Chapter 6 - Circle
Practice Set 6.1
1. Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of the circle.
Given:
In a circle with centre O,
AB = 12 cm
OP = 8 cm
To find:
Diameter of the circle
Solution:
AP = \( \frac{1}{2} \) AB [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
∴ AP = \( \frac{1}{2} \) x 12
∴ AP = \( \frac{12}{2} \)
∴ AP = 6 cm ….(i)
In ∆OPA,
∠OPA = 90° [seg OP ⊥ chord AB]
∆OPA is a right angled triangle
∴ OA² = OP² + AP² [Pythagoras theorem]
∴ OA² = 8² + 6² [From (i)]
∴ OA² = 64 + 36
∴ OA² = 100
∴ √(OA²) = √100 [Taking square root on both sides]
∴ OA = 10 cm
Radius (r) = 10 cm
∴ Diameter = 2r = 2 x 10
∴ Diameter = 20 cm
Ans: The diameter of the circle is 20 cm.
2. Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.
Given:
In a circle with centre O,
PQ = 24 cm
Diameter (d) = 26 cm
To find:
OR
Solution:
Radius (OP) = \( \frac{d}{2} \)
∴ Radius (OP) = \( \frac{26}{2} \)
∴ Radius (OP) = 13 cm ……(i)
PR = \( \frac{1}{2} \) PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
∴ PR = \( \frac{1}{2} \) x 24
∴ PR = \( \frac{24}{2} \)
∴ PR = 12 cm …..(ii)
In ∆ORP,
∠ORP = 90°
∆ORP is a right angled triangle
∴ OP² = OR² + PR² [Pythagoras theorem]
∴ 13² = OR² + 12² [From (i) and (ii)]
∴ 169 = OR² + 144
∴ OR² = 169 – 144
∴ OR² = 25
∴ √(OR²) = √25 [Taking square root on both sides]
∴ OR = 5 cm
Ans: The distance of the chord from the centre of the circle is 5 cm.
3. Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord.
Given:
In a circle with centre A,
AP = 34 cm
AM = 30 cm
To find:
PQ
Solution:
In ∆AMP
∠AMP = 90°
∴ ∆AMP is a right angled triangle
∴ AP² = AM² + PM² [Pythagoras theorem]
34² = 30² + PM²
∴ PM² = 34² – 30²
∴ PM² = 1156 – 900
∴ PM² = 256
∴ √(PM²) = √256 [Taking square root on both sides]
∴ PM = 16cm ……(i)
Now,
PM = \( \frac{1}{2} \) (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
∴ 16 = \( \frac{1}{2} \) (PQ) [From (i)]
∴ PQ = 16 x 2
∴ PQ = 32cm
Ans: The length of the chord of the circle is 32cm.
4. Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle.
Given:
In a circle with centre O,
OP = 41 units
PQ = 80 units
To find:
OM
Solution:
PM = \( \frac{1}{2} \) PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord]
∴ PM = \( \frac{1}{2} \) (80)
∴ PM = \( \frac{80}{2} \)
∴ PM = 40 Units ….(i)
In ∆OMP,
∠OMP = 90°
∆OMP is a right angled triangle
∴ OP² = OM² + PM² [Pythagoras theorem]
∴ 41² = OM² + 40² [From (i)]
∴ OM² = 41² – 40²
∴ OM² = 1681 – 1600
∴ OM² = 81
∴ √(OM²) = √81 [Taking square root on both sides]
∴ OM = 9 units
Ans: The distance of the chord from the centre of the circle is 9 units.
5. In figure 6.9, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ
Given:
Two concentric circles having centre O
To prove:
AP = BQ
Construction:
Draw seg OM ⊥ chord AB, A-M-B
Proof:
For smaller circle,
seg OM ⊥ chord PQ [Construction, A – P – M, M – Q – B]
∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
For bigger circle,
seg OM ⊥ chord AB [Construction]
∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ AP + PM = MQ + QB [A – P – M, M – Q – B]
∴ AP + MQ = MQ + QB [From (i)]
∴ AP = BQ
Hence Proved.
6. Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.
Given:
In a circle with centre O,
Diameter PQ bisects chords AB in points M
Diameter CD bisects chords AB in points N
To prove:
chord AB || chord CD.
Proof:
seg AM ≅ seg BM [Given, Diameter PQ bisects chords AB in points M]
∴ seg OM ⊥ chord AB [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P – M – O, O – N – Q]
∴ ∠OMA = 90° …..(i)
Also,
seg CN ≅ seg DN [Given, Diameter PQ bisects chords CD in points N]
∴ seg ON ⊥ chord CD [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P – M – O, O – N – Q]
∴ ∠ONC = 90° …..(ii)
Now,
∠OMA + ∠ONC = 90° + 90° [From (i) and (ii)]
∴ ∠OMA + ∠ONC = 180°
But,
∠OMA and ∠ONC form a pair of interior angles on lines AB and CD when seg MN is their transversal.
∴ chord AB || chord CD [Interior angles test]
Hence Proved.
Practice Set 6.2
1. Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle ?
Given:
In a circle with centre O,
Chord PQ = Chord RS = 16 cm
Radius OR = Radius OP = 10 cm
To find:
Distance of chords from the centre of the circle.
Solution:
PU = \( \frac{1}{2} \) PQ [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ PU = \( \frac{1}{2} \) x 16
∴ PU = 8 cm …(i)
In ∆OUP,
∠OUP = 90°
∆OUP is a right angled triangle
∴ OP² = OU² + PU² [Pythagoras theorem]
∴ 10² = OU² + 8² [From (i)]
∴ 100 = OU² + 64
∴ OU² = 100 – 64
∴ OU² = 36
∴ √(OU²) = √36 [Taking square root on both sides]
∴ OU = 6 cm
Now,
OT = OU [Congruent chords of a circle are equidistant from the centre.]
∴ OT = OU = 6cm
Ans: The distance of the chords from the centre of the circle is 6 cm.
2. In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of the chords.
Given:
In a circle with center O,
OA and OC are the radii and AB and CD are its congruent chords,
Radius OA = Radius OC = 13cm
OE = OF = 5 cm
To find:
Chord AB and Chord CD
Solution:
In ∆AFO,
∠AFO = 90°
∆AFO is a right angled triangle
∴ AO² = AF² + FO² [Pythagoras theorem]
∴ 13² = AF² + 5²
∴ 169 = AF² + 25
∴ AF² = 169 – 25
∴ AF² = 144
∴ (AF²) = √144 [Taking square root on both sides]
∴ AF = 12 cm …..(i)
Now,
AF = \( \frac{1}{2} \) AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = \( \frac{1}{2} \) AB [From (i)]
∴ AB = 12 x 2
∴ AB = 24 cm
chord AB ≅ chord CD
∵ AB = 24 cm
∴ CD = 24 cm
Ans: The lengths of the two chords are 24 cm each.
3. Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of ∠NPM.
Given:
In a circle with centre C
Chord PM ≅ Chord PN
To prove:
Ray PC is the bisector of ∠NPM.
Construction:
Draw seg CR ⊥ chord PN, P – R – N
and seg CQ ⊥ chord PM, P – Q – M
Proof:
Chord PM ≅ Chord PN [Given]
seg CR ⊥ chord PN
seg CQ ⊥ chord PM [Construction]
∴ segCR ≅ segCQ ….(i) [Congruent chords are equidistant from the centre]
In ∆PRC and ∆PQC,
∠PRC ≅ ∠PQC [Both are right angles]
seg CR ≅ seg CQ [From (i)]
seg PC ≅ seg PC [Common side]
∴ ∆PRC ≅ ∆PQC [Hypotenuse side test]
∴ ∠RPC ≅ ∠QPC [c. a. c. t.]
∴ ∠NPC ≅ ∠MPC [N – R – P, M – Q – P]
∴ Ray PC is the bisector of ∠NPM.
Hence Proved.
Practice Set 6.3
1. Construct D ABC such that ∠B = 100°, BC = 6.4 cm, ∠C = 50° and construct its incircle.
Solution:
Steps of construction:
(i) Construct ∆ABC of the given measurement.
(ii) Draw the bisectors of ∠B and ∠C. Let these bisectors intersect at point I.
(iii) Draw a perpendicular IM on side BC. Point M is the foot of the perpendicular.
(iv) With I as the centre and IM as radius, draw a circle which touches all the three sides of the triangle.
2. Construct ∆PQR such that ∠P = 70°, ∠R = 50°, QR = 7.3 cm and construct its circumcircle.
Solution:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° – 50°
∴ m∠Q = 110° – 50°
∴ m∠Q = 60°
Steps of construction:
(i) Construct ∆PQR of the given measurement.
(ii) Draw the perpendicular bisectors of side PQ and side QR of the triangle.
(iii) Name the point of intersection of the perpendicular bisectors as point C.
(iv) Join seg CP.
(v) With C as the centre and CP as radius, draw a circle which passes through the three vertices of the triangle.
3. Construct ∆ XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.
Solution:
Steps of construction:
(i) Construct ∆XYZ of the given measurement.
(ii) Draw the bisectors of ∠X and ∠Z. Let these bisectors intersect at point I.
(iii) Draw a perpendicular IM on side XZ. Point M is the foot of the perpendicular.
(iv) With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.
4. Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of the rhombus and its perimeter.
Solution:
Steps of construction:
(i) Construct ∆XYZ of the given measurement.
(ii) Draw the bisectors of ∠X and ∠Z. Let these bisectors intersect at point I.
(iii) Draw a perpendicular IM on side XZ. Point M is the foot of the perpendicular.
(iv) With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.
5. Construct ∆ DEF such that DE = EF = 6 cm, ∠F = 45° and construct its circumcircle.
Solution:
Steps of construction:
(i) Construct ∆DEF of the given measurement.
(ii) Draw the perpendicular bisectors of side DE and side EF of the triangle.
(iii) Name the point of intersection of perpendicular bisectors as point C.
(iv) Join seg CE.
(v) With C as centre and CE as radius, draw a circle which passes through the three vertices of the triangle.
Problem Set 6
1. Choose correct alternative answer and fill in the blanks.
(i) Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence the length of the chord is ………
(A) 16 cm
(B) 8 cm
(C) 12 cm
(D) 32 cm
Solution:
∴ OA² = AC² + OC²
∴ 10² = AC² + 6²
∴ AC² = 64
∴ AC = 8 cm
∴ AB = 2(AC)= 16 cm
OPTION (A) – 16 cm
(ii) The point of concurrence of all angle bisectors of a triangle is called the ……
(A) centroid
(B) circumcentre
(C) incentre
(D) orthocentre
OPTION (C) – incentre
(iii) The circle which passes through all the vertices of a triangle is called …..
(A) circumcircle
(B) incircle
(C) congruent circle
(D) concentric circle
OPTION (A) – circumcircle
(iv) Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is ….
(A) 12 cm
(B) 13 cm
(C) 14 cm
(D) 15 cm
Solution:
OA² = AC² + OC²
∴ OA² = 12² + 5²
∴ OA² = 169
∴ OA = 13 cm
OPTION (B) – 13 cm
(v) The length of the longest chord of the circle with radius 2.9 cm is …..
(A) 3.5 cm
(B) 7 cm
(C) 10 cm
(D) 5.8 cm
Solution:
Diameter is the longest chord of a circle.
∴ Diameter = 2 × radius
∴ Diameter = 2 × 2.9
∴ Diameter = 5.8
OPTION (D) – 5.8 cm
(vi) Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie.
(A) on the centre
(B) Inside the circle
(C) outside the circle
(D) on the circle
Solution:
l(OP) > radius
∴ Point P lies in the exterior of the circle.
OPTION (C) – outside the circle
(vii) The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is …..
(A) 2 cm
(B) 1 cm
(C) 8 cm
(D) 7 cm
Solution:
PQ = 8 cm, MN = 6 cm
∴ AQ = 4 cm, BN = 3 cm
∴ OQ² = OA² + AQ²
∴ 5² = OA² + 4²
∴ OA² = 25 – 16 = 9
∴ OA = 3 cm
Also,
ON² = OB² + BN²
∴ 5² = OB² + 3²
∴ OB = 4 cm
Now,
AB = OA + OB = 3 + 4 = 7 cm
OPTION (D) – 7 cm
2. Construct incircle and circumcircle of an equilateral ∆ DSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.
Solution:
Steps of construction:
(i) Construct ∆DPS of the given measurement.
(ii) Draw the perpendicular bisectors of side DP and side PS of the triangle.
(iii) Name the point of intersection of the perpendicular bisectors as point C.
(iv) With C as centre and CM as radius, draw a circle which touches all the three sides of the triangle.
(v) With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.
Now,
Radius of incircle = 2.2 cm
Radius of circumcircle = 4.4 cm
∴ \( \large \frac {Radius\, of\, circumcircle}{Radius\, of\, incircle}\) = \( \large \frac {4.4}{2.2}\)
∴ \( \large \frac {Radius\, of\, circumcircle}{Radius\, of\, incircle}\) = \( \large \frac {2}{1}\)
∴ \( \large \frac {Radius\, of\, circumcircle}{Radius\, of\, incircle}\) = 2 : 1
3. Construct ∆ NTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 110° and draw incircle and circumcircle of it.
Solution:
Steps of construction:
For incircle:
(i) Construct ∆NTS of the given measurement.
(ii) Draw the bisectors of ∠T and ∠S. Let these bisectors intersect at point I.
(iii) Draw a perpendicular IM on side TS. Point M is the foot of the perpendicular.
(iv) With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.
For circumcircle:
(i) Draw the perpendicular bisectors of side NT and side TS of the triangle.
(ii) Name the point of intersection of the perpendicular bisectors as point C.
(iii) Join seg CN.
(iv) With C as centre and CN as radius, draw a circle which passes through the three vertices of the triangle.
4. In the figure 6.19, C is the centre of the circle. seg QT is a diameter CT = 13, CP = 5, find the length of chord RS.
Given:
In a circle with centre C,
QT is a diameter
CT = 13 units
CP = 5 units
To find:
Length of chord RS
Construction:
Join points R and C
Solution:
CR = CT= 13 units …..(i) …[Radii of the same circle]
In ∆CPR,
∠CPR = 90°
∆CPR is a right angled triangle
∴ CR² = CP² + RP² …[Pythagoras theorem]
∴ 13² = 5² + RP² …[Given and From (i)]
∴ 169 = 25 + RP²
∴ RP² = 169 – 25
∴ RP² = 144
∴ \( \sqrt {RP²}\) = \( \sqrt {144}\) …[By Taking square root of both sides]
∴ RP = 12 cm ….(ii)
Now,
seg CP ⊥ chord RS …[Given]
∴ RP = \( \frac{1}{2} \) RS …[Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = \( \frac{1}{2} \) RS …[From (ii)]
∴ RS = 2 x 12
∴ RS = 24 units
Ans: The length of chord RS is 24 units.
5. In the figure 6.20, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP, then prove that AB = CD.
Given:
In a circle with centre P,
Chord AB and chord CD intersect on the diameter at the point E
∠AEP ≅ ∠DEP
To prove:
AB = CD
Construction:
Draw seg PM ⊥ chord AB, A – M – B
and seg PN ⊥ chord CD, C – N – D
Proof:
∠AEP ≅ ∠DEP …[Given]
∴ Seg EP is the bisector of ∠AED
∴ PM = PN …[Every point on the bisector of an angle is equidistant from the sides of the angle]
∴ chord AB ≅ chord CD …[Chords which are equidistant from the centre are congruent]
∴ AB = CD …[Length of congruent segments]
Hence Proved.
6. In the figure 6.21, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ ABC is an isosceles triangle.
Given:
In a circle with centre O,
diameter CD ⊥ chord AB, A – E – B
To prove:
∆ABC is an isosceles triangle
Proof:
Diameter CD ⊥ chord AB [Given]
∴ seg OE ⊥ chord AB [C – O – E, O – E – D]
∴ seg AE ≅ seg BE ……(i) …[Perpendicular drawn from the centre of the circle to the chord bisects the chord]
In ∆CEA and ∆CEB,
∠CEA ≅ ∠CEB …[Both are right angles]
seg AE ≅ seg BE …[From (i)]
seg CE ≅ seg CE …[Common side]
∴ ∆CEA ≅ ∆CEB …[S-A-S test]
∴ seg AC ≅ seg BC …[c. s. c. t.]
∴ ∆ABC is an isosceles triangle.
Hence Proved.