y2 + 12y + 36 = y2 +64 + 100 …[From (i) OP ⊥ XY or PQ ⊥ XY ∴ BE = (12 – x) cm ..[Tangents drawn from an external point are equal A chord of a circle of radius 10 cm subtends a right angle at its centre. PO = 20 … [radii Solution: BQ = and BP are tangents from B Now, in right angled ∆OBC (a) In the figure, a circle touches the sides of a quadrilateral ABCD (c) 4 cm ∴ r = 2 cm, Question 27. Now, in isosceles triangle POQwe have +++POQO++ PQ OQP =180c Equal sides subtend equal angles in isosceles triangle. => BC = 3 cm But ∠PBA = ∠ACB = 75° (Angles in the alternate segment) This is how you would get the mass of objects in a space shuttle, or something like it. If OA = 15 cm, then AP + AQ = There is a trick to measure the area of a circle. (b) 45° 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. In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ZPOQ = 70°, then ZTPQ is equal to [CBSE 2011] (a) 35° (b) 45° (d) 70° (c) 55° CR = CO We have, OP = 5 cm, OT = 13 cm (b) 45° OA² = AQ² + OQ² (c) 44° a + b + c = 15 ∠TOS + ∠OTS + ∠OST = 180° … [Angle-sum-property of a ∆ 90° = ∠ORQ + 75° Solution: ⇒ AC = \(\frac{1}{2}\) (48) = 24 cm A pair of tangents PQ and PR are drawn. (c) 24 cm (2012OD) EB = ED …(ii) Solution: PQ = 10 cm = x + y = 10 …(i) But AB = AC …[Given The length of chord AB parallel to XY and at a distance of 8 cm from P is (b) O is the centre of the circle. Solution: ⇒ OR = 2PR Similarly PT and PR are tangents Solution: AB = AE + EB = AE + AE = 2AE = 2x : As AS and AP are tangents to the circle from a point A (a) 4 cm AF + BF + CD = AE + BD + CE If OO’ is produced to meet the circle C (O’, r) at A and AT is a tangent to the circle C (O, r) such that O’Q ⊥ AT. (b) Let O be the centre of two concentric circles C1 and C2, whose radii are r1 = 4 cm and r2 = 5 cm. or 2 AD = AB + BC + CA, Question 14. (b) 3 cm ⇒ 22 – 8 = 2x ⇒ 2x = 14 = 2AD [From (i) and (ii)] In given figure, CP and CQ are tangents to a circle with centre O. (c) In the figure, two concentric circles of radii 3 cm and 5 cm with centre O In rt. AB2 = OA2 + OB2 …[Pythagoras’ theorem ∠RQS = 180° – 150o = 30°, Question 46. and AC = x + 6 = 13 cm. Solution: Similarly BP and BQ are tangents ∠BAC = ∠PBA = 75° PT is chord and OR is radius (c) 90° (c) 15 cm = ∠1 + ∠2 = 90° …[OD is equally inclined to the tangents In the figure, AP and BP are tangents to a circle with centre 0, such that AP = 5 cm and ∠APB = 60°. In the figure, PQ and PR are two tangents to a circle with centre O. Given: XY is a tangent at point P to the circle with centre O. At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. (d) a = 45°, b = 45° If AB is a diameter and ∠CAB = 30°. Thus, the angle between the tangents is 50°. ∠PTQ = 180° – (60° + 60°) = 180° – 120° = 60° ∠1 = \(\frac{120^{\circ}}{2}\) = 60° ∴ AC = CB … [⊥ from the centre bisects the chord, Question 22. Given: PT and PS are tangents from an external point P to the circle with centre O. AC2 = AB2 + BC2 …[Pythagoras’ theorem BP = BQ = 4 cm a + b + b + c + c + a = 30 XY is tangent and OP is the radius (iii) Now AE = AC – CE = AB – PE = 8 – 3 = 5 cm, Question 33. ⇒ ∠PRQ + ∠RPQ + ∠POR = 180°…[∆ Rule Measure: an action planned or taken to achieve a desired result. ⇒ ∠PRQ = \(\frac{240^{\circ}}{2}\) = 120°, Question 14. Solution: Also, OA ⊥ PA (d) 24.5 cm (b) 60° (d) 40° In the above figure, PQ is a chord of length 16 cm, of a circle of radius 10 cm. (a) 12 cm This happens with every point on the line XY except the point P. ∠BOC = 360 – 220° = 140°, Question 7. Use a calculator to add the numbers together to get the total square footage. ⇒ 24x = 144 – 64 ⇒ PT = \(\sqrt{144}\) Draw another common tangent through C which intersects AB at D, then DA = DC = DB ∠OPQ = 35° (b) 30° AE2 + ET2 = AT2 …(Pythagoras’ theorem In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. ∠OPR = 90° In right ∆OAQ, \(\frac{r}{3 r}=\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\) …[Let AO’ = O’X = OX = r ⇒ AO = r +r+ r = 3r In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. AP + AQ = 12+ 12 = 24 cm, Question 18. 2∠OAB = 50° (2011D, 2011OD, 2015 OD) (b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. ∴ ∠1 + ∠2 = 180° …(Co-interior angles ∴ ∠3 = ∠4 = 90° : Take a point Q on XY other than P and join to OQuestion Solution: Using a noise figure meter is the most straightforward way to measure noise figure. Let, PA = x cm, then AT = (12 – x) cm 82 + RO2 = (10)2 ∴ AB = \(2\left(\frac{10}{3}\right)=\frac{20}{3} \mathrm{cm}=6 \frac{2}{3}\) cm EA + EB = EC + ED …[Adding (i) & (ii) Also, OP is a bisector of line ∠APC ∠OQP = ∠POQ = 45°, Question 7. Let TP = x cm and TR = y cm (b) 6 cm (2016D) The object is therefore free to vibrate, and for a given stiffness of the spring the frequency of the vibrations enables the scientists to calculate the mass. (d) 6 cm (b) 60 cm Now ∠QPT = 60° … [Given Proof. (d) 25 cm But ∠B = 90° (given) ∠OTS = 60°/2 = 30° From P, at a distance of 8 cm AB is a chord drawn parallel to XY ∴ CQ = 38 – 27 = 11 cm = √(4 x 7) = 2√7 cm, Question 12. ∴ AB = AF + BF = x + 6 …(i) y + 5 = 8 OT ⊥ PQ …[Tangent is ⊥ to the radius through the point of contact (2011OD) If two tangents inclined at a angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to [NCERT Exemplar] => QP² = 169 – 25 = 144 = 12² Proof: In ∆OTP and AB + CD = (AS + DS) + (BQ + CQ) (a) 7 cm If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm, then the radius of the circle (in cm) is [CBSE 2013] y + 3 = 8 and ∠TBP = ∠TPB = y (say) (b) AD, AE and BC are the tangents to the circle at D, E and F respectively (c) 5.7 50° + 50° + ∠APB = 180° In the figure, if AP = PB, then ∴ ORDS is a square. To prove: x = 9 cm. (2012D) ∴ AOTP = AOSP …[R.H.S Mean absolute deviation is a way to describe variation in a data set. ∠ABQ = ∠BQR = 70° [alternate angles] ⇒ 2(x + y + z) = 30 AR = AP = 8 – r, Join OQ and OA BA = 4 cm ∴ ∠OTS = ∠OST = 30°, Question 37. Calculate the nominal equivalent resistance, tolerance (5%), minimum and maximum circuit resistance for the circuit shown in Figure 3, and record in Table 6. ∴ AC || DE, Question 47. BF = BD …(ii) RC = QC = 7 cm Solution: Proof. 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PT = PQ = 3.8 cm Prove that OT is the right bisector of line segment PQuestion (2015D) OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA BD = BF = 6 cm (b) Radius of the circle = 6 cm ∆TPQ is an isosceles ∆ and TO is the bisector of ∠PTQ , => BC² = 25 – 16 = 9 Let Radius OQ = r PD + QB = PA + QA = PQ, Question 25. So, TPQ is an isosceles triangle. ∠ABQ = \(\frac{1}{2}\) ∠AOQ = \(\frac{58^{\circ}}{2}\) = 29° In the figure, if AP = 10 cm, then BP = [Tangents drawn from an external point are equal Solution: AC = BC, Question 22. In rt. (c) 8 cm [∵ Lengths of tangents drawn from an external pt. then, DB = AB – AD Solution: PA = PB [Since, the length of tangents drawn from an ∠PBA = ∠PAB = θ [say] (c) 7 cm In ∆OPQ, OP = OQ Let AF = AE = x Draw rays, lines, & line segments. ⇒ ∠OPR + ∠POR + ∠ORP = 180° …[∆ Rule 60 = 30r ⇒ r = 2 cm, Question 16. ∴ ∠ABO = ∠ACO = 90° (d) 15 cm (5)² = AQ² + (3)² Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. The … = 2(EK) = 2(9) = 18 cm, Question 8. The given figure are classified as, (a) Figure (i), Figure (ii), Figure (iii), Figure (v) and Figure (vi) are Simple curves. (a) 4 If ∠PRQ = l and ∠OPQ = m, then find l + m. 0 votes. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. ∠B = 90° In rt. T & S ar(∆ABC) = ar(∆AOB) + ar(∆BOC) + ar (∆AOC) (b) In the figure, AP = PB To prove: PR = RQ In the figure, the sides AB, BC and CA of triangle ABC, touch a circle at P, Q and R respectively. ∠1 + ∠BAT = 90° 2∠1 = 180° – 60° = 120° (a) In the figure, APB is a tangent to the circle with centre O : Join OC (O’R)² = (RS)² + (O’S)² = (12)² + (5)² = 144 + 25 = 169 = (13)² PS = 2 PQ = 2 PR = 2 x 7.5 = 15 cm, Question 32. ∠OAY + ∠OED = 180° Radius of circle is 5 cm, Question 38. x = \(\frac{80}{24}=\frac{10}{3}\) cm In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. BC = 8 + 6 = 14 cm Solution: (d) 8 cm If EK = 9 cm, calculate the perimeter of AEDF (in cm). ∠OQP = 90° ∴ AB + CD = BC + AD (Hence proved), Question 25. 60° +∠OAB = 90° (2011D, 2012OD, 2013OD, 2014, 2015D & OD Similarly DH and DK are the tangent Solution: = 9 + 9 PT is tangent and OP is the radius. (b) In the figure, TP and TQ are the tangents drawn from T to the circle with centre O (b) 7.5 cm AF = AD In ∆PQR, Then calculate ∠BOC. BQ = BP = 3 cm In fact, this was a trick people used in the past to find the area under a curve on a graph (before technology gave us better methods). (d) 15 cm Write the measure of ∠OAB. (y + 6)2 = x2 + 102 5. Solution: Now in right ∆OQR (d) Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent If ∠OPQ = 35°, then FH = FM => ∠OPQ + ∠QPR = 90° To prove. ∠OPQ = \(\frac{1}{2}\)(∠QPR) ..[Tangents drawn from an external point are equal PA = PB …[Tangents drawn from an external point are equal Solution: OQ² = OP² + PQ² (Pythagoras Theorem) Similarly OP is radius and BC is tangents = (12 – 3) + (12 – 3) … [From (i), (ii) & (iii) (d) √119 cm (b) 7 cm PD = PA (tangents from P to the circle) In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. AP and AQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. To find: r = ? The VHA Office of Quality and Performance’s Performance Measurement Program (PMP) establishes a set of performance monitors that are consistent with the strategic goals of the VHA and designed to assure quality, access, and satisfaction with care provided in VHA facilities. If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that ∠QPR = 120°, prove that 2PQ = PO. By adding (i), (ii) & (iii) asked Dec 27, 2017 in Class IX Maths by navnit40 (-4,939 points) In the given figure, O is the centre of the circle. y + z = 8 cm …(ii) ∠OPQ + 50° = 90° Solution: => 2 ∠PBA = 150° 90° + ∠AOB + 90° + 50° = 360° OQ = BQ = 11 cm Solution: \(4 \sqrt{3 x(x+14)}\) = 2(2x + 28) (a) AD = AB + BC + CA Solution: BE = 12 – x = 12 – 7 = 5 cm OP, OQ and PQ are joined (c) Two circles of equal radii touch each other externally at P. OO’ produced meets at A OR = 5 cm, Question 15. For example, assume you are trying to buy flooring for a rectangular room with a nook. Solution: OPBQ is a square DR = DS Thus 29+OQP+0c =180c +OQP=45c Thus (b) is correct option. In ABOC, AB = \(10 \sqrt{2}\) cm, Question 5. Ellipsoid Now QR = PQ + PR = 4.5 + 4.5 = 9 cm, Question 26. OT = OS = r … [Given The distance of A from the centre of the circle is (a) 12 cm ∠ORP = 90° OF = OF …[Common Our mission is to provide a free, world-class education to anyone, anywhere. ∴ O lies on the bisector of ∠SAP => 150° + ∠OPQ = 180° 50°+ θ + θ = 180° (c) 80° OQ ⊥ PQ Proof: OQ ⊥ BC; OR ⊥ AC; OP ⊥ AB (2013D) AB = AC (2012OD) Solution: In ∆OQP, CF = CE (a) 11 cm (d) 4 cm Now in ∆POQ In the given figure, PA and PB are tangents to the circle with centre O. BC = 2 BP = 2 BQ = 2 AQ = 2 x 4 = 8 cm, Question 31. (c) 2 In the figure, the perimeter of ∆ABC is If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to = ED + DH + FH + EF Learn Insta try to provide online math tutoring for you. x + 5 = 12 => ∠BPA + ∠PBA = 150° Find ∠PCA. Join OA, O’B and O’O which passes through C => 2 (a + b + c) = 30 ∠1 + ∠OAB = 90° ∴ ∠PCA = 60°, Question 6. 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 [Tangent at any point of circle is perpendicular to radius throughly the point of contact] In the figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If the area of ∆ABC = 54 cm 2 then find the lengths of sides AB and AC. In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Question If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. In the given figure, PT is the tangent to the circle with centre O at P. PQ is a chord of the circle and ∠TPQ = 70°. ∠3 = 90° …[Given Two equal circles touch each other externally at C and AB is a common tangent to the circles. (d) 32.5 cm² (b) 4 cm Thus, quadrilateral PQOR is formed. (2013 OD, 2016 OD) A sub-group of quality measures are incorporated into the Five-Star Quality Rating System and used to determine scoring for the quality measures domain on Nursing Home Compare. (a) AC + AD = BD + CD Now AB + AC + BC = AE – BE + AD – CD + CF + BF 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 From A, AT is the tangent to the circle (O, r) OQ = 3 cm, PQ = 4 cm O’S = 5 cm and SR = 12 cm ∴ ∠ADO = ∠ACO …[Each 90° (b) 2 cm x + y + z = 15 …[∵ x + y = 12 ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 Length of chord AC = 2 BC = 2 x 3 = 6 cm, Question 47. (a) Given, PA and PB are tangent lines. Solution: 14 – 9 = AD ⇒ AD = 5 cm. From (ii) Similarly BP is tangent and OB is radius x = 7 cm Question 41. (b) In the figure, = 3 : 1 = \(\frac { 3 }{ 1 }\), Question 30. Find the perimeter (in cm) of a square circum scribing a circle of radius a cm. (d) Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°. (c) 30 cm² ∠OAP + ∠AOB + ∠OBP + ∠APB = 360° … [Quadratic rule ⇒ 2∠1 = 80° BF = BD …(ii) AF = a, BE = b and FC = c AQ + CQ = BR + CR ∠1 = ∠2 = 23 = 90° …[ Tangent is ⊥ to the radius B [through the point of contact AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it any point on the remaining part of the circle. ∴ CD = BD, Question 26. ∠OPT = 90° …[Tangent is ⊥ to the radius through the point of contact Similarly, ∠RQO = 15° PO is joined ∠AOB + ∠OAB + ∠OBA = 180° … [∆ rule => 144 = 25 + PQ² Draw rays, lines, & line segments. OA is radius and AP is the tangent => AR² = 25 – 9 = 16 = (4)² ∴ AF = AE … (i) ∠ATQ = 180° – (∠ABQ + ∠BAT) (2013OD) But AP = AQ = 12 cm (tangents from A to the circle) AP = 12 cm ∠TQP = ∠TPQ = 60° ∠5 + ∠7 = 360° – 180° = 180° AC || BC => 110° + ∠PTQ = 180° ∠1 + ∠1 + 60° = 180° Solution: => ∠QOP + 120° = 180° If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF. Given, ∠POQ = 130° AE ⊥ DE (c) 5 cm PQ and RS are the tangents drawn to the circles Solution: PQ = PT (tangents from P to the circle) 2∠OTS = 180° – 120° Therefore, Reflex ∠POQ = 2 ∠PRQ. Find the length of side AD. => ∠Q = 180° – (70° + 70°) = 40°. Then AO : AO’ = BC = 6 cm, AB = 8 cm Then length of chord BC which touches the inner circle at P is equal to (d) 50° OP² = OQ² + QP² => ∠OPQ = 90° – 60° = 30° (2016OD) OQ ⊥ AB and OQ bisects AB In probability and statistics, a probability mass function (PMF) is a function that gives the probability that a discrete random variable is exactly equal to some value. But R is mid-point of AB (a) 5 cm 2(x + y + z) = 30 AP = AQ Chord BC touches the inner circle at P If the lengths of sides AB, BC, and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF. (b) 6 cm Draw a tangent AB to the inner circle But ∠PAB + ∠PBA = 180° – 30° = 150° In rt. 12y = 164 – 36 = 128 ⇒ y = \(\frac{128}{12}=\frac{32}{3}\) ∠POQ + ∠QPO = 90° In ARSQ, ∠RSQ + ∠QRS + ∠RQS = 180° … [∆ Rule BC = DB + CD = 6 + 9 = 15 cm …(iii) (2015D) Solution: OB = 3 cm ∠OPQ = ∠OPT – ∠QPT => ∠APB = 90°, Question 41. In ∆PAB, ∠PAB + ∠PBA + ∠APB = 180° …[Angle-sum-property of a ∆ In the figure, AB is the diameter of a circle with centre O and AT is a tangent. 180° + 26 + 28 = 180° + 180° … [From (iii) ∠1 + ∠2 + 23+ 24 + 25 + 26+ 27 + ∠8 = 360° …(Complete angles (a) √7 cm An engineer can measure the noise figure over a certain frequency range, and the analyzer can display the system gain together with the noise figure to help the measurement. (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 => ∠PBA = 75° (a) 60° Related Surface Area Calculator | Volume Calculator. CF = 10 – x = 10 – 7 = 3 cm, Question 31. EC² = 25 – 9 = 16 EC² = 5² – 3² [OC = radius = 5 cm, OE = AE – AO = 8 – 5 = 3 cm] (a) 30 cm To prove: OP ⊥ XY The radius of the circle is (a) 20 cm 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: (b) 134° ∴ ∠POQ = 120° …[∠POQ = 180o – (30° + 30°) 2∠1 = 110° ⇒ ∠1 = 55° In the figure, two equal circles, with centres 0 and O’, touch each other A at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. In the larger circle, AB is a chord and OC ⊥ AB. Prove that: ∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360° AD = BD (a) 30° Hope given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS are helpful to complete your math homework. => 60° + 90° + ∠OPQ = 180° Solution: Relationship between length of an arc and its degree measure. (b) If figure, DE and DF are tangents to the circle drawn from D. ∴ ∠OCB = 90° … above theorem Area of (∆ABC), Question 51. 60 = r(5 + 12 + 13) x = 10 – 3 = 7 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. Similarly in right ∆RSO’ OC² = OE² + EC² Now, in right angled ∆OBC, by using Pythagoras theorem, RC = 11 – 4 = 7 cm A noise figure meter also has limitations. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS, Mark the correct alternative in each of the following : By adding (i) to (iv) (2015D) Solution: (c) 40° BP = BQ … (ii) We need to prove that: Solution: Solution: Another tangent at C intersects the line I at D and m at E. Prove that ∠DOE = 90°. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm). Now AO : AO’ = 3r : r (a) ∆ABC is circumscribed of circle with centre O In figure, if O P ∥ R S, ∠ O P Q = 1 1 0 o and ∠ Q R S = 1 3 0 o, then ∠ P Q R is equal to: (c) 35° (b) 60° ∠1 = ∠2 … [From (i) & (ii) Solution: ∠CBA = 180° – 30° – 90° = 60° If ∠QPR = 46°, then calculate ∠QOR. ∴PQ || RS, Question 19. ⇒ (5)2 + PT2 = (13)2 OQ ⊥ PQ O’Q ⊥ AT In the figure, a circle is inscribed in a ∆ABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. Identify points, lines, line segments, rays, and angles. Let r be the radius of the incircle (2016D) Solution: Const. (c) AQ = QC ∠POQ = 180° – 60° = 120°, Question 24. BP = √127 cm, Question 23. Solution: (c) 3 cm ⇒ ∠PRQ = ∠PQR …[∵ Angles opposite equal sides are equal A chord RS is drawn parallel to the tangent PQuestion Find ∠RQS. ∠POQ = ∠OQP In ॥gm, opposite sides are equal ∴ x = 7 … [As side of ∆ cannot be -ve (c) 14 cm AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm ∠1 + ∠1 = 180° – 70° In the figure, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. => ∠OPQ + ∠QPB = 90° Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively. In ∆AO’D and ∆AOC (a) \(\frac { 3\surd 3 }{ 2 }\) cm BQ = BC – CQ = 7 – 3 = 4 cm In ∆ABC, Then, ∠QAD = ∠QBD …(i) [c.p,c.t.] => 625 = r² + 576 AB and AC are tangents CR = 11 – 4 = 7 cm AB is joined (d) 90° ar(∆ABC) = ar(∆BOC) + ar(∆AOC) + ar(∆AOB), Question 30. ⇒ ∠POR = 30° (a) 30° PA = PB = 12 cm …(i) Solution: From the given figure below, find the azimuths for the bearings A: 242 29 Measures: N02 34'26"E. Rebar Found/Cap PLS 1256 27.00 Ofset Found/Cap 5756 Measures: N97 0531 260.51 Dirt Drive 182.54 29 272 54 29" 87° 05 31 267°05'31 From the given figure below, find the azimuths for the bearings B: 182° 34' 26" 358° 33' 34" 2° 34' 2° 34' 26" … But OP = OQ Solution: OT = OS …[radii of same circle ∠PAB = ∠PBA … (i) …(Angles opposite to equal sides Also QD is perpendicular to AB and QD bisects AB. ∠AOB = 180° – 80° = 100°, Question 13. (d) 3√3 cm ∠1 + ∠TPQ = 90° OP ⊥ PQ In the figure, APB is a tangent to a circle with centre O at point P. If ∠QPB = 50°, then the measure of ∠POQ is (a) 50° (d) 62\(\frac { 1 }{ 2 }\)° If ∠AOQ = 58°, find ∠ATQ. ⇒ ∠ORP = 90° 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. Similarly PE = EC and DP = DB Solution: (b) 18 cm In the given figure, PQ and PR are two tangents to a circle with centre O. Solution: ⇒ ∠1 = 40° (d) None of these (b) In the figure, Area of ∆ABC = 54 cm2 …[Given Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B. Joint: OA, OF, OE, OB and OC RN = MR = z cm ∴ APAB is an equilateral triangle (a) 60° If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to ∠QOR = 360° – 90° – 46° – 90° = 134°, Question 42. OQ is radius and AQB is the tangent AE = DE (side of square) Adding (i), (ii) and (iii), we get 90° + 23 = 180° Solution: In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. Then OT = (y + 6) cm So, when the degree measure of the angle at the centre is 1°, then area of sector is 1 / 360 ° ×πr 2 . Question 42. If EK = 9 cm, then the perimeter of ∆EDF is [CBSE 2012] Proof: ∠1 = 90° … (i) (c) 55° Solution: Const. ∴ AB = CD (Hence proved), Question 32. (d) 1.9 (c) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. ∠1 = ∠2 (b) QR Solution: TP = TQ = EK + EM (d) 180° (b) 9 => AQ² = 25 – 9 = 16 = (4)² 2AP = Perimeter of ∆ In the given figure, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. 60 = r(AB + BC + AC) (a) AC = AB (2015D) AB + AB = BC + BC … [From (i) In ATOS, ⇒ 2x + 2y + 2z = 10 + 8 + 12 AB + CQ = AC + BQ (a) In ∆DEF (b) 50° (b) 6 cm In the given figure, AB and AC are tangents to the circle with centre o such that ∠BAC = 40°. (AP + BP) + (CR + DR) = AS + BQ + CQ + DS Similarly PT = PR (d) In the figure, ∆ABC is the circumscribed a circle Solution: If SQ = 6 cm and QR = 4 cm, then OR = In Figure, common tangents AB and CD to the two circles with lo, centres O1 and O2 intersect at E. Prove that AB = CD. (2013OD) But OP is the bisector of ∠AOB ∴ \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}=\frac{1}{3}\), Question 49. R is mid-point of PT Solution: Hence, AB = AP = 5 cm …[∵ All sides of an equilateral A are equal, Question 36. ∠OPQ + ∠OQP + ∠POQ = 180° Now PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26 cm, Question 29. ∠ABC = 180° – (75° + 75°) = 180° – 150° = 30°, Question 28. (c) In the figure, PT is the tangent to the circle with centre O. r = 7 Mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean. (c) 140° (d) 3√2 cm (d) 15 (∠1 + ∠2 + 23 + 24) + 26 + 28 = 180° + 180° (a) In the figure, quadrilateral PQRS is circumscribed a circle Solution: QL = QM = y cm Hence AOB is a diameter of the circle with centre O. In the figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°. ∠BAT = 90° …. OQ and O’S are the radii of these circles and (2011OD, 2012OD, 2013D, 2014OD, 2015D) Solution: (a) In the figure, two equal circles touch, each other externally at T OP ⊥ AP (c) 60° (2015OD) AC = AE + CE = x + 9 …(ii) But ∠OPQ = 30°, Question 35. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. ∴ x + y = 12 cm …(i) But AP = PB (given) = 180 – (29 + 90) = 180° – 119° = 61°, Question 20. …[Pythagoras’ theorem In the figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. CD = AD = 4 cm = AR + BQ + CR => ∠COD = 180° – ∠AOB = 180° – 125° = 55°, Question 52. (b) AB + CD = BC + AD => 2r = 14 – 10 = 4 Diagonals of the octagon would be separated by (constructable) angles of 45 degrees. Similarly DC and DB are tangents to the second circle from D (b) 45° Solution: Solution: OA = 5 AFDE is a square (d) 3 cm Solution: In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°.
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