Circle

From figure,

AB is the chord which is at a distance 12 cm so,

OC = 12 cm and OA = radius = 13 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 13² = 12² + AC²

⇒ AC² = 13² - 12²

⇒ AC² = 169 - 144 = 25

⇒ AC = $\sqrt{25}$ = 5 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = 5 cm.

AB = AC + CB = 5 + 5 = 10 cm.

Hence, length of chord = 10 cm.

From figure,

AB is the chord and radius = OA = 25 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = $\frac{48}{2}$ = 24 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 25² = OC² + 24²

⇒ OC² = 25² - 24²

⇒ OC² = 625 - 576 = 49

⇒ OC = $\sqrt{49}$ = 7 cm.

Hence, chord is at a distance of 7 cm from the center.

From figure,

AB is the chord which is at a distance 3 cm from the center so OC = 3 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = $\frac{8}{2}$ = 4 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ OA² = 3² + 4²

⇒ OA² = 9 + 16

⇒ OA² = 25

⇒ OA = $\sqrt{25}$ = 5 cm.

Hence, radius = 5 cm.

Diameter = 20 cm,

∴ Radius = $\frac{\text{Diameter}}{2}$ = 10 cm.

From figure,

AB is the chord which is at a distance 6 cm so,

OC = 6 cm and OA = radius = 10 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 10² = 6² + AC²

⇒ AC² = 10² - 6²

⇒ AC² = 100 - 36 = 64

⇒ AC = $\sqrt{64}$ = 8 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = 8 cm.

AB = AC + CB = 8 + 8 = 16 cm.

Hence, length of chord = 16 cm.

From figure,

AB is the chord which is at a distance 6 cm from the center so OC = 6 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = $\frac{16}{2}$ = 8 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ OA² = 6² + 8²

⇒ OA² = 36 + 64

⇒ OA² = 100

⇒ OA = $\sqrt{100}$ = 10 cm.

Radius = 10 cm,

∴ OD = 10 cm.

From figure,

In right angle triangle ODF,

⇒ OD² = OF² + DF² (By pythagoras theorem)

⇒ DF² = OD² - OF²

⇒ DF² = 10² - 8²

⇒ DF² = 100 - 64 = 36

⇒ DF = $\sqrt{36}$ = 6 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord

∴ DE = DF + FE = 6 cm + 6 cm = 12 cm.

Hence, the length of chord which is at a distance of 8 cm from the center of the circle = 12 cm.

(i) From figure,

AB and CD are chords of length 8 cm and 6 cm, respectively.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AE = BE = $\frac{8}{2}$ = 4 cm and,

CF = FD = $\frac{6}{2}$ = 3 cm.

Radius = OA = OC = 5 cm.

In right angle triangle OAE,

⇒ OA² = AE² + OE² (By pythagoras theorem)

⇒ OE² = OA² - AE²

⇒ OE² = 5² - 4²

⇒ OE² = 25 - 16 = 9

⇒ OE = $\sqrt{9}$ = 3 cm.

In right angle triangle OCF,

⇒ OC² = CF² + OF² (By pythagoras theorem)

⇒ OF² = OC² - CF²

⇒ OF² = 5² - 3²

⇒ OF² = 25 - 9 = 16

⇒ OF = $\sqrt{16}$ = 4 cm.

Distance between two chords = EF = OF - OE = 4 - 3 = 1 cm.

Hence, distance between two chords = 1 cm.

(ii) From figure,

AB and CD are chords of length 8 cm and 6 cm respectively.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AE = BE = $\frac{8}{2}$ = 4 cm and,

CF = FD = $\frac{6}{2}$ = 3 cm.

Radius = OA = OC = 5 cm.

In right angle triangle OAE,

⇒ OA² = AE² + OE² (By pythagoras theorem)

⇒ OE² = OA² - AE²

⇒ OE² = 5² - 4²

⇒ OE² = 25 - 16 = 9

⇒ OE = $\sqrt{9}$ = 3 cm.

In right angle triangle OCF,

⇒ OC² = CF² + OF² (By pythagoras theorem)

⇒ OF² = OC² - CF²

⇒ OF² = 5² - 3²

⇒ OF² = 25 - 9 = 16

⇒ OF = $\sqrt{16}$ = 4 cm.

Distance between two chords = EF = OF + OE = 4 + 3 = 7 cm.

Hence, distance between two chords = 7 cm.

(i) Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AM = BM = $\frac{24}{2}$ = 12 cm.

In right angle triangle OAM,

⇒ OA² = OM² + AM² (By pythagoras theorem)

⇒ OA² = 5² + 12²

⇒ OA² = 25 + 144

⇒ OA² = 169

⇒ OA = $\sqrt{169}$ = 13 cm.

Hence, radius = 13 cm.

(ii) From figure,

OC = radius = 13 cm.

In right angle triangle OCN,

⇒ OC² = ON² + CN² (By pythagoras theorem)

⇒ CN² = OC² - ON²

⇒ CN² = 13² - 12²

⇒ CN² = 169 - 144 = 25

⇒ CN = $\sqrt{25}$ = 5 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ ND = CN = 5 cm.

CD = CN + ND = 10 cm.

Hence, CD = 10 cm.

Given,

AB = 8 cm, EC = 3 cm

Let radius OB = OC = r

OE = (r - 3) cm.

Since, chord AB is bisected by OE so OE ⊥ AB (As straight line drawn from the centre of circle to bisect a chord, is perpendicular to it.)

Now in right ∆OBE,

⇒ OB² = BE² + OE²

⇒ r² = 4² + (r – 3)²

⇒ r² = 16 + r² – 6r + 9

⇒ r² - r² + 6r = 16 + 9

⇒ 6r = 25

⇒ r = $\frac{25}{6} = 4\frac{1}{6}$ cm.

Hence, radius = $4\frac{1}{6}$ cm.

In the figure, chords AB ∥ CD and O is the centre of the circle.

Radius of the circle = 15 cm

Length of AB = 24 cm and CD = 18 cm.

Join OA and OC.

AB = 24 cm and OM ⊥ AB.

∴ AM = MB = $\frac{24}{2}$ = 12 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle OAM,

⇒ OA² = OM² + AM² (By pythagoras theorem)

⇒ OM² = OA² - AM²

⇒ OM² = 15² - 12²

⇒ OM² = 225 - 144

⇒ OM² = 81

⇒ OM = $\sqrt{81}$ = 9 cm.

Similarly ON ⊥ CD

CN = ND = $\frac{18}{2}$ = 9 cm

Similarly In right ∆CNO,

⇒ OC² = ON² + CN² (By pythagoras theorem)

⇒ ON² = OC² - CN²

⇒ ON² = 15² - 9²

⇒ ON² = 225 - 81

⇒ ON² = 144

⇒ ON = $\sqrt{144}$ = 12 cm.

MN = OM + ON = 9 + 12 = 21 cm.

Hence, MN = 21 cm.

Let OE = x cm.

From figure,

In right angle triangle OCF,

⇒ OC² = OF² + CF² (By pythagoras theorem)

⇒ OC² = (x + 3)² + 2²

⇒ OC² = x² + 9 + 6x + 4

⇒ OC² = x² + 6x + 13

Since, radius = OA = OC.

∴ OA² = OC² = x² + 6x + 13.

In right angle triangle OAE,

⇒ OA² = OE² + AE²

⇒ x² + 6x + 13 = x² + 5²

⇒ x² - x² + 6x = 25 - 13

⇒ 6x = 12

⇒ x = $\frac{12}{6}$ = 2 cm.

⇒ OC² = x² + 6x + 13

⇒ OC² = 2² + 6(2) + 13

⇒ OC² = 4 + 12 + 13

⇒ OC² = 29

⇒ OC = $\sqrt{29}$ cm.

Diameter = 2 × radius = 2 × $\sqrt{29} = 2\sqrt{29}$ cm.

Hence, diameter = $2\sqrt{29}$ cm.

From figure,

OA = radius = r cm.

BD = DC = $\frac{24}{2}$ = 12 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle ABD,

$$\begin{aligned}\Rightarrow AB^2 = BD^2 + AD^2 \\[1em] \Rightarrow (12\sqrt{5})^2 = 12^2 + AD^2 \\[1em] \Rightarrow 720 = 144 + AD^2 \\[1em] \Rightarrow AD^2 = 720 - 144 \\[1em] \Rightarrow AD^2 = 576 \\[1em] \Rightarrow AD = \sqrt{576} = 24 \text{ cm}.\end{aligned}$$

OD = AD - OA = (24 - r) cm.

In right angle triangle OBD,

⇒ OB = radius = r cm.

⇒ OB² = OD² + BD²

⇒ r² = (24 - r)² + 12²

⇒ r² = 576 + r² - 48r + 144

⇒ r² - r² + 48r = 720

⇒ 48r = 720

⇒ r = $\frac{720}{48}$ = 15 cm.

Hence, radius = 15 cm.

From figure,

OA = radius = r cm.

BD = DC = $\frac{6}{2}$ = 3 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle ABD,

⇒ AB² = AD² + BD²

⇒ 6² = AD² + 3²

⇒ AD² = 36 - 9

⇒ AD² = 27

⇒ AD = $\sqrt{27}$ = $3\sqrt{3}$ cm.

OD = AD - AO = $(3\sqrt{3} - r)$ cm.

In right angle triangle OBD,

⇒ OB = radius = r cm.

⇒ OB² = OD² + BD²

⇒ r² = $(3\sqrt{3} - r)$² + 3²

⇒ r² = 27 + r² - $6\sqrt{3}$r + 9

⇒ r² - r² + $6\sqrt{3}$r = 36

⇒ $6\sqrt{3}$r = 36

⇒ r = $\frac{36}{6\sqrt{3}} = 2\sqrt{3}$ cm.

Hence, radius = $2\sqrt{3}$ cm.

Given,

AM = 18 cm and MB = 8 cm

From figure,

AB = AM + MB = 18 + 8 = 26 cm

Radius of the circle = $\frac{26}{2}$ = 13 cm.

Let CD is the shortest chord drawn through M.

∴ CD ⊥ AB

From figure,

OM = AM - AO = 18 - 13 = 5 cm

OC (radius) = OA = 13 cm

Now in right ∆OMC,

OC² = OM² + MC² (By pythagoras theorem)

13² = 5² + MC²

MC² = 13² - 5²

MC² = 169 - 25 = 144

MC = $\sqrt{144}$ = 12 cm.

M is Mid-Point of CD (As perpendicular from center to the chord bisects it.)

CD = 2 × MC = 2 × 12 = 24 cm.

Hence, length of shortest chord = 24 cm.

Let side BC = 4 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ BM = MC = 2 cm.

Given,

Diameter = 5 cm, radius = $\frac{5}{2}$ = 2.5 cm.

In right ∆OBM,

⇒ OB² = BM² + OM²

⇒ (2.5)² = 2² + OM²

⇒ 6.25 = 4 + OM²

⇒ OM² = 2.25

⇒ OM = $\sqrt{2.25}$ = 1.5 cm.

Similarly in right ∆OAN,

⇒ OA² = AN² + ON²

⇒ (2.5)² = 2² + ON²

⇒ 6.25 = 4 + ON²

⇒ ON² = 2.25

⇒ ON = $\sqrt{2.25}$ = 1.5 cm.

From figure,

⇒ MN = OM + ON = 1.5 + 1.5 = 3 cm.

⇒ AB = DC = MN = 3 cm.

⇒ Area = length × breadth = 4 cm × 3 cm = 12 cm².

Hence area of rectangle = 12 cm².

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AC = CB = $\frac{30}{2}$ = 15 cm.

From figure,

In right triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 25² = OC² + 15²

⇒ 625 = OC² + 225

⇒ OC² = 400

⇒ OC = $\sqrt{400}$ = 20 cm.

In right triangle O'AC,

⇒ O'A² = O'C² + AC² (By pythagoras theorem)

⇒ 17² = O'C² + 15²

⇒ 289 = O'C² + 225

⇒ O'C² = 64

⇒ O'C = $\sqrt{64}$ = 8 cm.

Distance between centers = OO' = OC + O'C = 20 + 8 = 28 cm.

Hence, distance between their centres = 28 cm.

In the figure, AB and CD are the two chords of a circle with center O. M and N are mid-points of AB and CD, respectively and MN is the line joining the mid-points of two chords and passing through center O.

Since, the straight line drawn from the centre of a circle to bisect a chord is perpendicular to the chord,

∴ OM ⊥ AB and ON ⊥ CD.

So,

∠OMA = ∠OMB = 90° and ∠ONC = ∠OND = 90°

Since, ∠OMA = ∠OND = 90° (Alternate angles) and,

∠OMB = ∠ONC = 90° (Alternate angles)

Hence, proved that AB || CD.

Since, AB || CD and ∠OMA = ∠OMB = 90°

From figure,

∠OMA = ∠OND = 90° (Alternate angles are equal)

∠OMB = ∠ONC = 90° (Alternate angles are equal)

∴ ON ⊥ CD or MN ⊥ CD

We know that,

The perpendicular to a chord from the center of the circle bisects the chord.

∴ NC = ND.

Hence, proved that diameter is perpendicular to other chord and bisects it.

From figure,

AD, BE and CF are medians of the triangle.

Let G be the centroid of triangle ABC.

Triangle ABC is an equilateral triangle,

∴ AB = BC = CA and ∠ABC = ∠BAC = ∠BCA = 60°

In △BFC and △BEC,

⇒ BC = BC (Common Side)

⇒ ∠FBC = ∠ECB = 60°.

⇒ BF = EC (As F is mid-point of AB and E is mid-point of AC and AB = AC.)

△BFC ≅ △BEC (By SAS axiom.)

∴ BE = CF (By C.P.C.T.) .........(1)

Now, in △ABE and △ABD

AB = AB (Common Side)

∠BAE = ∠ABD = 60°

BD = AE (As D is mid-point of BC and E is mid-point of AC and BC = AC.)

△ABE ≅ △ABD (By SAS axiom.)

∴ BE = AD (By C.P.C.T.) ............. (2)

From equation 1 and 2, we get:

⇒ AD = BE = CF

⇒ $\frac{2}{3}AD = \frac{2}{3}BE = \frac{2}{3}CF$

We know that G (the centroid) of the triangle divides the median in a 2 : 1 ratio.

∴ GA = GB = GC.

So, we can say that G is equidistant from the three vertices A. B and C.

G is circumcentre of ΔABC.

Hence, proved that the centroid and circumcentre are coincident.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AD = DB

We can say that D is mid-point of AB.

Since, BC is diameter and O is center so, OB = OC = radius.

We can say that O is mid-point of BC.

In △ABC,

Since, D is mid-point of AB and O is mid-point of BC

By mid-point theorem,

⇒ OD || AC and OD = $\frac{1}{2}AC$

⇒ AC = 2OD.

Hence, proved that CA = 2OD.

Let AB = AC = x

Given,

OM ⊥ AB and ON ⊥ AC

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AM = MB = $\frac{x}{2}$

and

AN = NC = $\frac{x}{2}$

∴ MB = NC ..........(1)

Since, equal chords of a circle are equidistant from the centre,

∴ ON = OM = y (let).

Let radius of circle be r.

From figure,

OQ = OP = r

QN = OQ - ON = r - y

PM = OP - OM = r - y

∴ QN = PM ..........(2)

In △QNC and △PMB,

NC = MB [From (1)]

QN = PM [From (2)]

∠QNC = ∠PMB (Both equal to 90°)

△QNC ≅ △PMB by SAS axiom.

∴ PB = QC (By C.P.C.T.)

Hence, proved that PB = QC.

Draw OM perpendicular to BC.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

So, in the smaller circle, M is mid-point of BC so,

BM = MC = x (let)

Similarly, in larger circle M is mid-point of AD so,

AM = MD = y (let)

From figure,

AB = AM - BM = (y - x)

CD = MD - MC = (y - x)

∴ AB = CD.

Hence, proved that AB = CD.

Draw perpendiculars from O to AB and CD.

In △OME and △ONE,

∠OME = ∠ONE = 90°

∠OEM = ∠OEN (As OE bisects ∠AED)

OE = OE (Common side)

∴ △OME ≅ △ONE (By A.A.S. axiom)

∴ OM = ON (By C.P.C.T.C.)

Hence, chords AB and CD are equidistant from the center of circle.

In the same circle, chords equidistant from the centre are equal.

∴ AB = CD.

Hence, proved that AB = CD.

Draw OM ⊥ AB and ON ⊥ CD,

In △OAM and △ODN,

OA = OD (Radius of circle)

∠AOM = ∠DON (Vertically opposite angles are equal)

∠OMA = ∠OND (Both equal to 90°)

∴ △OAM ≅ △ODN (By A.S.A. axiom).

∴ ND = AM (By C.P.C.T.) ........(1)

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

so, equation 1 can be written as,

⇒ $\frac{\text{CD}}{2} = \frac{\text{AB}}{2}$

⇒ AB = CD.

Hence, proved that AB = CD.

Draw ON ⊥ CD and OM ⊥ AB. Join OE.

(i) Since, equal chords are equidistant from the center of the circle,

∴ ON = OM.

In △ONE and △OME,

ON = OM

∠ONE = ∠OME (Both equal to 90°)

OE = OE (Common side)

∴ △ONE ≅ △OME (By R.H.S. congruence rule).

∴ NE = ME = y (let) (By C.P.C.T.) ........(1)

Let AB = CD = x.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CN = ND = $\frac{CD}{2} = \frac{x}{2}$ and

AM = MB = $\frac{AB}{2} = \frac{x}{2}$.

From figure,

AE = AM + ME = $\frac{x}{2} + y$

CE = CN + NE = $\frac{x}{2} + y$

Hence, proved that AE = CE.

(ii) From figure,

BE = ME - MB = $y - \frac{x}{2}$

DE = NE - ND = $y - \frac{x}{2}$

Hence, proved that BE = DE.

arc APB = arc CQD (Given)

If two arcs are equal then chords are also equal.

∴ AB = CD = x (let).

Ratio = $\frac{AB}{CD} = \frac{x}{x} = \frac{1}{1}$

Hence, AB : CD = 1 : 1.

The figure of the circle is shown below:

Given,

OC bisects ∠AOB.

∴ ∠AOC = ∠BOC

Since, equal arcs subtend equal angles at center.

∴ AC = BC

∴ C is mid-point of AB.

Hence, proved that OC bisects the arc AB.

The figure of the circle is shown below:

Let C be the mid-point of arc AB.

∴ AC = BC.

Since, equal arcs subtend equal angles at center.

∴ ∠AOC = ∠BOC.

Hence, proved that ∠AOC = ∠BOC.

Given AB = CD

Since, in a circle, equal chords cut off equal arcs.

∴ arc AB = arc CD

Subtracting arc BD from both sides we get,

⇒ arc AB - arc BD = arc CD - arc BD

⇒ arc AD = arc CB.

Hence, proved that arc AD = arc CB.

If P and Q are any two points on a circle, then the line segment PQ is called a chord.

Hence, Option 3 is the correct option.

If P is a point on circle and O is center then OP = r but since, P is in the interior of the circle, so it will be smaller than the radius.

OP < r

Hence, Option 4 is the correct option.

Circumference of circle = 2πr

Since, π = $\frac{22}{7}$ and radius is always a positive no.

∴ Circumference of a circle must be a positive real number.

Hence, Option 1 is the correct option.

Let OM be the distance of AB from the center of the circle.

Since diameter = 34 cm, so radius = $\frac{34}{2}$ = 17 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AM = MB = 15 cm.

In right triangle OAM,

⇒ OA² = AM² + OM² (By pythagoras theorem)

⇒ OM² = OA² - AM²

⇒ OM² = 17² - 15²

⇒ OM² = 289 - 225

⇒ OM² = 64

⇒ OM = $\sqrt{64}$ = 8 cm.

Hence, Option 4 is the correct option.

Let r be the radius of the circle.

From figure,

AC = AO + OC = r + r = 2r.

In right △ABC,

⇒ AC² = AB² + BC²

⇒ (2r)² = 12² + 16²

⇒ 4r² = 144 + 256

⇒ 4r² = 400

⇒ r² = $\frac{400}{4}$ = 100

⇒ r = $\sqrt{100}$ = 10 cm.

Hence, Option 3 is the correct option.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AC = CB = $\frac{8}{2}$ = 4 cm.

In right △OAC,

⇒ OA² = OC² + AC²

⇒ 5² = OC² + 4²

⇒ OC² = 5² - 4²

⇒ OC² = 25 - 16

⇒ OC² = 9

⇒ OC = $\sqrt{9}$ = 3 cm.

From figure,

CD = OD - OC = 5 - 3 = 2 cm.

Hence, Option 1 is the correct option.

A secant is defined as a line that intersects a circle at exactly two distinct points.

∴ Statement 1 is true.

A radius is a line segment connecting the center of a circle to any point on its circumference.

By definition, all points on the circumference are equidistant from the center, meaning all radii have the same length.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.

We know that,

Chords of equal length are equidistant from centre.

∴ Assertion (A) is true.

Also,

The chords which are equidistant from the centre of a circle are equal in length.

∴ Reason (R) is true.

Assertion is just the converse of reason.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

We know that,

Arcs of equal lengths subtends equal angles at the center of the circle.

∴ Assertion (A) is true.

We know that,

The chords of equal length cut off equal arcs.

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

Given,

AB is the diameter of the circle with center O and, Chord AC = Chord AD.

We know that in a circle, equal chords subtend equal arcs.

⇒ arc AC = arc AD = x (let) ........................(1)

∴ Reason (R) is true.

AB is a diameter of the circle.

As we know that the diameter of a circle divides the circle into two equal arcs.

⇒ arc ACB = arc ADB.

⇒ arc AC + arc BC = arc AD + arc DB.

From equation (1), we get :

⇒ x + arc BC = x + arc DB.

⇒ arc BC = arc DB.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

(i) Given,

AB bisects PQ.

∴ OM bisects PQ.

Since, the straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is perpendicular to the chord,

∴ OM ⊥ PQ.

In right △OMP,

⇒ OP² = OM² + PM² (By pythagoras theorem)

⇒ 15² = 9² + PM²

⇒ PM² = 15² - 9²

⇒ PM² = 225 - 81

⇒ PM² = 144

⇒ PM = $\sqrt{144}$ = 12 cm.

PQ = 2PM = 24 cm.

Hence, PQ = 24 cm.

(ii) From figure,

AM = AO + OM = 15 + 9 = 24 cm.

In right △APM,

⇒ AP² = AM² + PM² (By pythagoras theorem)

⇒ AP² = 24² + 12²

⇒ AP² = 576 + 144

⇒ AP² = 720

⇒ AP = $\sqrt{720}$ = $12\sqrt{5}$ cm.

Hence, AP = $12\sqrt{5}$ cm.

(iii) From figure,

MB = OB - OM = 15 - 9 = 6 cm.

In right △MPB,

⇒ BP² = PM² + MB² (By pythagoras theorem)

⇒ BP² = 12² + 6²

⇒ BP² = 144 + 36

⇒ BP² = 180

⇒ BP = $\sqrt{180}$ = $6\sqrt{5}$ cm.

Hence, BP = $6\sqrt{5}$ cm.

Draw OM ⊥ QR.

In right △OQM,

⇒ OQ² = OM² + QM²

⇒ 10² = OM² + 6²

⇒ OM² = 100 - 36

⇒ OM² = 64

⇒ OM = $\sqrt{64}$ = 8 cm.

In right △POM,

⇒ PO² = OM² + PM²

⇒ 17² = 8² + PM²

⇒ PM² = 289 - 64

⇒ PM² = 225

⇒ PM = $\sqrt{225}$ = 15 cm.

From figure,

PQ = PM - QM = 15 - 6 = 9 cm.

Hence, PQ = 9 cm.

From figure,

Since, OM ⊥ AB, it bisects it. (As perpendicular from center to chord bisects it)

∴ AM = MB = $\frac{48}{2}$ = 24 cm.

In right △AOM,

⇒ AO² = OM² + AM² (By pythagoras theorem)

⇒ AO² = 10² + 24²

⇒ AO² = 100 + 576

⇒ AO² = 676

⇒ AO = $\sqrt{676}$ = 26 cm.

Radius = 26 cm.

Since, ON ⊥ CD, it bisects it. (As perpendicular from center to chord bisects it)

∴ CN = ND = $\frac{20}{2}$ = 10 cm.

In right △CNO,

⇒ OC² = ON² + NC² (By pythagoras theorem)

⇒ 26² = ON² + 10²

⇒ ON² = 676 - 100

⇒ ON² = 576

⇒ ON = $\sqrt{576}$ = 24 cm.

Hence, the chord of length 20 cm is at a distance of 24 cm from the centre.

From figure,

PM = MQ = $\frac{6}{2}$ = 3 cm.

In right △CMP,

⇒ CP² = CM² + PM² (By pythagoras theorem)

⇒ 5² = CM² + 3²

⇒ CM² = 25 - 9

⇒ CM² = 16

⇒ CM = $\sqrt{16}$ = 4 cm.

In right △DMP,

⇒ DP² = DM² + PM² (By pythagoras theorem)

⇒ 4² = DM² + 3²

⇒ DM² = 16 - 9

⇒ DM² = 7

⇒ DM = $\sqrt{7}$ = 2.65 cm.

CD = CM + MD = 4 + 2.65 = 6.65 cm.

Hence, CD = 6.65 cm.

From C, D draw CM, DN perpendiculars to AB.

From figure,

MCDN is a rectangle.

∴ MN = CD (Opposite sides of rectangle are equal).

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ MP = $\frac{1}{2}$AP and NP = $\frac{1}{2}$PB

From figure,

MN = MP + PN

= $\frac{1}{2}$AP + $\frac{1}{2}$PB

= $\frac{1}{2}$(AP + PB)

= $\frac{1}{2}$AB

∴ CD = $\frac{1}{2}$AB [∵ MN = CD]

⇒ AB = 2CD

Hence, proved that AB = 2CD.

Let M be the point of intersection of line CD and line APQB.

(i) Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ In circle with center D,

PM = MQ .........(1)

and

In circle with center C,

AM = MB .........(2)

Subtracting equation (1) from (2),

⇒ AM - PM = MB - MQ

⇒ AP = QB

Hence, proved that AP = QB.

(ii) Let AP = QB = x.

From figure,

AQ = AB - QB = AB - x

BP = AB - AP = AB - x.

∴ AQ = BP.

Hence, proved that AQ = BP.

In NOMP,

∠P = 90° (As chords intersect at right angles)

∠M = ∠N = 90° (Straight lines from center bisecting the chord are perpendicular to it.)

∠O = 360° - (∠M + ∠N + ∠P)
= 360° - (90° + 90° + 90°)
= 90°

Since, equal chords are equidistant from center,

∴ OM = ON.

Since, all angles = 90° and all sides are equal.

Hence, NOMP is a square.

As chords AB and AC are equidistant from the center, so AB = AC. [∵ equal chords are equidistant from center]

Since, angle in a semicircle = 90°.

∠B = ∠C

In △ABD and △ACD,

∠B = ∠C (Both equal to 90°)

AD = AD (Common)

AB = AC (Proved above)

∴ △ABD ≅ △ACD (By R.H.S. congruence rule).

∴ ∠BAD = ∠CAD and ∠BDA = ∠CDA (By C.P.C.T.)

Hence, proved that AD bisects ∠BAC and ∠BDC.