I’m Up and Down, and Round and Round
Ganita Manjari Part 1 · Chapter 5 · NCERT Solutions
Welcome to Encyclohub’s complete NCERT solutions for Class 9 Maths – Chapter 5: I’m Up and Down, and Round and Round (Ganita Manjari Part 1).
All questions and solutions on circles, chords, and cyclic quadrilaterals are presented clearly so you can understand every concept step by step. These solutions are useful for homework, revision and exam preparation.
Study Tip:
Draw a neat figure for every circle question before you start — most chord and cyclic-quadrilateral proofs become obvious once the radii and perpendiculars are marked.
Exercise Set 5.1
Draw ΔABC with AB = 5 cm, ∠A = 70°, and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Given: AB = 5 cm, ∠A = 70°, ∠B = 60°
To Construct: ΔABC and its circumcircle.
Steps of Construction:
(i) Draw a line segment AB = 5 cm.
(ii) Construct an angle of 70° at point A and an angle of 60° at point B.
(iii) Let the arms of these two angles intersect at point C.
(iv) Join AC and BC to obtain ΔABC.
(v) Draw the perpendicular bisectors of sides AB, BC, and AC.
(vi) Let the perpendicular bisectors intersect at O. Then O is the circumcentre of ΔABC.
(vii) With centre O and radius OA, draw a circle passing through A, B, and C. This is the circumcircle of ΔABC.
Now, ∠C = 180° − (70° + 60°) = 50°
Since all the angles of ΔABC are less than 90°, the triangle is acute-angled.
Therefore, the circumcentre of an acute-angled triangle lies inside the triangle.
Draw ΔABC with AB = 5 cm, ∠A = 100°, and AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Given: AB = 5 cm, ∠A = 100°, AC = 4 cm
To Construct: ΔABC and its circumcircle.
Steps of Construction:
(i) Draw a line segment AB = 5 cm.
(ii) Construct an angle of 100° at point A.
(iii) With A as centre and radius 4 cm, draw an arc cutting an arm of ∠A at C.
(iii) Join BC to obtain ΔABC.
(iv) Draw the perpendicular bisector of AB, BC, and AC.
(v) Let the perpendicular bisectors intersect at O. Then O is the circumcentre of ΔABC.
(vi) With centre O and radius OA, draw a circle passing through A, B, and C. This is the circumcircle of ΔABC.
Since ∠A = 100° > 90°
Therefore, ΔABC is an obtuse-angled triangle.
Therefore, the circumcentre of an obtuse-angled triangle lies outside the triangle.
Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.
Given: AB = 6 cm, BC = 7 cm, CA = 7 cm
To Construct: ΔABC, its circumcircle and circumcentre O.
Steps of Construction:
(i) Draw a line segment AB = 6 cm.
(ii) With A as centre and radius 7 cm, draw an arc.
(iii) With B as centre and radius 7 cm, draw another arc intersecting the first arc at C.
(iv) Join AC and BC to obtain ΔABC.
(v) Draw the perpendicular bisector of AB, BC, and CA.
(vi) Let the perpendicular bisectors intersect at O. Then O is the circumcentre of ΔABC.
(vii) With centre O and radius OA, draw a circle passing through A. The circle will also pass through B and C. This is the circumcircle of ΔABC.
Measuring the lengths OA, OB, and OC, we found that
OA = OB = OC ≈ 4 cm
What is the least possible radius of a circle through two points A and B?
For any two points A and B, infinitely many circles can pass through both points. The radius of these circles depends on the position of the centre.
The least possible radius occurs when the centre of the circle is the midpoint of AB. In this case, AB becomes the diameter of the circle.
Therefore, the least possible radius is AB2.
Hence, the least possible radius of a circle through two points A and B is half the length of AB.
Exercise Set 5.2
Show that the triangle formed by a chord and the centre of the circle is isosceles.
Let AB be a chord of a circle with centre O. Join OA and OB.
Since OA and OB are radii of the same circle,
OA = OB.
Therefore, ΔOAB has two equal sides. Hence, ΔOAB is an isosceles triangle.
Thus, the triangle formed by a chord and the centre of a circle is isosceles.
Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Let the two isosceles triangles be ΔOAB and ΔOCD, where base AB = base CD.
Since OA, OB, OC, and OD are radii of the same circle,
OA = OC and OB = OD
Also, AB = CD (given)
Therefore, by the SSS congruence ΔOAB ≅ ΔOCD.
Hence, two such isosceles triangles having equal base lengths are congruent.
Exercise Set 5.3
Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
(Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)
GIven: A circle with chord AB and centre C. CM ⊥ AB, i.e., ∠CMA = ∠CMB = 90°.
To Prove: CM bisects AB, i.e., AM = BM.
Proof:
In ΔCMA and ΔCMB,
CA = CB (radii of the same circle)
CM = CM (common)
∠CMA = ∠CMB = 90° (CM ⊥ AB)
Therefore, by the RHS congruence ΔCMA ≅ ΔCMB.
Hence, corresponding sides are equal,
AM = BM.
Therefore, the perpendicular from the centre of a circle to a chord bisects the chord.
An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.
Given: An isosceles triangle ABC inscribed in a circle, with AB = AC.
To Prove: The altitude from A to BC passes through the centre O of the circle.
Proof: Let AD be the altitude from A to BC, meeting BC at D.
In ΔABD and ΔACD,
AB = AC (given)
AD = AD (common side)
∠ADB = ∠ADC = 90°
By the RHS congruence ΔABD ≅ ΔACD.
Hence, corresponding sides are equal,
BD = DC.
Thus, D is the midpoint of chord BC.
Since AD⊥BC, AD is the perpendicular bisector of the chord BC.
We know that the perpendicular bisector of a chord passes through the centre of the circle.
Therefore, the altitude AD passes through the centre of the circle O.
Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
Given: AB = 6 cm and CD = 8 cm are the two parallel chords on opposite sides of the centre O of the circle.
To find: Distance MN between the midpoints of the chords.
Construction: Let M and N be the midpoints of chords AB and CD, respectively.
Join OA, OC, OM, and ON.
Since the perpendicular from the centre of a circle to a chord bisects the chord,
AM = AB2 = 62 = 3 cm
and
CN = CD2 = 82 = 4 cm
In ΔOAM,
By the Baudhāyana-Pythagoras theorem,
OA2 = OM2 + AM2
52 = OM2 + 32
25 = OM2 + 9
OM2 = 25 – 9
OM2 = 16
OM = 4 cm.
Similarly, in ΔOCN,
By the Baudhāyana-Pythagoras theorem,
OC2 = ON2 + CN2
52 = ON2 + 42
25 = ON2 + 16
ON2 = 25 – 16
ON = 3 cm.
Since the chords are on opposite sides of the centre,
MN = OM + ON
MN = 4 + 3 = 7 cm.
Hence, the distance between the midpoints of the chords is 7 cm.
Exercise Set 5.4
Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.
Given: AB and FG are two chords of a circle with centre O such that AB = FG. E and H are midpoints of AB and FG. CE ⊥ AB and CH ⊥ FG.
To Prove: CE = CH
Since E is the midpoint of chord AB,
AE = AB2 ……… (i)
Similarly, H is the midpoint of chord FG,
FH = FG2 ……….. (ii)
Given that AB = FG ……. (iii)
Therefore, from (i), (ii), and (iii),
AE = FH …………. (iv)
Now, in right-angled ΔCEA,
By the Baudhāyana–Pythagoras theorem,
CA2 = CE2 + AE2
r2 = CE2 + AE2 ……… (∵ CA = r)
AE2 = r2 – CE2 ……….. (v)
Similarly, in right-angled ΔCHF,
By the Baudhāyana–Pythagoras theorem,
CF2 = CH2 + FH2
r2 = CH2 + FH2 ……… (∵ CF = r)
FH2 = r2 – CH2 ………….. (vi)
From (iv), (v) and (vi),
r2 – CE2 = r2 – CH2
CE2 = CH2
Taking the square root,
CE = CH
Hence, equal chords of a circle are equidistant from the centre.
Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
Given: AB and GF are two chords of a circle with centre O. Both chords are equidistant from the centre i.e. CE = CH. CE ⊥ AB and CH ⊥ GH.
To Prove: AB = GF
Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
AE = BE
or AE = AB2 …….. (i)
Also, GH = FH
or GH = GF2 …….. (ii)
In ΔCEA and ΔCHG,
CA = CG ……… (radii of the same circle)
∠CEA = ∠CHG ………… (each 90°)
CE = CH ………. (given)
Thus by the RHS congruence ΔCEA ≅ ΔCHG.
So, AE = GH
AB2 = GF2 ……… [using (i) and (ii)]
AB = GF
Hence, the chords of a circle which are equidistant from the centre are equal.
Solve the previous question using the Baudhāyana–Pythagoras theorem.
Given: AB and GF are two chords of a circle with centre O. Both chords are equidistant from the centre i.e. CE = CH. CE ⊥ AB and CH ⊥ GH.
To Prove: AB = GF
Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
AE = BE
AE = AB2
2AE = AB …….. (i)
Also, GH = FH
GH = GF2
2GH = GF …….. (ii)
Now, in right-angled ΔCEA,
By the Baudhāyana–Pythagoras theorem,
CA2 = CE2 + AE2
r2 = CE2 + AE2
CE2 = r2 – AE2 ……… (iii)
Similarly, in right-angled ΔCHG,
By the Baudhāyana–Pythagoras theorem,
CG2 = CH2 + GH2
r2 = CH2 + GH2
CH2 = r2 – GH2 ……… (iv)
CE = CH ………. (v)
From (iii), (iv) and (v),
r2 – AE2 = r2 – GH2
AE2 = GH2
Taking the square root,
AE = GH
Multiplying both sides by 2,
2AE = 2GH
AB = GF ………. [using (i) and (ii)]
Hence, the chords of a circle which are equidistant from the centre are equal.
Exercise Set 5.5
Find the length of the chord of a circle where the radius is 7 cm and the perpendicular distance is 6 cm.
Let AB be the chord of a circle with centre C. Let CD⊥AB, where D is the midpoint of AB. CD = 6 cm.
Join CA and CB.
Radius of circle, CA = CB = 7 cm.
Since the perpendicular from the centre to a chord bisects the chord,
AD = BD
In right-angled ΔCDB,
By the Baudhāyana–Pythagoras theorem,
CB2 = CD2 + BD2
72 = 62 + BD2
49 = 36 + BD2
BD2 = 49 – 36
BD2 = 13
BD = √13 cm
Therefore,
AB = 2BD
AB = 2√13 cm
Hence, the length of the chord is 2√13 cm.
Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√r2−d2.
Let AB be the chord, C the centre of the circle, and CD the perpendicular from C to AB.
CD = d and CA = r ……… (given)
Since the perpendicular from the centre to a chord bisects the chord, D is the midpoint of AB.
Therefore,
AD = AB2
In right-angled ΔCDA, by the Baudhāyana–Pythagoras theorem,
CA2 = CD2 + AD2
r2 = d2 + AD2
AD2 = r2 – d2
AD = √r2−d2
Hence,
AB = 2AD
AB = 2√r2−d2
Therefore, the length of the chord is 2√r2−d2.
In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.
We know that if the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length = 2√r2−d2.
Let the distance of the chord AB and CD from the centre be 2x and x.
Thus,
AB = 2√r2−(2x)2 = 2√r2−4x2
CD = 2√r2−x2
Clearly, CD ≠ 2AB.
Hence, knowing only that one chord is at twice the distance from the centre as another is not sufficient to conclude that one chord is twice the length of the other.
Exercise Set 5.6
In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
Given: Radius of the circle = 12 cm
∠AOB = 60°
To Find: Length of chord AB
Join OA and OB.
Since OA = OB = 12 cm (radii of the same circle),
ΔAOB is an isosceles triangle.
∴ ∠OAB = ∠OBA ………. (angles opposite to equal sides of an isosceles triangle are equal)
In ΔAOB,
∠AOB + ∠OAB + ∠OBA = 180° …………. (sum of angles of a triangle)
60° + ∠OAB + ∠OAB = 180°…………. (∠OAB = ∠OBA)
2∠OAB = 180° – 60°
∠OAB = 120°2 = 60°
∠OAB = ∠OBA = ∠AOB = 60°
Therefore, ΔAOB is an equilateral triangle.
So, AB = OA = OB = 12 cm
Hence, the length of the chord AB is 12 cm.
Let A and B be two points on a circle with centre O.
(i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
(iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
(i) No. Angles subtended by the same chord of a circle in the same segment are equal.
Therefore, ∠AXB = ∠AYB.
Hence, there cannot be two points X and Y on the same side of AB such that ∠AXB and ∠AYB are different.
(ii) No. If AB is a diameter, then the angle subtended by it at any point on the circle is 90°.
Thus, if X and Y lie on opposite sides of AB,
∠AXB = ∠AYB = 90°.
Therefore, equal angles do not necessarily imply that X and Y lie on the same side of the circle.
(iii) Yes. The points A, B, and X determine a circle.
Since ∠AXB = ∠AYB, both angles stand on the same chord AB.
By the converse of the theorem “angles in the same segment of a circle are equal”, the points A, B, X, and Y are concyclic.
Therefore, the circle through A, B, and X also passes through Y.
Find x in Fig. 5.26.
Here ∠ADC = 100°
Therefore, the angle subtended by the arc ABC at the centre, reflex ∠AOC = 2 × 100° = 200°.
Since, ∠AOC + reflex ∠AOC = 360°
∠AOC + 200° = 360°
∠AOC = 360° – 200° = 160°
Now, ∠AOC is the angle subtended by arc ADC at the centre of the circle.
Since the angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle,
∴ x = ∠AOC2 = 160°2 = 80°.
End of Chapter Exercises
In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
Given: AB is a chord of a circle with centre O and radius 13 cm such that OA = 13 cm and OC = 5 cm. Draw OC⊥AB. Join OA.
To find: The length of the chord AB.
In right ΔOCA, by Baudhāyana-Pythagoras theorem:
OA2 = OC2 + AC2
(13)2 = (5)2 + AC2
169 = 25 + AC2
AC2 = 169 – 25
AC2 = 144
AC = 12 cm
Since the perpendicular from the centre of a circle to a chord bisects the chord, AC = BC.
AB = 2 × AC = 2 × 12 = 24 cm.
Therefore, the length of the chord is 24 cm.
An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
Angle subtended by the arc at the centre = 70°
Since the angle subtended by an arc of a circle at the centre of the circle is double the angle subtended by the arc on the circumference.
Angle subtended at a point on the circle = 12 × 70° = 35°
Therefore, the measure of the angle is 35°.
The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
Given: AB is a chord, and AC is a diameter of a circle with centre O such that AB = 24 cm and AC = 26 cm. Draw OM⊥AB.
To find: The distance between the centre and the chord, i.e. OM.
Radius of the circle, OA = OB = 262 = 13 cm.
Since the perpendicular from the centre of a circle to a chord bisects the chord,
AM = BM = 242 = 12 cm
In right ΔOMA, by Baudhāyana-Pythagoras theorem:
OA2 = OM2 + AM2
132 = OM2 + 122
169 = OM2 + 144
OM2 = 169 – 144
OM2 = 25
OM = 5 cm
Therefore, the distance from the centre of the circle to the chord is 5 cm.
A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
Given: AB is a chord of a circle with centre O and radius 15 cm such that OA = 15 cm and OC = 9 cm. Draw OC⊥AB. Join OA.
To find: The length of the chord AB.
In right ΔOCA, by Baudhāyana-Pythagoras theorem:
OA2 = OC2 + AC2
(15)2 = (9)2 + AC2
225 = 81 + AC2
AC2 = 225 – 81
AC2 = 144
AC = 12
Since the perpendicular from the centre of a circle to a chord bisects the chord, AC = BC.
AB = 2 × AC = 2 × 12 = 24 cm.
Therefore, the length of the chord is 24 cm.
Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Let AB be a chord of a circle with centre O. Let M be the midpoint of AB.
Join OA, OB, and OM.
In ΔOAM and ΔOBM,
OA = OB ………. (radii of the circle)
AM = MB ………. (M is the midpoint of AB)
OM = OM ………. (common)
∴ ΔOAM ≅ ΔOBM ………. (by SSS congruence rule)
∠OMA = ∠OMB ……….. (by c.p.c.t)
Also,
∠OMA + ∠OMB = 180° ……… (linear pair)
∠OMA + ∠OMA = 180°
2∠OMA = 180°
∠OMA = 90°
∴ ∠OMA = ∠OMB = 90°
Hence, OM⊥AB.
Therefore, OM is the perpendicular bisector of chord AB, and it passes through the centre O.
Hence proved.
The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.
Given: AB is the diameter of the circle, and C is a point on the circumference.
To find: ∠ACB
The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.
Since AB is the diameter, the angle subtended by AB at the centre is ∠AOB = 180°
Therefore, ∠ACB = 12 × 180° = 90°
Hence, the angle subtended by the diameter at any point on the circle is a right angle.
ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?
Given: ABCD is a cyclic quadrilateral inscribed in a circle with centre O such that ∠A = 75° and ∠B = 110°.
To find: ∠C and ∠D.
In a cyclic quadrilateral, the sum of opposite angles is 180°,
∠A + ∠C = 180°
∠B + ∠D = 180°
Since ∠A = 75°,
Therefore, ∠C = 180° – 75° = 105°
Also, ∠B = 110°,
Therefore, ∠D = 180° – 110° = 70°
Hence, ∠C = 105° and ∠D = 70°.
Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.
Given: Quadrilateral PQRS is inscribed in a circle with centre O such that ∠P = (2x + 10)°, ∠R = (3x − 20)°.
To find: ∠P, ∠R and x.
Since PQRS is a cyclic quadrilateral, the sum of opposite angles is 180°.
∠P + ∠R = 180°
(2x + 10) + (3x − 20) = 180
5x – 10 = 180
5x = 180 + 10
5x = 190
x = 1905 = 38
Now,
∠P = 2 × 38 + 10 = 76 + 10 = 86°
∠R = 3 × 38 − 20 = 114 – 20 = 94°
Thus, ∠P = 86° and ∠R = 94°.
The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
Given: AB is a chord of a circle with centre O such that AB = 16 cm. OC ⊥ AB and OC = 6 cm.
To find: Radius of the circle, OA.
Since the perpendicular from the centre of a circle to a chord bisects the chord,
AC = BC = 162 = 8 cm
In right ΔOCA, by Baudhāyana-Pythagoras theorem:
OA2 = OC2 + AC2
(OA)2 = (6)2 + (8)2
(OA)2 = 36 + 64
(OA)2 = 100
OA = 10 cm
Therefore, the radius of the circle is 10 cm.
A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Let ABCD be a cyclic quadrilateral with sides AB = BC = 5 units and CD = AD = 12 units.
Join BD.
In ΔABD and ΔCBD,
AB = CB …….. (each 5 units)
AD = CD ……. (each 12 units)
BD = BD ……. (common)
ΔABD ≅ ΔCBD ……. (by SSS congruence rule)
∠BAD = ∠BCD ………. (by c.p.c.t)
∠BAD + ∠BCD = 180° ………. (sum of opposite angles of cyclic quadrilateral is 180°)
∴ ∠BAD = ∠BCD = 180°2 = 90°
Thus, ΔABD and ΔCBD are right-angled congruent triangles with equal area.
Area of quadrilateral ABCD = 2 × area of ΔABD
= 2 × 12 × 5 × 12
= 60 sq. units.
Therefore, the area of quadrilateral ABCD is 60 sq. units.
Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?
The best way to find out is to look at the angles:
(i) If all four angles are less than 90°, the circumcentre lies inside the quadrilateral.
(ii) If one angle is greater than 90°, the circumcentre lies outside.
(iii) If one angle is 90°, the circumcentre lies at the midpoint of the corresponding diagonal.
Thus, examining the angles of the cyclic quadrilateral is the easiest way to determine the position of its circumcentre.
When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
Given: Two equal chords AB and CD of a circle with centre O, intersect at point P inside the circle.
To Prove: AP = CP and BP = DP
Join OP. Draw OL ⊥ AB and OM ⊥ CD.
Since the perpendicular from the centre of a circle to a chord bisects the chord,
AL = BL and CM = DM.
Since equal chords are equidistant from the centre,
OL = OM.
In ΔOPL and ΔOMP,
OL = OM ……….. (proved above)
∠OLP = ∠OMP ……….. (each 90°)
OP = OP ……….. (common)
ΔOPL ≅ ΔOMP ………. (by RHS congruence rule)
LP = MP ………. (by c.p.c.t)
Since
AB = CD
12AB = 12CD
AL = CM
Adding LP on both sides,
AL + LP = CM + LP
AL + LP = CM + MP ……….. (∵ LP = MP)
AP = CP
Also, AB – AP = CD – CP
BP = DP
Hence, proved.
Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
(Hint: Is it a circumcircle of a suitable triangle?)
Steps:
(i) Draw a line segment AB = 6 cm.
(ii) Construct the perpendicular bisector of AB, meeting AB at C.
(iii) On the perpendicular bisector, mark a point O such that CO = 3 cm. Join OB.
(iv) With O as centre and OB as radius, draw a circle.
Since O lies on the perpendicular bisector of AB,
OA = OB.
Thus, the circle passes through A and B, and CO = 3 cm is the distance of chord AB from the centre.
Hence, the required circle is constructed.
Show that a rectangle is the only parallelogram that can be inscribed in a circle.
Given: ABCD is a cyclic parallelogram inscribed in a circle.
To prove: ABCD is a rectangle.
Since opposite angles of a parallelogram are equal,
∠A = ∠C
Also, the sum of opposite angles of a cyclic quadrilateral is 180°,
∠A + ∠C = 180°
∠A + ∠A = 180° ………. (∵ ∠A = ∠C)
2∠A = 180°
∠A = 180°2 = 90°
Thus, ∠A = ∠C = 90°
Also, opposite angles of a parallelogram are equal, so
∠B = ∠D = 90°
Thus, all the angles of ABCD are right angles.
Hence, ABCD is a rectangle.
Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
Given: ABCD is a rectangle inscribed in a circle with diagonals AC and BD intersecting at O.
To prove: O is the centre of the circle.
Since ABCD is a rectangle, its diagonals are equal and bisect each other.
Therefore,
AC = BD
and AO = OC = AC2, BO = OD = BD2
Since AC = BD,
AO = BO = CO = DO.
Thus, O is equidistant from the four vertices A, B, C, D.
Hence, O is the centre of the circle.
Therefore, the point of intersection of the diagonals of a rectangle inscribed in a circle is the centre of the circle.
Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Let AB be any chord of fixed length in a circle with centre O, and let M be its midpoint.
The perpendicular from the centre to a chord bisects the chord.
Hence, OM⊥AB
Since all the chords have the same length, their perpendicular distances from the centre are equal. Therefore, the midpoint M of every such chord is at the same distance from O.
Hence, the locus of the midpoints of all chords of a fixed length is a circle with centre O.
In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.
Given: AB and AC are two congruent chords of a circle with centre O.
To prove: O lies on the angle bisector of ∠BAC.
Join OB and OC.
In ΔOAB and ΔOAC,
AB = AC ………. (given)
OB = OC ………. (radii of the same circle)
OA = OA ……….. (common)
ΔOAB ≅ ΔOAC ………. (by SSS congruence rule)
∠BAO = ∠CAO ……… (by c.p.c.t)
Thus, AO bisects ∠BAC.
Hence, the centre of the circle lies on the angle bisector of ∠BAC.
Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
Let AB and CD be the two chords of a circle with centre O such that AB = 10 cm and CD = 24 cm.
Draw perpendiculars OM and ON to AB and CD, respectively.
Join OB and OD.
OB = OD = r …….. (radii of the circle)
Since the perpendicular from the centre of a circle to a chord bisects the chord.
AM = MB = 102 = 5 cm
CN = ND = 242 = 12 cm
Let ON = x cm
Therefore, OM = (x + 7) cm and NM = 7 cm
In ΔOND, by Baudhāyana-Pythagoras theorem:
OD2 = ON2 + ND2
r2 = x2 + (12)2
r2 = x2 + 144 ………… (i)
In ΔOMB, by Baudhāyana-Pythagoras theorem:
OB2 = OM2 + MB2
r2 = (x + 7)2 + (5)2
r2 = x2 + 14x + 49 + 25
r2 = x2 + 14x + 74 ………. (ii)
From (i) and (ii),
x2 + 144 = x2 + 14x + 74
144 = 14x + 74
14x = 144 – 74
14x = 70
x = 7014 = 5
Therefore,
r2 = (5)2 + (12)2
r2 = 25 + 144 = 169
r = √169
r = 13 cm
Hence, the radius of the circle is 13 cm.
A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.
Let ABCDEF be a regular hexagon inscribed in a circle with centre O and radius r.
Since a regular hexagon divides the circle into six equal parts,
∠AOB = 360°6 = 60°
Also, OA = OB = r
Therefore, ΔAOB is an equilateral triangle.
Hence, AB = r
So, the length of each side of the hexagon is r.
Now, let OM⊥AB. Since the perpendicular from the centre of a circle to a chord bisects the chord,
AM = r2
In right-angled ΔAOM, by Baudhāyana-Pythagoras theorem:
OA2 = OM2 + AM2
r2 = OM2 + (r2)2
r2 = OM2 + r24
OM2 = r2 – r24
OM2 = 3r24
OM = √32r
Thus, the side of the hexagon is r, and the distance of each side from the centre is √32r.
A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
Given: MN is a diameter in a quadrilateral MNOP inscribed in a circle with centre B.
To find: Relation between ∠MOP and ∠MNP.
Join MO and PN.
We know that the angle at the centre is twice the angle at the circumference standing on the same chord.
Therefore,
∠MBP = 2∠MNP ………. (i)
∠MBP = 2∠MOP ………. (ii)
From (i) and (ii),
2∠MNP = 2∠MOP
∴ ∠MNP = ∠MOP.
Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
Given: ABCD is a cyclic quadrilateral with side CD produced to a point E.
To prove: ∠EDA = ∠ABC
∠EDA + ∠CDA = 180° (linear Pair) …….. (i)
∠ABC + ∠CDA = 180° (sum of opposite angles of a cyclic quadrilateral is 180°) ………. (ii)
From (i) and (ii),
∠EDA + ∠CDA = ∠ABC + ∠CDA
∠EDA = ∠ABC.
“There is no chord of a circle that is longer than its diameter.” How do you justify this statement?
Let AB be a diameter of a circle with centre O and radius r. Let CD be any other chord.
Since AB is a diameter, it passes through the centre O. Therefore, the distance of AB from the centre is 0.
The chord CD, being any other chord, is at some positive distance from the centre.
We know that the chord nearer to the centre of a circle is longer.
Therefore, AB > CD
Hence, the diameter is the longest chord of a circle.
Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
Let the circle have centre O and radius r, and let A be a fixed point inside the circle with OA = d (d < r).
Let PQ be any chord through A, and let M be its midpoint. Since the perpendicular from the centre to a chord bisects the chord, OM⊥PQ.
In ΔOMA (right-angled at M, or M coinciding with A),
OM ≤ OA, since OA is the hypotenuse of right ΔOMA when M ≠ A, and OM = OA when M = A (i.e. when PQ ⊥ OA at A itself).
Now, the length of a chord at perpendicular distance OM from the centre is 2√r2−OM2.
Since chord-length decreases as the distance from the centre increases, the chord is shortest when OM is largest.
The largest possible value of OM, subject to M lying on chord PQ through A, is OM = OA = d, which happens exactly when M = A, i.e. when PQ passes through A and OA⊥PQ.
Hence, the shortest chord through A has length 2√r2−d2, and it is the chord perpendicular to OA at A.
How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
In ΔAOC,
OA = OC (radii of the same circle)
Therefore, ΔAOC is an isosceles triangle.
Hence, ∠OAC = ∠OCA = b
Similarly, in ΔAOB,
OA = OB (radii of the same circle)
Therefore, ΔAOB is an isosceles triangle.
Hence, ∠OAB = ∠OBA = a
Now,
∠BAC = ∠OAB + ∠OAC = (a + b)
Also, in ΔABC,
∠ABC + ∠BAC + ∠ACB = 180° ………. (sum of angles of a triangle)
a + (a + b) + b = 180°
2a + 2b = 180°
2(a + b) = 180°
(a + b) = 90°
Therefore, ∠BAC = 90°
Hence, the angle in a semicircle is always a right angle.
In a circle, two chords CC′ and DD′ are drawn perpendicular to a diameter AB. Prove that the segment MM′ joining the midpoints of the chords CD and C′D′ is perpendicular to AB.
Given: AB is a diameter of a circle with centre O. CC′ and DD′ are two chords perpendicular to AB.
M and M′ are the midpoints of CD and C′D′, respectively.
To prove: MM′⊥AB.
Proof:
Since O is the centre of the circle and AB is the diameter, O lies on AB.
Since CC′⊥AB and DD′⊥AB, the perpendiculars from the centre O to the chords CC′ and DD′ bisect them.
Therefore, the perpendicular from O to CC′ passes through its midpoint, and the perpendicular from O to DD′ passes through its midpoint.
Since M is the midpoint of CD and M′ is the midpoint of C′D′, the segment MM′ joins corresponding midpoints of the two chords.
Hence, MM′ is parallel to CC′ and DD′.
But CC′⊥AB.
Therefore, MM′⊥AB.
Hence proved.
How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
In ΔAOB,
OA = OB (radii of the same circle)
Therefore, ΔAOB is an isosceles triangle.
Hence, ∠OAB = ∠OBA = p
Similarly, in ΔBOC,
OB = OC (radii of the same circle)
Therefore, ∠OBC = ∠OCB = q
In ΔCOD,
OC = OD (radii of the same circle)
Therefore, ∠OCD = ∠ODC = u
In ΔAOD,
OA = OD (radii of the same circle)
Therefore, ∠OAD = ∠ODA = v
Since the sum of the angles of a quadrilateral is 360°,
∠A + ∠B + ∠C + ∠D = 360°
(p + v) + (p + q) + (q + u) + (u + v) = 360°
2p + 2q + 2u + 2v = 360°
2(p + q + u + v) = 360°
p + q + u + v = 180°
Now, for the opposite angles,
∠A + ∠C = (p + v) + (q + u) = 180°
∠B + ∠D = (p + q) + (u + v) = 180°
Hence, the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
Chapter 5 – Quick Revision
- Equal chords of a circle are equidistant from the centre, and chords equidistant from the centre are equal.
- The perpendicular from the centre of a circle to a chord bisects the chord, and the perpendicular bisector of a chord always passes through the centre.
- The circumcentre of a triangle lies inside an acute-angled triangle, outside an obtuse-angled triangle, and at the midpoint of the hypotenuse of a right-angled triangle.
- The diameter is the longest chord of a circle; a chord closer to the centre is longer than one farther away.
- The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.
- Angles in the same segment of a circle, subtended by the same chord, are equal. The angle in a semicircle is a right angle.
- In a cyclic quadrilateral, opposite angles are supplementary (sum to 180°), and the exterior angle at a vertex equals the interior opposite angle.
- A rectangle is the only parallelogram that can be inscribed in a circle, and its diagonals intersect at the centre.