**Chapter 6 - Circle**

**Practice Set 6.1**

**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**

**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**

**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**

**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 cmRadius 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. **