{"id":828,"date":"2020-09-15T10:22:06","date_gmt":"2020-09-15T04:52:06","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=828"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-4-ex-4-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2<\/h2>\n

In Questions 1 to 3, choose the correct option and give justification.<\/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
\nA) 7 cm.
\nB) 12 cm.
\nC) 15 cm.
\nD) 24.5 cm.
\nSolution:
\nA) 7 cm.
\n\"KSEEB
\nIn \u22a5\u2206OPQ, \u2220P = 90\u00b0
\n\u2234 As per Pythagoras theorem,
\nOP2<\/sup> + PQ2<\/sup> = OQ2<\/sup>
\nOP2<\/sup> + (24)2<\/sup> = (25)2<\/sup>
\nOP2<\/sup> + 576 = 625
\nOP2<\/sup> = 625 – 576
\nOP2<\/sup> = 49
\n\u2234 OP = 7 cm.<\/p>\n

Question 2.
\nIn the following 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
\nA) 60\u00b0
\nB) 70\u00b0
\nC) 80\u00b0
\nD) 90\u00b0
\nSolution:
\nB) 70\u00b0
\n\"KSEEB
\n\u2220POQ =110\u00b0 (given)
\nOP \u22a5PT, OQ \u22a5QT ( \u2235 Angle between radius and tangent)
\n\u2234 \u2220OPT = \u2220OQT = 90\u00b0
\nNow, OPTQ is a quadrilateral.
\n\u2234 \u2220O + \u2220P + \u2220Q + \u2220T = 360\u00b0
\n110\u00b0 + 90\u00b0 + 90\u00b0 + \u2220T = 360\u00b0
\n290\u00b0 + \u2220T = 360\u00b0
\n\u2234\u2220T = 360\u00b0 – 290\u00b0
\n\u2234 \u2220T = 70\u00b0
\n\u2234 \u2220PTQ = 70\u00b0<\/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 an angle of 80\u00b0, then \u2220POA is equal to
\nA) 50\u00b0
\nB) 60\u00b0
\nC) 70\u00b0
\nD) 80\u00b0
\nSolution:
\nA) 50\u00b0
\n\"KSEEB
\nOA, OB are radii,
\nPA, PB are tangents
\n\u2234 \u2220PAO = 90\u00b0, \u2220PBO = 90\u00b0
\nIn quadrilateral OAPB,
\n\u2220BOA = 180\u00b0 – 80\u00b0 = 100\u00b0
\nBut\u00a0\"\\angle
\n\"=\\frac{1}{2}
\n\u2234\u2220POA = 50\u00b0.<\/p>\n

Question 4.
\nProve that the tangents drawn at the ends of a diameter of a circle are parallel.
\nSolution:
\nData: The tangents drawn at the ends of a diameter of a circle are parallel.
\n\"KSEEB
\nAOB is the p diameter in the circle with centre ‘O’.
\nTangent PAQ is at the endpoint A, and
\nTangent RBS is A at the endpoint B.
\nOA is the radius,
\nPAQ is tangent
\n\u2234 \u2220PAO = 90\u00b0
\nOB is the radius,
\nRBS is tangent.
\n\u2234 RBO = 90\u00b0
\nNow, \u2220PAO + \u2220RBO = 90\u00b0 + 90\u00b0 = 180\u00b0
\nAB intersects straight lines PAQ and RBS.
\nSum of a pair of interior angles is 180\u00b0. Hence these lines are parallel to each other.
\n\u2234 PQ || RS proved.<\/p>\n

Question 5.
\nProve that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
\n\"KSEEB
\nSolution:
\nData: Perpendicular at the point of contact to the tangent to a circle passes through the centre.
\nPQ is the tangent to circle with centre \u2019O’.
\nLet the perpendicular drawn at the point P is \u2220RPQ.
\n\u2220RPQ = 90\u00b0 …………. (i)
\nRadius drawn to circle at the point of contact is perpendicular.
\n\u2234 \u2220OPQ = 90\u00b0 …………. (ii)
\nFrom eqn. (i) and eqn. (ii), we have
\n\u2220RPQ = \u2220OPQ = 90\u00b0 .
\nThis is a contradiction, because
\n\u2220RPQ is the part of \u2220OPQ.
\n\u2234 \u2220RPQ < \u2220OPQ
\n\u2234 \u2220RPQ \u2260\u2220OPQ.
\n\u2234 The perpendicular at the point of contact to the tangent to a circle passes through the centre.<\/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.
\n\"KSEEB
\nSolution:
\nData: OB is the radius, BA is the tangent.
\n\u2234 \u2220OBA = 90\u00b0, OA = 5 cm. (data)
\nLength of tangent, AB = 4 cm.
\n\u2234 Radius, OB =?
\nIn \u22a5\u2206OBA, \u2220OBA = 90\u00b0
\n\u2234 OB2<\/sup>\u00a0+ BA2<\/sup>\u00a0= OA2<\/sup>\u00a0.
\nOB2<\/sup>\u00a0+ (4)2<\/sup>\u00a0= (5)2<\/sup>
\nOB2<\/sup>\u00a0+ 16 = 25
\nOB2<\/sup> = 25 – 16
\nOB2<\/sup>\u00a0= 9
\n\u2234 OB = 3 cm.
\n\u2234 Radius of circle, OB = 3 cm.<\/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.
\n\"KSEEB
\nSolution:
\nRadius of big circle,
\nOA = OB = 5 cm.
\nRadius of small circle, OP = 3 cm.
\nAngle between radius and tangent is 90\u00b0.
\n\u2234 \u2220OPA = \u2220OPB = 90\u00b0
\n( \u2235 Chord AB is tangent to small circle.)
\nNow, in \u22a5\u2206 OPA, \u2220OPA = 90\u00b0
\nOP2<\/sup> + AP2<\/sup> = OA2<\/sup>
\n(3)2<\/sup> + AP2<\/sup> = (5)2<\/sup>
\n9 + AP2<\/sup> = 25
\n\u2234 AP2<\/sup> = 25 – 9
\nAP2<\/sup> = 16
\n\u2234 AP = 4 cm.
\nSimilarly, in \u22a5\u2206OPB, PB = 4 cm.
\n\u2234 Length of chord, AB = AP + PB = 4 + 4
\n\u2234 Chord, AB = 8 cm.<\/p>\n

Question 8.
\nA quadrilateral ABCD is drawn to circumscribe a circle (see the fig. given below). Prove that AB + CD = AD + BC.
\n\"KSEEB
\nSolution:
\nData: In the quadrilateral, ABCD is drawn to circumscribe a circle.
\nTo Prove: AB + CD = AD + BC
\nThe lengths of tangents drawn from an extremal point to a circle are equal.
\n\u2234 Tangents AP and AS are drawn from point A.
\n\u2234 AP = AS
\nTangents BP and BQ are drawn from point B.
\n\u2234 BP = BQ
\nTangents CQ, CR are drawn from point C.
\n\u2234 CQ = CR.
\nTangents DR, DS are drawn from point A.
\n\u2234 DR = DS
\nL.H.S.: AB + CD = AP + PB + CR + RD
\n= AS + BQ + CQ + DS
\n= AS + SD + BQ + CQ
\n\u2234 AB + CD = AD + BC
\n\u2234 L.H.S. = R.H.S.<\/p>\n

Question 9.
\nIn the following figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X’Y’ at B. Prove that \u2220AOB = 90\u00b0.
\n\"KSEEB
\nSolution:
\nData: XY and X\u2019Y\u2019 are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X\u2019Y\u2019 at B
\nTo Prove: \u2220AOB = 90\u00b0.
\nTangent XY || Tangent X\u2019Y\u2019
\n\u2234 \u2220PAB + \u2220QBA = 180\u00b0 (\u2235 interior angles)
\n\"\\frac{1}{2}
\n\u2220OAB + \u2220OBA = 90\u00b0
\nNow, in AOAB,
\n\u2220AOB + \u2220OAB + \u2220OBA = 180\u00b0
\n\u2220AOB + 90\u00b0 = 180\u00b0
\n\u2234 \u2220AOB = 180\u00b0 \u2013 90\u00b0
\n\u2234 \u2220AOB = 90\u00b0.<\/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.
\nSolution:
\n\"KSEEB
\nTo Prove: \u2220APB + \u2220AOB = 180\u00b0
\nOA is radius, PA is tangent.
\n\u2234\u2220PAO = 90\u00b0
\nOB is radius, PB is tangent.
\n\u2234 \u2220PBO = 90\u00b0
\nNow OAPB is a quadrilateral.
\n\u2234\u2220PAO + \u2220PBO = 90\u00b0 + 90\u00b0 = 180\u00b0
\nSum of four angles of a quadrilateral is 360\u00b0
\n\u2234 \u2220PAO + \u2220PBO + \u2220APB + \u2220AOB = 360\u00b0
\n180\u00b0 + \u2220APB + \u2220AOB = 360\u00b0
\n\u2220APB + \u2220AOB = 360\u00b0 \u2013 180\u00b0
\n\u2234 \u2220APB + \u2220AOB = 180\u00b0
\nIf the sum of two angles is equal to 180\u00b0, then they are supplementary angles.
\n\u2234 \u2220APB and \u2220AOB are supplementary angles.<\/p>\n

Question 11.
\nProve that the parallelogram circumscribing a circle is a rhombus.
\nSolution:
\n\"KSEEB
\nTo prove: Parallelogram ABCD is a rhombus.
\nABCD is a parallelogram in which \u2018O\u2019 is the centre of the interior circle drawn in the parallelogram.
\nTangents AC and BD are drawn which intersect at \u2018OP\u2019.
\nSides of the parallelogram AB. BC. CD and AD touch at the points P, Q, R, and S respectively.
\nOP, OS joined.
\nWe have OP \u22a5 AB, OS \u22a5 AD.
\nIn \u22a5\u2206 OPB and \u2206OSD,
\n\u2220OPB = \u2220OSD =90\u00b0
\nOB = OD (tangent is bisected)
\nOP = OS (Radii)
\n\u2234 \u2206OPB = \u2206OSD
\n\u2234 PB = SD \u2192 (1)
\nAP = AS \u2192 (2) (\u2235 tangent of the external point)
\nBy combining eqn. (1) and Eqn. (2),
\nAP + PB = AS + SC
\nAB = AD
\nSimilarly, AB = BC = CD = DA
\n\u2234 Parallelogram ABCD is a rhombus<\/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 the figure given). Find the sides AB and AC.
\nSolution:
\n\"KSEEB
\nTo Prove: Side AB = ? and Side AC = ?
\nBD = 8 cm. DC = 6 cm.
\nBE = 8 cm. CF = 6 cm.
\nAE = AF = x cm.
\nIn \u2206ABC,
\na = BC = BD + DC =8 + 6 =14 cm.
\nb = CA = (x + 6) cm.
\nc = AB = (x + 8) cm.
\n\\(S=\\frac{a+b+c}{2} \\quad=\\frac{14+(x+6)+(x+8)}{2}\\)
\n\\(=\\frac{2 x+28}{2}\\)
\n= (x + 14) cm.
\nArea of \u2206ABC = \\(=\\sqrt{s(s-a)(s-b)(s-c)}\\)
\n\\(=\\sqrt{(x+14) \\times x \\times 8 \\times 6}\\)
\n\\(=\\sqrt{48 x \\times(x+14)} c m^{2}\\) …………….. (i)
\nArea of \u2206ABC
\n= \u2206OBC + \u2206OCA + \u2206OAB W.
\n\\(=\\frac{1}{2} \\times 4 \\times a+\\frac{1}{2} \\times 4 \\times b+\\frac{1}{2} \\times 4 \\times c\\)
\n– 2(a + b + c)
\n= 2 \u00d7 2S
\n= 4S
\n= 4(x +14) cm2<\/sup>………………. (ii)
\nFrom eqn. (i) and eqn. (ii),
\n\\(\\sqrt{48 x \\times(x+14)}=4(x+14)\\)
\n48x \u00d7 (x + 14) = 16 x (x + 14)2
\n3x = x + 14
\n\u2234 x = 7 cm.
\nAB = c = x + 8 = 7 + 8 = 15 cm.
\nAC = b = x + 6 = 7 + 6 = 13 cm.<\/p>\n

Question 13.
\nProve that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
\n\"KSEEB
\nSolution:
\nData: Circle with centre \u2018O\u2019 is circumscribed in a quadrilateral ABCD.
\nTo Prove: \u2220AOD + \u2220BOC = 180\u00b0
\nSides of the quadrilateral AB, BC, CD, and DA touches at the points P, Q, R, and S respectively.
\nOA. OB, OC, OD, and OP, OQ, OR, OS are joined.
\nOA bisects \u2220POS.
\n\u22201 =\u22202
\n\u22203 = \u22204
\n\u22205 = \u22206
\n\u22207 = \u22208
\n2(\u22201 + \u22204 + \u22205 + \u22208) = 360\u00b0
\n(\u22201 + \u22208) (\u22204 + \u22205) = 180\u00b0
\n\u2234 \u2220AOD = \u2220BOD = 180\u00b0
\n\u2234 Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 4 Circles Ex 4.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2. Karnataka SSLC Class 10 Maths Solutions Chapter 4 Circles Exercise 4.2 In Questions 1 to 3, choose the correct option …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/828"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=828"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/828\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}