Question

A chord of a circle is \[12{\text{ }}cm\] in length and its distance from the centre is\[8{\text{ }}cm\]. Find the length of the chord of the same circle which is at a distance of \[6{\text{ }}cm\] from the centre.

A. \[{\text{30 }}cm\]

B. \[{\text{24 }}cm\]

C. \[{\text{16 }}cm\]

D. \[{\text{18 }}cm\]

Answer 1

Hint: To solve this problem first we must measure the circle radius by applying the pythagoras theorem so for it we will assume the\[12{\text{ }}cm\] chord midpoint further after having the radius we will simply evaluate the necessary length of other chord by applying pythagoras theorem again.

Complete step-by-step answer:

Since a \[12{\text{ }}cm\] chord is \[8{\text{ }}cm\] away from the middle of the circle.
Suppose the center of the circle is O, and E is the midpoint of AB. Therefore both AEO and BEO are triangles with right angles.
So AE = Length of AB chord$/2$
and EO = \[8{\text{ }}cm\] ( given)
Let R be the radius of the circle
By applying pythagoras theorem in right angle triangle ∆ AEO we get
$\sqrt {{6^2} + {8^2}} = \sqrt {(36 + 64)} = \sqrt {100} = 10cm$
So the radius of circle is $10cm$
Now the chord CD is \[\;6cm\] from O.
Let us assume F is the midpoint of CD , so CFO and DFO both are right angled triangles.
FO = \[\;6cm\] ( given)
and CO=DO= $10cm$ ( radius of circle)
so CF=DF
By applying the pythagoras theorem in right angle triangle CFO we get
CF = \[ + \]$\sqrt {{{10}^2} - {6^2}} = \sqrt {(100 - 36)} = \sqrt {64} = 8cm$
Therefore the length of chord CD = CF \[ + \] DF = \[2{\text{ }} \times \]CF\[ = {\text{ }}2{\text{ }} \times {\text{ }}8{\text{ }} = {\text{ }}16{\text{ }}cm.\]

Hence the correct answer is option C.

Note: A circle chord is a part of a straight line whose endpoints both lie on the circle. A chord 's infinite line extension is a secant line, or literally secant line. More generally, a chord is a line segment that joins two points , for example an ellipse, at any curve.