Solution:

Draw seg OA.

Now, AB = 12 cm ... (Given)

∴ \(\displaystyle \text {AP} = \frac{1}{2} \times \text{AB}\)
... (Perpendicular drawn from the centre of a circle to the chord bisects the chord)

∴ \(\displaystyle \text {AP} = \frac{1}{2} \times 12\)

∴ \(\displaystyle \text {AP} = 6\) cm ... (i)

Now, in right \(\triangle \text{OPA}\),

 \(\displaystyle \text {OA}^2 = \text {OP}^2 + \text {AP}^2\) ... (Pythagoras’ Theorem)

∴ \(\displaystyle \text {OA}^2 = 8^2 + 6^2\)

∴ \(\displaystyle \text {OA}^2 = 64 + 36\)

∴ \(\displaystyle \text {OA}^2 = 100\)

∴ \(\displaystyle \text {OA} = \sqrt{100}\)

∴ \(\displaystyle \text {OA} = 10\)

∴ r = 10 cm ... (ii)

∴ d = 2r = 2 × 10 = 20 cm... (iii)

∴ The diameter of the circle is 20 cm.

Solution:

 d = 26 cm ... (Given)

 r = \(\displaystyle \frac{d}{2}\) = \(\displaystyle \frac{26}{2}\) = 13 cm ... (i)

Now, OP \(\perp\) AB

∴ \(\displaystyle \text {AP} = \frac{1}{2} \times \text{AB}\)
... (Perpendicular drawn from the centre of a circle to the chord bisects the chord)

But, AB = 24 cm ... (Given)

 AP = \(\displaystyle \frac{AB}{2}\) = \(\displaystyle \frac{24}{2}\) = 12 cm ... (ii)

Now, in right \(\triangle \text{OPA}\),

 \(\displaystyle \text {OP}^2 + \text {AP}^2 = \text {OA}^2\) ... (Pythagoras’ Theorem)

∴ \(\displaystyle \text {OP}^2 + 12^2 = 13^2\)

∴ \(\displaystyle \text {OP}^2 + 144 = 169\)

∴ \(\displaystyle \text {OP}^2 = 169 - 144\)

∴ \(\displaystyle \text {OP}^2 = 25\)

∴ \(\displaystyle \text {OP} = \sqrt{25}\)

∴ \(\displaystyle \text {OP} = 5\)  cm... (iii)

∴ The distance of the chord from the centre is 5 cm.

Solution:

 r = 34 cm ... (Given)

 OP = 30 cm ... (Given)

Now, OP \(\perp\) AB

∴ \(\displaystyle \text {AP} = \frac{1}{2} \times \text{AB}\)
... (Perpendicular drawn from the centre of a circle to the chord bisects the chord)

Now, in right \(\triangle \text{OPA}\),

 \(\displaystyle \text {OP}^2 + \text {AP}^2 = \text {OA}^2\) ... (Pythagoras’ Theorem)

∴ \(\displaystyle 30^2 + \text {AP}^2 = 34^2\)

∴ \(\displaystyle 900 + \text {AP}^2 = 1156\)

∴ \(\displaystyle \text {AP}^2 = 1156 - 900\)

∴ \(\displaystyle \text {AP}^2 = 256\)

∴ \(\displaystyle \text {AP} = \sqrt{256}\)

∴ \(\displaystyle \text {AP} = 16\)  cm... (i)

∴ AB = 2 × AP = 2 × 16 = 32 cm ... (ii)

∴ The length of the chord is 32 cm.

Solution:

 PQ = 80 units ... (Given)

And OM \(\perp\) PQ ... (Given)

∴ \(\displaystyle \text{PM} = \frac{1}{2} \times \text{PQ}\)

... (Perpendicular drawn from the centre of a circle to the chord bisects the chord)

∴ \(\displaystyle \text{PM} = \displaystyle \frac{1}{2} \times 80\)

∴ \(\displaystyle \text{PM} = \displaystyle \text{ 40 units}\) ... (i)

Now, in right \(\triangle \text{OMP}\),

 \(\displaystyle \text {OP}^2 = \text {OM}^2 + \text {PM}^2\) ... (Pythagoras’ Theorem)

∴ \(\displaystyle 41^2 = 40^2 + \text {OM}^2\)

∴ \(\displaystyle 1681 = 1600 + \text {OM}^2\)

∴ \(\displaystyle 1681 - 1600 = \text {OM}^2\)

∴ \(\displaystyle 81 = \text {OM}^2\)

i.e. \(\displaystyle \text {OM}^2 = 81\)

∴ \(\displaystyle \text {OM} = \sqrt{81}\)

∴ \(\displaystyle \text {OM} = 9\)  units... (ii)

∴ The distance of the chord from the centre of the circle is 9 units.

Draw OM ⊥ AB.

A perpendicular drawn from the centre of a circle on its chord, bisects the chord. ... (Theorem)

∴ In the larger circle, OM bisects the chord AB.

∴ \(\displaystyle \text{AM} = \text{BM}\) ... (i)

and in the smaller circle, OM bisects the chord PQ.

∴ \(\displaystyle \text{PM} = \text{QM}\) ... (ii)

∴ Subtracting (ii) from (i),

 \(\displaystyle \text{AM} - \text{PM} = \text{BM} - \text{QM}\)

∴ \(\displaystyle \text{AP} = \text{BQ}\)

Proof:

Given:

AB is a diameter of a circle with centre O.
Diameter AB bisects chords PQ and MN in points X and Y respectively.

To Prove:

Chord PQ \(\text{| |}\) Chord MN

Proof:

X is the midpoint of chord PQ. ... (Given)
Diameter AB bisects chord PQ at point X.
∴ seg OX \(\perp\) chord PQ ... (Theorem)
∴ \(\angle \text{OXP} = 90^\circ\) ... (i)

Also, Y is the midpoint of chord MN. ... (Given)
∴ seg OY \(\perp\) chord MN ... (Theorem)
∴ \(\angle \text{OYM} = 90^\circ\) ... (ii)

Adding (i) and (ii):
 \(\angle \text{OXP} + \angle \text{OYM} = 90^\circ + 90^\circ\)
∴ \(\angle \text{OXP} + \angle \text{OYM} = 180^\circ\)
∴ Chord PQ \(\text{| |}\) Chord MN ... (Interior angles test)

∴ If a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.