Question

on the given diagram O' is the centre of the circle and AB is parallel to CD,
AB=24 cm and distance between the chords AB and CD is 17 cm. If the radius
of the circle is 13 cm, find the length of the chord CD.​

Answer 1

The length of the chord CD is 10 cm.

Step-by-step explanation:

Given:

O is the centre of the circle and AB is parallel to CD, such that

AB = 24 cm

PQ = 17 cm

OB = OD = radius = 13 cm

To Find:

CD = ?

Solution:

Theorem :

Perpendicular drawn from the centre to the chord bisect the chord

OP ⊥ AB  ....Given

AP = PB        ...........Theorem

$PB=\dfrac{1}{2}AB$

Substituting the values we get

$PB=\dfrac{1}{2}24=12\ cm$

Now in right triangle OPB , by Pythagoras theorem

$(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}$

Substituting the values we get

$OB^{2}=OP^{2}+PB^{2}\\\\OP^{2}=13^{2}-12^{2}=25\\\\OP=\sqrt{25}=5\ cm$

Now,

PQ = 17 cm    .......Given

$PQ=OP+OQ$ ..........Line Addition Property

Substituting the values we get

$OQ=17-5=12\ cm$

Now in right triangle OPD , by Pythagoras theorem

$OD^{2}=OQ^{2}+QD^{2}$

Substituting the values we get

$QD^{2}=13^{2}-12^{2}=25\\\\QD=\sqrt{25}=5\ cm$

Also

CQ = QD         ............Theorem

$CD =2\times QD=2\times 5 = 10\ cm$

Therefore,

The length of the chord CD is 10 cm.