in the given figure the measure of oqp is

= (5)2 + (12)2 (2014OD) ∴ OT bisects PQ Prove that: AB + CD = BC + DA. (2011D) ∆TPQ is an isosceles ∆ and TO is the bisector of ∠PTQ , Solution: ∴ ∆AO’D ~ ∴AOC …(AA similarity ⇒ (5)2 + PT2 = (13)2 The tangents at P and Q intersect at a point T. Find the length of TP. In the figure, there are two concentric, circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. OQ = BQ = 11 cm If EK = 9 cm, then the perimeter of ∆EDF is [CBSE 2012] (c), Question 51. Also, AE = 8 cm, Join OC (b) 70° In right angled ∆PQO, ∆OSP, Solution: (d) 3√2 cm ∠RQS = 180° – 150o = 30°, Question 46. (c) 60° ∆OPT, ∠OPQ = ∠OPT – ∠QPT ∠1 = ∠2 = 23 = 90° …[ Tangent is ⊥ to the radius B [through the point of contact ABCD is a quadrilateral circumscribe a circle and its sides AB, BC, CD and DA touch the circle at P, Q, R and S respectively If ∠QPR= 46, then ∠QOR equals (c) AB + CD = AC + BC ∠TQP = ∠TPQ = 60° Proof. If PA ⊥ PB, then find the length of each tangent. x = \(\frac{80}{24}=\frac{10}{3}\) cm ∠OPQ = OQP = 40° Similarly, QD = BD = 3 cm …(iii) BP = BQ = 27 cm …[Tangents drawn from an external point (c) 8.5 cm Solution: (b) 7 cm In figure, a quadrilateral ABCD is drawn to circum- DA scribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, RA and S respectively. (c) 24 cm (d) 60° Solution: (c) 2 (c) In the figure, PA and PB are the tangents to the circle with centre O => 60° + ∠POQ = 180° ∴ ∠OTS = ∠OST = 30°, Question 37. DR = 5 cm (y + 6)2 = x2 + 102 Draw another common tangent through C which intersects AB at D, then DA = DC = DB (a) Firstly, draw a circle of radius 5 cm having centre O. (2014OD) (a) 30° ∴ OP ⊥ XY … [Shortest side is ⊥, Question 43. => 2r = 14 – 10 = 4 ∴ AB = x + 8 = 15 cm ∠OAB = 90° – 60° = 30°, Question 3. In the given figure, PQ and PR are two tangents to a circle with centre O. (b) 6 cm If PR = 7.5 cm, then PS is equal to PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°, then ∠OPQ is QR = RC 3) Given a wet soil with q v = 0.18, r b = 1300 kg m-3, Q = G A t, and the average daily net radiation is R n = 150 W m-2, find the ratio G /R n if the increase in average soil temperature to 1 m depth over a month’s time is 10 o C. From problem 2, we know that Cv. or 6.67 cm or 6.6 cm, Question 48. ∴ r = 2 cm, Question 27. ∠P + ∠A + ∠B = 180° In ∆PAB, ∠PAB + ∠PBA + ∠APB = 180° …[Angle-sum-property of a ∆ Given: OD = 3 cm; OE = 3 cm; OF = 3 cm ar(∆ABC) = 54 cm2 OP = 5 cm QR is the common tangent Question 52. OP ⊥ AP ∴∠QOR + 15° + 15° = 180° ∠BAC = ∠PBA = 75° (a) √7 cm Solution: AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm RC = 11 – 4 = 7 cm BC = 8 + 6 = 14 cm = EK + EM DA and DC are tangents to the first circle from D OP ⊥ PQ (a) Let PQ be the tangent from Q to the circle with O as centre In quadrilateral OPTQ, Solution: The radius of the circle is [NCERT Exemplar] In the figure, AB is the diameter of a circle with centre O and AT is a tangent. Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively. Given: PT and PS are tangents from an external point P to the circle with centre O. (ii) ∠AOB + ∠COD = 180° ∠5 + ∠7 = 360° – 180° = 180° CD = CE …(iii) Similarly BO, CO and DO are the bisectors of (d) 80° Given that ∠ACB = 56°, calculate (i)∠CAB (ii)∠OAC. Proof: ∠1 = 90° … (i) Ellipsoid O’R = 13 cm A circle with centre O is inscribed in ∆ABC. PT = PR = 3.8 cm (b) 6 cm Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle. (b) 6 cm OA² = OB² + BA² (Pythagoras Theorem) = (3)² + (4)² = 9 + 16 = 25 = (5)² In right ∆OPQ, y + 5 = 8 Solution: ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 In the figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. (a) 12 cm ∠1 = \(\frac{120^{\circ}}{2}\) = 60° Solution: In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. OP = OQ = OE = 5 cm … [Radius of the circle AP = 12 cm ∠A + ∠B + ∠Q = 180° ∠OAB = ∠OBA … [Angles opposite equal sides (d) None of these Question 42. Solution: In the Figure Given Below 'O' is the Centre of the Circle. Simple curve is a curve that does not cross itself. PA = AE = x cm …[Tangent drawn from an external point Solution: In the figure, if PR is tangent to the circle at P and Q is the centre of the circle, then ∠POQ = ∠OPT + ∠POT = 180° – 90° = 90°, Question 20. BQ = 29 – 18 = 11 cm (2014OD) (b) a = 35°, b = 55° Solution: ∴ Perimeter = 4(AB) (d) 32.5 cm² => a = 35° [sum of cointerior is 180°] ∠B = 90° ∠BOC + ∠DOC = 180° 125° + ∠DOC = 180° ∠DOC = 18 Then, AB + AB = BC + BC … [From (i) BD = BF = 6 cm A chord of a circle of radius 10 cm subtends a right angle at its centre. BC = 6 cm, AB = 8 cm z = 15 – 12 = 3 Area of (∆ABC), Question 51. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Now, in ∆ABQ, ∠8 = ∠1 …(i) (c.p.c.t.) PQ = PT (tangents from P to the circle) Also, OA ⊥ AP = 9 + 9 If EK = 9 cm, calculate the perimeter of AEDF (in cm). AB + CD = AD + BC In the figure, a ∆ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively 6 cm 9 cm of lengths 6 cm and 9 cm. Solution: In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. But ∠OPQ = 30°, Question 35. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is [CBSE 2012] Similarly DC and DB are tangents to the second circle from D => ∠BAO = 90° – 65° = 25°, Question 50. OA ⊥ AP BC = DB + CD = 6 + 9 = 15 cm …(iii) => ∠POA = \(\frac { 1 }{ 2 }\) x 100° = 50°, Question 5. => QP = 12 cm Answer/ Explanation. (5)² = AQ² + (3)² AB and AC are tangents OB = 3 cm In Figure 19.2, Degree measure of PQR = x° The degree measure of a semicircle in 180° and that of a major arc is 360° minus the degree measure of the corresponding minor arc. Radius of circle is 5 cm, Question 38. EC² = 25 – 9 = 16 Adding (iv) and (v) In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. But ∠OPQ + ∠OQP = 180° – 120° = 60° ADCO and BDCO’ are squares AF + BF + CD = AE + BD + CE ∠OAP = ∠OBP = 90° QC = AC = 3 cm …(ii) OA² = OR² + AR² (b) 134° 1st method: ⇒ ∠PRQ = \(\frac{240^{\circ}}{2}\) = 120°, Question 14. ∴ OC ⊥ AB => 65° + ∠BAO = 90° Now in right ∆OPQ, (a) 3 cm Solution: Let OAPB be a sector of a circle with centre O and radius ‘ r ’ as shown in the figure. ⇒ ∠PAB + ∠PAB + 60° = 180° Solution: OT = OS …[radii of same circle Radius of the circle = 7 cm, Question 3. then, DB = AB – AD AB2 = 2(10)2 \(\frac{1}{2}=\frac{P Q}{P O}\) ∴ 2PQ = PO, Question 33. = (PA – AC) + (PB – BD) (b) 30° Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm PR = 12 cm = x + z = 12 …(iii) ∠ORQ = 90° – 75o = 150 Thus, the angle between the tangents is 50°. …(ii) D and AE are tangents to the circle from A (d) 62\(\frac { 1 }{ 2 }\)° AQ = BR …. (c) 4 cm From (i) and (ii), we get 12r = 24 BF = BD …(ii) Solution: (d) AB = BC (a) In the figure, two equal circles touch, each other externally at T ∠ORP = 90° The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. (a) 4 cm ⇒ 10 + z = 15 … [From (i) To prove. Proof: AF = AE ..(i) ∠PCA = ∠CBA …[Angle in the alternate segment In the given figure, PA and PB are tangents to the circle with centre O. ∴ Reflex ∠POQ = 2∠PRO ⇒ ∠3 + ∠3 = 180°…[From (i) => AQ² = (5)² – (3)² Similarly OP is radius and BC is tangents ∆OQP, cos 60° = \(\frac{PQ}{PO}\) Similarly BP and BQ are tangents AB = 12 cm …[Given Solution: But ∠POQ = ∠PQO = 90° (OP ⊥ PQ) If ∠APB = 60°, then calculate ∠OAB, (2011D) TP = TQuestion .. [Tangents drawn from an external point Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F F respectively. Now, we draw a chord AC of circle C2, which touches the circle C1 at B. (a) 18 cm AB = a + a = 2a Then, ∠ACB = Most radioactive sources produce gamma rays, which are of various energies and intensities. To prove: OP ⊥ XY (b) Radius of the circle = 6 cm Solution: Join AE, AF Another tangent at C intersects the line I at D and m at E. Prove that ∠DOE = 90°. Then, ∠QAD = ∠QBD …(i) [c.p,c.t.] Hence, ∠QAB = 70° [from Eq. Find the perimeter (in cm) of a square circum scribing a circle of radius a cm. ⇒ ∠3 + 24 = 180° … [Linear pair ∴ 75° + 75° + ∠RQS = 180° ∠POQ = 180°- (40° + 40°) = 180° – 80° = 100°, Question 27. (d) 8 cm AC =11 cm But AB = 29 cm Solution: Chord BC touches the inner circle at P (b) If figure, DE and DF are tangents to the circle drawn from D. In ATOS, OT ⊥ PT ∠AOB + 80o = 180° Proof: ∠ACO = 90° … [Tangent is ⊥ to the radius through the point of contact (c) 3 = 2AD [From (i) and (ii)] In rt. Let ∠TOP = θ …[Tangent is ⊥ to the radius through the point of contact CR = CO So, TPQ is an isosceles triangle. OA = 6 cm, OB = 3 cm and AP = 10 cm (a) √7 cm If AB is a diameter and ∠CAB = 30°. Solution: (c) 24 cm BC = BQ + QC (d) 3√3 cm In the figure, PQ and PR are two tangents to a circle with centre O. If the radius DA of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD, then calculate the length of CD. OP ⊥ AB ∴ AB = CD (Hence proved), Question 32. PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to (b) A circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively then the sum of two opposite sides is equal to the sum of other two opposite sides (b) In the figure, two circles with centre O and O’ touch each other externally But AP = PB (given) (d) 180° Now, In ∆OPR, AFDE is a square OA = OC …[Radius a = 35°, b = 55°, Question 34. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then calculate the length of AP (in cm). ∴ ∠opq = ∠oqp = 40° in ∆opq, ⇒ ∠poq + ∠opq + ∠oqp = 180° ⇒ ∠poq + 40° + 40° = 180° ∠poq = 180° – 80° = 100°. In ∆BOC, ∠3 + ∠4 + ∠8 = 180° …(v) Let PL = PN = x cm x = -14 or x = 7 Solution: Then calculate ∠BOC. AC = (x + 6) cm Solution: ∠PAB = ∠PBA … (i) …(Angles opposite to equal sides 2θ = 180° – 50° = 130° (d) 45° (b) 40° AC = BC (2015D) (b) In the figure, AD = 7 cm, Question 21. OR = 5 cm, Question 15. (a) PA and PB are the tangents to the circle from P and ∠APB = 80° Const. ∴ QM = 3 cm, RN = 5 cm, PL = 7 cm, Question 40. (c) 5 cm => 144 = 25 + PQ² ∠ABQ = \(\frac{1}{2}\) ∠AOQ = \(\frac{58^{\circ}}{2}\) = 29° (a) 60° ∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360° CQ = CR … (iii) (2014 D) In Figure, a right triangle ABC, circumscribes a circle of radius r. If AB and BC are of lengths 8 cm and 6 cm respectively, find the value of r. (2012OD) ∠OQP = 900 (b) Let O be the centre of two concentric circles C1 and C2, whose radii are r1 = 4 cm and r2 = 5 cm. (a) In the figure given below, P and Q are centers of two circles intersecting at B and C. ACD is a straight line. (b) 45° Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Now, in isosceles triangle POQwe have +++POQO++ PQ OQP =180c Equal sides subtend equal angles in isosceles triangle. ⇒ ∠OAB = 25°, Question 12. (d) In the figure, PQ is diameter XPY is tangent to the circle with centre O and radius 5 cm (2011D, 2012OD, 2013OD, 2014, 2015D & OD ∠1 = ∠2 … [From (i) & (ii)

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