Chord Properties of a Circle

Given: Length of the chord AC = 16 cm.

Radius of the circle (r) = 10 cm

Diameter of the circle = 10 x 2 = 20 cm.

Draw OB ⊥ AC, where O is the center of the circle. Join OA.

B is the midpoint of AC, as OB is perpendicular to the chord AC.

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

= $\frac{1}{2} \times (16)$

= 8 cm.

In Δ OAB, ∠B = 90°

Using Pythagoras theorem,

∴ OA² = OB² + AB²

⇒ (10)² = OB² + (8)²

⇒ 100 = OB² + 64

⇒ OB² = 100 - 64

⇒ OB² = 36

⇒ OB = $\sqrt{36}$

⇒ OB = 6 cm.

Hence, the distance of the chord from the center of the circle is 6 cm.

From figure,

Given: Length of the chord AC = 4.8 cm.

Radius of the circle (r) = 2.5 cm

Distance of the chord from the center of the circle = OB.

B is the midpoint of AC, as OB is perpendicular to the chord AC.

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

= $\frac{1}{2} \times (4.8)$

= 2.4 cm.

In Δ OAB, ∠B = 90°

Using Pythagoras theorem,

∴ OA² = OB² + AB²

⇒ (2.5)² = OB² + (2.4)²

⇒ 6.25 = OB² + 5.76

⇒ OB² = 6.25 - 5.76

⇒ OB² = 0.49

⇒ OB = $\sqrt{0.49}$

⇒ OB = 0.7 cm.

Hence, the distance of the chord from the center of the circle is 0.7 cm.

From figure,

Given: Radius of the circle (r) = 40 cm

AC is the chord and OB is the perpendicular distance from center.

B is the midpoint of AC, as OB is perpendicular to the chord AC.

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

In Δ OAB, ∠B = 90°

Using Pythagoras theorem,

∴ OA² = OB² + AB²

⇒ (40)² = (24)² + AB²

⇒ 1600 = 576 + AB²

⇒ AB² = 1600 - 576

⇒ AB² = 1024

⇒ AB = $\sqrt{1024}$

⇒ AB = 32 cm

Length of the chord = AC = 2(AB)

= 2(32)

= 64 cm.

Hence, the length of the chord is 64 cm.

From figure,

AC is the chord and OB is the perpendicular distance of chord from the center.

Length of the chord AC = 48 cm.

B is the midpoint of AC, as OB is perpendicular to the chord AC.

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

= $\frac{1}{2} \times 48$

= 24 cm.

In Δ OAB, ∠B = 90°

Using Pythagoras theorem,

∴ OA² = OB² + AB²

⇒ OA² = (7)² + 24²

⇒ OA² = 49 + 576

⇒ OA² = 625

⇒ OA = $\sqrt{625}$ = 25 cm.

Hence, radius of circle is 25 cm.

From figure,

AB is the chord of length 16 cm which is at a distance 15 cm from the center so OC = 15 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 = 15² + 8²

⇒ OA² = 225 + 64

⇒ OA² = 289

⇒ OA = $\sqrt{289}$ ​ ⇒ OA = 17 cm

Radius = 17 cm,

∴ OD = 17 cm.

From figure,

In right angle triangle ODF,

⇒ OD² = OF² + DF² (By pythagoras theorem)

⇒ DF² = OD² - OF²

⇒ DF² = 17² - 8²

⇒ DF² = 289 - 64 = 225

⇒ DF = $\sqrt{225}$ = 15 cm.

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

DF = FE = 15 cm

From figure,

DE = DF + FE = 15 cm + 15 cm = 30 cm.

Hence, the length of chord which is at a distance of 8 cm from the center of the circle = 30 cm.

Let AB and CD be chords of length 16 cm and 30 cm respectively.

From figure,

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\frac{AB}{2} = \frac{16}{2}$ = 8 cm.

CE = $\frac{CD}{2} = \frac{30}{2}$ = 15 cm.

From figure,

OA = OC = radius = 17 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 17² = OE² + 15²

⇒ 289 = OE² + 225

⇒ OE² = 289 - 225

⇒ OE² = 64

⇒ OE = $\sqrt{64}$ = 8 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 17² = OF² + 8²

⇒ 289 = OF² + 64

⇒ OF² = 289 - 64

⇒ OF² = 225

⇒ OF = $\sqrt{225}$ = 15 cm.

From figure,

⇒ EF = OE + OF = 8 + 15 = 23 cm.

Hence, distance between the chords = 23 cm.

Let AB = 18 cm and CD = 80 cm be chords on same side of the center of the circle.

OE ⊥ CD and OF ⊥ AB

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\frac{AB}{2} = \frac{18}{2}$ = 9 cm.

CE = $\frac{CD}{2} = \frac{80}{2}$ = 40 cm.

From figure,

OA = OC = radius = 41 cm.

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ 41² = OE² + 40²

⇒ 1681 = OE² + 1600

⇒ OE² = 1681 - 1600

⇒ OE² = 81

⇒ OE = $\sqrt{81}$ = 9 cm.

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ 41² = OF² + 9²

⇒ 1681 = OF² + 81

⇒ OF² = 1681 - 81

⇒ OF² = 1600

⇒ OF = $\sqrt{1600}$ = 40 cm.

From figure,

⇒ EF = OF - OE = 40 - 9 = 31 cm.

Hence, distance between the chords = 31 cm.

From figure,

AB = 1.4 cm and CD = 4 cm

OF ⊥ AB and OE ⊥ CD.

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\frac{AB}{2} = \frac{1.4}{2}$ = 0.7 cm.

CE = $\frac{CD}{2} = \frac{4}{2}$ = 2 cm.

EF = 3.9 cm and OA = OC = r cm.

Let OE = x and OF = 3.9 - x

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ OC² = x² + 2²

⇒ OC² = x² + 4 .....(1)

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ OA² = (3.9 - x)² + O.7²

⇒ OA² = (3.9 - x)² + 0.49 ....(2)

From (1) and (2), we get :

⇒ x² + 4 = (3.9 - x)² + 0.49

⇒ x² + 4 = x² - 7.8x + 15.21 + 0.49

⇒ x² + 4 = x² - 7.8x + 15.7

⇒ 4 = 15.7 - 7.8x

⇒ 7.8x = 15.7 - 4

⇒ 7.8x = 11.7

⇒ x = $\frac{11.7}{7.8}$

⇒ x = 1.5

Substituting value of x in equation (1), we get :

⇒ OC² = (1.5)² + 4

⇒ OC² = 2.25 + 4

⇒ OC² = 6.25

⇒ OC = $\sqrt{6.25}$ = 2.5 cm.

Hence, radius of the circle = 2.5 cm.

AB = 8 cm and CD = 6 cm

OF ⊥ AB and OE ⊥ CD

We know that,

Perpendicular from the center to the chord, bisects it.

∴ AF = $\frac{AB}{2} = \frac{8}{2}$ = 4 cm.

CE = $\frac{CD}{2} = \frac{6}{2}$ = 3 cm.

EF = 1 cm and OA = OC = r cm.

Let OF = x and OE = x + 1

In right-angled triangle OCE,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OC² = OE² + CE²

⇒ r² = (x + 1)² + 3²

⇒ r² = (x + 1)² + 9 .....(1)

In right-angled triangle OAF,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OF² + AF²

⇒ r² = x² + 4²

⇒ r² = x² + 16 ....(2)

From (1) and (2),

⇒ x² + 16 = (x + 1)² + 9

⇒ x² + 16 = x² + 2x + 1 + 9

⇒ x² + 16 = x² + 2x + 10

⇒ 16 = 2x + 10

⇒ 2x = 16 - 10

⇒ 2x = 6

⇒ x = $\frac{6}{2}$

⇒ x = 3.

Substituting value of x in equation (2), we get :

⇒ r² = 3² + 16

⇒ r² = 9 + 16

⇒ r² = 25

⇒ r = $\sqrt{25}$ = 5 cm.

Hence, radius of the circle = 5 cm.

Given,

PQ = PR = 25 cm

In an isosceles triangle, the perpendicular from a vertex between equal sides bisects the opposite side.

Thus,

S is the mid-point of QR.

∴ QS = $\frac{QR}{2} = \frac{14}{2}$ = 7 cm.

PS is perpendicular bisector of QR and perpendicular from center bisects the chord.

Thus, centre of the circle O lies on PS. Let radius of circle OQ be r.

In right-angled triangle PSQ,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ PQ² = PS² + SQ²

⇒ 25² = PS² + 7²

⇒ 625 = PS² + 49

⇒ PS² = 625 - 49

⇒ PS² = 576

⇒ PS = $\sqrt{576}$ = 24 cm

OS = PS - OP = 24 - r

In right-angled triangle OSQ,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OQ² = OS² + SQ²

⇒ r² = (24 - r)² + 7²

⇒ r² = r² - 48r + 576 + 49

⇒ 48r = 625

⇒ r = $\frac{625}{48}$ = 13.02 cm ≈ 13 cm

Hence, radius of the circle = 13 cm.

Given,

AB = AC = $12\sqrt{5}$ cm

In an isosceles triangle, the perpendicular from a vertex between equal sides bisects the opposite side.

Thus,

L is the mid-point of BC.

∴ BL = $\frac{BC}{2} = \frac{24}{2}$ = 12 cm.

AL is perpendicular bisector of BC and perpendicular from center bisects the chord.

Thus, centre of the circle O lies on AL. Let radius of circle OB be r.

In right-angled triangle ALB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AL² + BL²

⇒ ($12\sqrt{5}$)² = AL² + 12²

⇒ 720 = AL² + 144

⇒ AL² = 720 - 144

⇒ AL² = 576

⇒ AL = $\sqrt{576}$ = 24 cm

OL = AL - AO = 24 - r

In right-angled triangle OBL,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OB² = OL² + BL²

⇒ r² = (24 - r)² + 12²

⇒ r² = r² - 48r + 576 + 144

⇒ 48r = 720

⇒ r = $\frac{720}{48}$ = 15 cm

Hence, radius of the circle = 15 cm.

Let ABC be equilateral triangle inscribed in circle.

AB = BC = AC

Draw AD ⊥ BC.

Since, in an equilateral triangle the perpendicular from a vertex bisects the opposite side.

Thus,

D is the mid-point of BC.

∴ BD = $\frac{BC}{2} = \frac{9}{2}$ = 4.5 cm.

Since, the chord BC is bisected at point D, and perpendicular from center bisects the chord.

Centre of the circle O lies on AD. Let radius of circle (OA) be r.

In right-angled triangle ADB,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AD² + BD²

$$\begin{aligned}\Rightarrow 9^2 = AD^2 + \Big(\frac{9}{2}\Big)^2 \\[1em] \Rightarrow 81 = AD^2 + \Big(\frac{81}{4}\Big) \\[1em] \Rightarrow AD^2 = 81 - \frac{81}{4} \\[1em] \Rightarrow AD^2 = \frac{324 - 81}{4}\\[1em] \Rightarrow AD^2 = \frac{243}{4} \\[1em] \Rightarrow AD = \sqrt{\frac{243}{4}} = \frac{9\sqrt{3}}{2}.\end{aligned}$$

We know that,

In an equilateral triangle, the centroid and circumcentre coincide.

Since AD is a median, and centroid divides median in 2:1 ratio,

AO : OD = 2 : 1

∴ Radius (AO) = $\frac{2}{3} \times AD$

= $\frac{2}{3} \times \frac{9\sqrt{3}}{2} = 3\sqrt3$ cm.

Hence, the radius of the circle = $3\sqrt{3}$ cm.

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

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

BL = LC = x (let)

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

AL = LD = y (let)

From figure,

AB = AL - BL = (y - x)

CD = LD - LC = (y - x)

∴ AB = CD.

Hence, proved that AB = CD.

Draw OM ⊥ QR.

Since, perpendicular from center bisects the chord.

Thus, QM = $\frac{QR}{2} = \frac{12}{2}$ = 6 cm.

In right △OQM,

By pythagoras theorem,

⇒ 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.

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.

Draw LM through O perpendicular to AB and CD.

In △ OLB and △ OMC :

OB = OC [radii of same circle]

∠OLB = ∠OMC = 90°

∠OBL = ∠OCM [Alternate interior angles are equal]

∴ △ OLB ≅ △ OMC [By A.A.S. axiom]

Since the triangles are congruent, their corresponding parts are equal.

OL = OM

Chords equidistant from the center of a circle are equal in length.

∴ AB = CD.

Hence, proved that AB = CD.

Join OQ.

Radius = OQ = OB = x (let)

⇒ OR = OB - RB = (x - 4) cm.

We know that,

A straight line drawn from the center of a circle to bisect a chord, is perpendicular to the chord.

Since, PR = QR, thus R is mid-point of PQ.

∴ OR ⊥ PQ

In right-angled triangle OQR,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OQ² = OR² + QR²

⇒ x² = (x - 4)² + 8²

⇒ x² = x² + 4² - 2 × x × 4 + 64

⇒ x² = x² + 16 - 8x + 64

⇒ x² - x² + 8x = 80

⇒ 8x = 80

⇒ x = $\frac{80}{8}$ = 10 cm.

Hence, radius of circle = 10 cm.

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 and CA ∥ OD.

From C draw CL perpendicular to AB and from D drawn DM perpendicular to AB.

From figure,

LCDM is a rectangle.

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

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

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

From figure,

LM = LP + PM

$$\begin{aligned}\Rightarrow LM = \frac{1}{2} AP + \frac{1}{2} PB \\[1em] \Rightarrow LM = \frac{1}{2} (AP + PB) \\[1em] \Rightarrow LM = \frac{1}{2} AB \\[1em] \therefore CD = \frac{1}{2} AB \\[1em] \Rightarrow AB = 2CD.\end{aligned}$$

Hence, proved that AB = 2CD.

Let diameter POQ bisect chords AB and CD at L and M respectively.

Then, OL ⟂ AB and OM ⟂ CD

∴ ∠ALO = ∠OMD = 90°

These two angles are alternate interior angles, where PQ is transversal intersecting chords AB and CD.

Since, these angles are equal

∴ AB ∥ CD.

Hence, the two chords are parallel.

Let AB and AC be two chords.

AOD be a diameter such that ∠BAO = ∠CAO.

OL ⟂ AB and OM ⟂ AC.

In △OLA and △OMA,

OA = OA [common side]

∠OLA = ∠OMA = 90°

∠LAO = ∠MAO [AO bisects ∠A]

∴ △OLA ≅ △OMA [By A.A.S. rule]

Then,

OL = OM (By C.P.C.T.C.)

AB = AC [Chords which are equidistant from centre are equal]

Hence, proved that the two chords are equal.

Given: AB and CD are two equal chords of a circle with center O.

In Δ AOB and Δ COD,

OA = OC (Radii of a circle)

OB = OD (Radii of a circle)

AB = CD (Given)

By SSS congruency criterion,

Δ AOB ≅ Δ COD

∴ ∠AOB = ∠COD (By C.P.C.T.C.)

Hence, equal chords of a circle subtend equal angles at the centre.

In OLPM,

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

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

∠O = 360° - (∠L + ∠M + ∠P)

= 360° - (90° + 90° + 90°)

= 90°.

Since, equal chords are equidistant from center,

∴ OL = OM.

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

Thus, OLPM is a square.

Hence, proved that OLPM is a square.

AB is a chord of a circle with centre O.

Let CD be the perpendicular bisector of AB.

∠ACD = 90° [CD ⊥ AB]

Line joining the center of a circle to the midpoint of a chord is perpendicular to the chord.

∠ACO = 90° [Since C is the midpoint of AB, OC ⊥ AB]

∴ ∠ACD = ∠ACO which is wrong

∴ CD must pass through O.

Hence, the perpendicular bisector of a chord of a circle always passes through the centre.

We know that,

The perpendicular bisector of a chord passes through the centre of the circle.

∴ l bisects AB, l ⟂ AB

Given, AB ∥ CD

If a line is perpendicular to one of two parallel lines, it is also perpendicular to the other.

Now, l ⟂ AB and AB ∥ CD

∴ l ⟂ CD

So, l passes through the centre O and is perpendicular to CD, thus it must bisect CD.

Hence, l is the perpendicular bisector of 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.

Let POQ be a diameter, bisecting chord AB at L.

Join OA and OB

In △OLA and △OLB:

OA = OB [Both are radii of the same circle]

AL = LB [PQ bisects the chord AB]

OL = OL [common side]

△OLA ≅ △OLB [By the SSS rule]

Since the triangles are congruent, their corresponding parts must be equal:

∠AOL = ∠BOL

This proves that the diameter PQ bisects ∠AOB, which is the angle subtended by the chord AB at the center O.

Hence, proved that the diameter also bisects the angle subtended by the chord at the centre.

(i) Given,

L and M are the mid-points of two equal chords AB and CD.

We know that,

A straight line drawn from the center of a circle to bisect a chord, which is not a diameter, is at right angles to the chord.

∴ OL ⊥ AB and OM ⊥ CD.

Since, AB and CD are equal chords and perpendicular drawn from center to equal chords are equal in length.

∴ OL = OM

Thus, in triangle OLM,

∴ ∠OLM = ∠OML (Angles opposite to equal sides are equal)

Hence, proved that ∠OLM = ∠OML.

(ii) We have,

∠OLM = ∠OML

∠OLA - ∠ALM = ∠OMC - ∠CML

Since, ∠OLA and ∠OMC both equal to 90°.

∴ ∠ALM = ∠CML.

Hence, proved that ∠ALM = ∠CML.

Let AB = AC = x.

Given,

OM ⊥ AC and OL ⊥ AB.

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

∴ AM = MC = $\frac{x}{2}$

and

AL = LB = $\frac{x}{2}$

∴ MC = LB ......(1)

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

∴ OM = OL = y (let).

Let radius of circle be r.

From figure,

OQ = OP = r

QM = OQ - OM = r - y

PL = OP - OL = r - y

∴ QM = PL .....(2)

In △QMC and △PLB,

MC = LB [From (1)]

QM = PL [From (2)]

∠QMC = ∠PLB (Both equal to 90°)

△QMC ≅ △PLB by SAS axiom.

∴ PB = QC (By C.P.C.T.)

Hence, proved that PB = QC.

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 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.

Let the chord be AB and the perpendicular from the center O meet the chord at M.

In the right-angled triangle △OMA:

Using the Pythagorean theorem:

OA² = AM² + OM²

17² = AM² + 8²

289 = AM² + 64

AM² = 289 - 64

AM² = 225

AM = $\sqrt{225}$ = 15 cm

Since the perpendicular from the center bisects the chord.

Length of chord = 2 × AM = 30 cm.

Hence, option 3 is the correct option.

Let the chord be AB and the perpendicular from the center O meet the chord at M.

Since the perpendicular from the center bisects the chord AB.

AM = $\frac{AB}{2} = \frac{40}{2}$ = 20 cm

In the right-angled triangle △OMA:

Using the Pythagorean theorem:

OA² = AM² + OM²

OA² = 20² + 15²

OA² = 225 + 400

OA² = 625

OA = $\sqrt{625}$ = 25 cm.

Hence, option 4 is the correct option.

The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Since,

AC = BC [C is the midpoint of AB]

O is the centre.

Thus, OC is perpendicular to AB.

∠ACO = 90°.

Hence, option 3 is the correct option.

Given,

OP = OQ

CD = 6 cm

Chords equidistant from the centre of a circle are equal.

AB = CD

∴ AB = 6 cm

Hence, option 1 is the correct option.

Let the chord be AB and the perpendicular from the center O meet the chord at M.

Since the perpendicular from the center bisects the chord AB.

AM = $\frac{70}{2}$ = 35 cm

In the right-angled triangle △OMA:

Using the Pythagorean theorem:

OA² = AM² + OM²

37² = 35² + OM²

OM² = 1369 - 1225

OM² = 144

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

Hence, option 4 is the correct option.

1. Given,

d = 40 m

r = $\frac{40}{2}$ = 20 m

The circumference of the circular park is = 2πr

= 2 × $\frac{22}{7}$ × 20

= 125.6 m

Hence, option (a) is the correct option.

2. Since poles A, B, and C are at equal distances on the boundary, they divide the total circumference into three equal arcs.

The distance along the boundary (arc BC) = $\frac{\text{Circumference}}{3} = \frac{125.6}{3}$ = 41.8 m

From the figure, going clockwise from B to C means the child travels the major arc, not the minor arc.

Major arc = Total circumference − minor arc

= 125.6 − 41.87 = 83.73 m

Hence, option (b) is the correct option.

3. Poles A, B, and C are at equal distances on the boundary.

Since, arc BA = arc AC = arc CB.

This means the chords AB, BC, and CA are equal in length.

A triangle with three equal sides is an equilateral triangle.

Hence, option (c) is the correct option.

4. In an equilateral triangle, the perpendicular bisector, median, and altitude are all the same line.

Thus, perpendicular from B on AC, will pass through the center, centroid and circumcenter of the circle.

Hence, option (d) is the correct option.

5. Radius of inscribed circle = 20 m

The relationship between the side of an equilateral triangle and the circumradius is:

Side length = $r\sqrt3 = 20\sqrt3.$

Hence, option (b) is the correct option.

We know that,

Equal chords of a circle are equidistant from the centre.

Reason (R) is true.

Since, AB = AC [Chords of circle are equal]

∴ OD = OE = 4 cm

Assertion (A) is true.

Both A and R are true.

Hence, option 3 is the correct option.

The longer the chord, the closer it is to the center.

The shorter the chord, the farther it is from the center.

Assertion (A) is false.

If two chords are equidistant from the center, they must be equal in length, but they do not have to be necessarily parallel.

Reason (R) is false.

Both A and R are false.

Hence, option 4 is the correct option.

Let AB and CD are two equal chords of a circle with center O.

In Δ AOB and Δ COD,

OA = OC (Radii of a circle)

OB = OD (Radii of a circle)

AB = CD (Given)

By SSS congruency criterion,

Δ AOB ≅ Δ COD

Using corresponding parts of congruent triangles,

∠AOB = ∠COD

Assertion (A) is true.

Through three non-collinear points, exactly one circle can pass.

Through three collinear points, no circle can be drawn.

Reason (R) is false.

A is true, R is false.

Hence, option 1 is the correct option.

The statement is false because there is an inverse relationship between the chord length and its distance from the center :

The longer the chord, the closer it is to the center.

The shorter the chord, the farther it is from the center.

Hence, option 2 is the correct option.

Let AB be chord of length = 10 cm

Radius OA = 13 cm

Let OD be the perpendicular distance from the center to the chord, it bisects the chord into two equal parts:

AD = DB = $\frac{10}{2}$ = 5 cm

In triangle OAD,

According to the Pythagorean Theorem:

OA² = OD² + AD²

13² = OD² + 5²

169 = OD² + 25

OD² = 169 - 25

OD² = 144

OD = $\sqrt{144}$ = 12 cm

The shortest distance between the chord and the center is 12 cm.

Hence, option 1 is the correct option.

Given,

Chord AB = 12 cm

CD bisects chord AB at P

AP = PB = $\frac{12}{2}$ = 6 cm

CP = 3 cm

OC = OA = R

OP = OC - CP = R - 3

In triangle OPA,

Apply the Pythagorean Theorem :

OA² = OP² + AP²

R² = (R - 3)² + 6²

R² = R² - 6R + 9 + 36

R² = R² - 6R + 45

6R = 45

R = $\frac{45}{6}$ = 7.5 cm

Hence, option 3 is the correct option.

Let the center of the circular park be O and positions of Amit, Vinay and Sukrit be A, V and S respectively.

The radius of the circle is 5 m.

AV = VS = 6 m

Draw VM ⊥ AS.

In an isosceles triangle, the perpendicular from a vertex between equal sides bisects the opposite side.

∴ AM = MS

Thus,

Let OM = x m and MS = y m

VM = OV - OM = (5 - x) m.

In right-angled triangle VMS,

⇒ VS² = VM² + MS²

⇒ 6² = (5 - x)² + y²

⇒ 36 = (5 - x)² + y²

⇒ y² = 36 - (5 - x)² ....(1)

In right-angled triangle OMS,

⇒ OS² = OM² + MS²

⇒ 5² = x² + y²

⇒ y² = 25 - x²....(2)

From (1) and (2), we get :

⇒ 25 - x² = 36 - (5 - x)²

⇒ 25 - x² = 36 - (25 - 10x + x²)

⇒ 25 - x² = 36 - 25 + 10x - x²

⇒ 25 - x² = 36 - 25 + 10x - x²

⇒ 25 = 11 + 10x

⇒ 10x = 14

⇒ x = 1.4

Substituting value of x in equation (2):

⇒ y² = 25 - (1.4)²

⇒ y² = 23.04

⇒ y = $\sqrt{23.04}$ = 4.8 m

From figure,

⇒ AS = MS + AM

⇒ AS = 2MS

⇒ AS = 2(4.8) = 9.6 m

Hence, the distance between Amit and Sukrit is 9.6 m.

Let AB be a diameter of the circle and O be the centre.

By definition, it passes through the center, so AB = OA + OB = r + r = 2r.

Let CD be any other chord of the circle that does not pass through the center.

∴ OC = OD = r

The sum of the lengths of any two sides must be greater than the length of the third side.

OC + OD > CD

2r > CD

AB > CD

This proves the length of the diameter is always greater than the length of any other chord that does not pass through the center.

Hence, the diameter is the greatest chord of a circle.