{"id":28718,"date":"2021-07-08T18:15:23","date_gmt":"2021-07-08T12:45:23","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28718"},"modified":"2021-07-08T18:15:23","modified_gmt":"2021-07-08T12:45:23","slug":"ncert-solutions-for-class-10-maths-chapter-10-ex-10-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-10-ex-10-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 10 Circles Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFrom a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
\n(a) 7 cm Sol.
\n(b) 12 cm
\n(c) 15 cm
\n(d) 24.5 cm
\nSolution:
\n\"NCERT
\nTherefore, radius of the circle is 7 cm
\nSo, option (A) of the give question is right.<\/p>\n

Question 2.
\nIn figure, if TP and TQ are the two tangents to a circle with centre O so that \u2220POQ = 110\u00b0, then \u2220PTQ is equal to
\n(a) 60\u00b0
\n(b) 70\u00b0
\n(c) 80\u00b0
\n(d) 90\u00b0
\nSolution:
\n\"NCERT
\n\u2220OPT = 90\u00b0
\n\u2220OQT = 90\u00b0
\n\u2220POQ = 110\u00b0
\nTPOQ is a quadrilateral,
\n\u2234 \u2220PTQ + \u2220POQ = 180\u00b0 \u21d2 \u2220PTQ + 110\u00b0 = 180\u00b0
\n\u21d2 \u2220PTQ = 180\u00b0- 110\u00b0 = 70\u00b0
\nHence, correct option is (b).<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIf tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80\u00b0, then \u2220POA is equal to
\n(a) 50\u00b0
\n(b) 60\u00b0
\n(c) 70\u00b0
\n(d) 80\u00b0
\nSolution:
\nIn AOAP and AOBP
\nOA = OB [Radii]
\nPA = PB
\n\"NCERT
\n[Lengths of tangents from an external point are equal]
\nOP = OP [Common]
\n\u2234 \u2206OAP \u2245 \u2206OBP [SSS congruence rule]
\n\u2220AOB + \u2220APB = 180\u00b0 \u21d2 \u2220AOB + 80\u00b0 = 180\u00b0
\n\u21d2 \u2220AOB = 180\u00b0 – 80\u00b0 = 100\u00b0
\nFrom eqn. (i), we get
\n\u21d2 \u2220POA = \\(\\frac { 1 }{ 2 }\\) x 100\u00b0 = 50\u00b0
\nHence, correct option is (a)<\/p>\n

Question 4.
\nProve that the tangents drawn at the ends of a diameter of a circle are parallel.
\nSolution:
\nAB is a diameter of the circle, p and q are two tangents.
\nOA \u22a5 p and OB \u22a5 q
\n\u22201 = \u22202 = 90\u00b0
\n\u21d2 p || q \u22201 and \u22202 are alternate angles]
\n\"NCERT<\/p>\n

Question 5.
\nProve that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
\nSolution:
\nXY tangent to the circle C(0, r) at B and AB \u22a5 XY. Join OB.
\n\u2220ABY = 90\u00b0 [Given]
\n\u2220OBY = 90\u00b0
\n[Radius through point of contact is perpendicular to the tangent]
\n\u2234 \u2220ABY + \u2220OBY = 180\u00b0 \u21d2 AB Oiscollinear
\n\u2234 AB passes through centre of the circle.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nThe length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
\nSolution:
\nOA = 5 cm, AP = 4 cm OP = Radius of the circle
\n\u2220OPA = 90\u00b0 [Radius and tangent are perpendicular]
\n\"NCERT<\/p>\n

Question 7.
\nTwo concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
\nSolution:
\nRadius of larger circle = 5 cm Radius of smaller circle = 3 cm
\nOP \u22a5 AB
\n[Radius of circle is perpendicular to the tangent]
\nAB is a chord of the larger circle
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nA quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.
\nSolution:
\nAP = AS … (i)
\n[Lengths of tangents from an external point are equal]
\nBP = BQ … (ii)
\nCR = CQ … (iii)
\nDR = DS … (iv)
\n\"NCERT
\nAdding equations (i), (ii), (iii) and (iv), we get
\nAP + BP + CR + DR = AS + BQ + CQ + DS
\n\u21d2 (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
\n\u21d2 AB + CD = AD + BC
\nHence proved.<\/p>\n

Question 9.
\nIn figure, XY and X’Y’ are two parallel tangents to a circle , x with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that \u2220AOB = 90\u00b0.
\n\"NCERT
\nSolution:
\nTo Prove : \u2220AOC = 90\u00b0
\nJoin OC
\nXY and X’Y’ are two tangents
\nXY || X’Y’ (given) and AB is transversal … (i)
\n\u2220PAC + \u2220QBC = 180\u00b0 (Sum of int \u2220s)
\nIn \u2206PAO and \u2206CAO, we have
\nPA = AC (Tangent drawn from exterior point)
\nAO = AO (common)
\nPO = QC (radii of same circle)
\n\u2206 \u2206PAO \u2245 \u2206CDO
\n\u21d2 \u22201 = \u22202
\nSimilarly \u22203 = \u22204
\nNow, \u2220PAC + \u2220QBC = 180\u00b0
\n\u21d2 \u22201 + \u22202 + \u22203 + \u22204 = 180\u00b0
\n[\u2235 \u22201 = \u22202 and \u22204 = \u22203]
\n\u21d2 2(\u22202 + \u22203) = 180\u00b0
\nor \u22202 + \u22203 = 90\u00b0
\nNow, In \u2206 OAB,
\n\u2220AOB + \u22202 + \u22203= 180\u00b0 [put \u22202 + \u22203 = 90\u00b0]
\nor \u2220AOB+ 90\u00b0= 180\u00b0
\n\u21d2 \u2220AOB = 90\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nProve that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
\n\"NCERT
\nSolution:
\nGiven : Two tangents PA and PB are drawn from the point P to the circle with centre O. A and B are the points of contact of two tangents PA and PB respectively.
\nTo Prove : \u2220APB + \u2220AOB = 180\u00b0
\nConstruction: Join OA and OB
\nProof : Since the tangent PA is drawn to the circle at A with centre O, we have;
\nOA \u22a5 PA \u21d2 \u2220OAP = 90\u00b0
\nSimilarly, since the tangent PB is drawn to the circle at B with centre O. we have,
\nOB \u22a5 PB \u21d2 OBP = 90\u00b0
\nWe know the sum of four angles of a quadrilateral is 360\u00b0
\nSc, in the quadrilateral, OA PB, we have
\n\u2220OAP + \u2220APB + \u2220OBP + \u2220AOB = 360\u00b0
\n\u21d2 \u2220APB + \u2220AOB = 360\u00b0 (\u2220OAP + \u2220OBP) = 360\u00b0 – 180\u00b0
\n\u21d2 \u2220APB + \u2220AOB = 180\u00b0 Hence Proved.<\/p>\n

Question 11.
\nProve that the parallelogram circumscribing a circle is a rhombus.
\nSolution:
\nGiven: ABCD is a || gm AB, BC, CD and DA are touching a circle.
\nTo Prove: AB = BC = CD = DA
\nProof : Since the length of two tangents drawn from an external point to a circle are equal, we have
\n\"NCERT
\nAP = AS, BP = BQ, CR = CQ and DR = DS
\nAB + CD = AP + BP + CR + DR
\nAD + BC = AS + BQ + CQ + DS
\n= AS + DS + BQ + CQ
\n= AD + BC
\n\u21d2 AB + CD = AD + BC
\n\u21d2 2AB = 2AD
\nw AB = AD
\nBut AB= CD and BC = AD
\nAB = BC = CD = DA
\nHence || gm ABCD is a rhombus.<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nA triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.
\n\"NCERT
\nSolution:
\nLet the sides AB, BC and AC touch the in circle at E, D and F respectievly. Join the centre O of the circle with A, B, C, D, E and F.
\nThe point D divide BC in two parts CD = 6 cm and BD = 8 cm
\nSince tangent to a circle from an external point are equal CD = CF = 6 cm and DB = BE = 8 cm.
\n\"NCERT
\nAF = AE = x cm (say)
\nOD= OE = OF = 4 [radii of the incircle]
\nNow, Area of \u2206AOC = \\(\\frac { 1 }{ 2 }\\) (6 + x) x 4
\n= (6 + x) \u00d7 2 = (12 + 2x) sq. cm. … (i)
\nSimilarly Area of \u2206AOB = \\(\\frac { 1 }{ 2 }\\) (8 + x) \u00d7 4
\n= (16 + 2x) sq. cm … (ii)
\nArea of ABOC = \\(\\frac { 1 }{ 2 }\\) (6 + 8) x 4 = 28 sq. cm … (iii)
\nAdding equations (i), (ii) and (iii), we have
\nArea of \u2206ABC = [(12 + 2x) + (16 + 2x) + 28] sq.
\n= [56 + 4x] sq. cm … (iv)
\nAgain perimeter of \u2206ABC = AB + BC + CA = (x + 8) + (6 + 8) + (6 + x)
\n= (28 + 2x) cm
\n\"NCERTAB = x + 8 = 7 + 8 = 15 cm
\nAC = x + 6 = 7 + 6 = 13 cm.<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nProve that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
\nSolution:
\nGiven : A circle with centre O touches the side, PQ, QR, RS and PS of quadrilateral at the points A, B, C and D respectively.
\n\"NCERT
\nTo Prove : \u2220PQQ + \u2220ROS = 180\u00b0
\nand \u2220POS + \u2220QOR = 180\u00b0
\nConstruction: Join OA, OB, OC and OD.
\nProof: In \u2206POD and \u2206POA .
\nPA = FD [tangents from expemal point]
\nOP = OP [common]
\n\u2220ODP = \u2220ODA [each 90\u00b0]
\n\u2206POD = \u2206POA [RHS congruency]
\n\u22201 = \u22202 [C.P.C.T]
\nSimilarly we have
\n\u22203= \u22204; \u22205 = \u22206 and \u22207 = \u22208
\nNow \u22201 + \u22202 + \u22203+ \u22204 + \u22205 +
\n\u22206 + \u22207 + \u22208 = 360\u00b0
\nUsing the above result
\n\u22202 + \u22202 + \u22203+ \u22203+ \u22206 + \u22206 + \u22207 + \u22207 = 360\u00b0
\nor 2(\u22202 + \u22203)+ (\u22206 + \u22207)
\n\\(\\frac{360^{\\circ}}{2}\\) = 180\u00b0
\nand
\n\u22201 + \u22201 + \u22204 + \u22204 + \u22205 + \u22205 + \u22208 + \u22208 = 360\u00b0
\nor 2(\u22201 + \u22204 + \u22205 + \u22208) = 360\u00b0
\nor \u22201 + \u22204 + \u22205 + \u22208 = \\(\\frac{360^{\\circ}}{2}\\)
\nor (\u22201 + \u22208) + (\u22204 + \u22205) = 180\u00b0
\nor \u2220POS +\u2220QOR = 180\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2 Question 1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of …<\/p>\n

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NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2 Question 1. 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