perpendicular from the center divides the chord in two halves. Given, OP$=6$cm
Thus, In $△OAP$, Using Pythagoras theorem
$O{A}^2=A{P}^2+O{P}^2$
${10}^2=A{P}^2+6^2$
$A{P}^2=64$
$AP=8$ cm
Thus, $AB=2AP=16$ cm
Now, new chord CD is drawn with length $8$ cm. Draw a perpendicular on CD from O to cut CD at N.
Now, In $△ONC$
$O{C}^2=N{C}^2+O{N}^2$
${10}^2=4^2+O{N}^2$
$ON=\sqrt{84}$ cm