{"id":2502,"date":"2020-09-08T10:42:03","date_gmt":"2020-09-08T05:12:03","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2502"},"modified":"2021-07-02T15:03:43","modified_gmt":"2021-07-02T09:33:43","slug":"kseeb-solutions-class-9-maths-chapter-12-ex-12-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-12-ex-12-5\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.5"},"content":{"rendered":"

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.5<\/h2>\n

Question 1.
\nIn Fig., A, B, and C are three points on a circle with centre O such that \u2220BOC = 30\u00b0 and \u2220AOB = 60\u00b0. If D is a point on the circle other than the arc ABC, find \u2220ADC.
\n\"KSSEB
\nSolution:
\n\u2220AOB + \u2220BOC = 60\u00b0 + 30\u00b0 = 90\u00b0
\n\u2234 \u2220AOC = 90\u00b0 (Angle subtended at the centre)
\n\u2220ADC = ?
\nThe angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
\n\u2234 Angle subtended at the centre
\n\u2220AOC= 2 \u00d7 \u2220ADC
\n90\u00b0 = 2 \u00d7 \u2220ADC 90
\n\u2234 \u2220ADC = \\(\\frac{90}{2}\\) = 45\u00b0
\n\u2234 \u2220ADC = 45\u00b0.<\/p>\n

Question 2.
\nA chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
\n\"KSSEB
\nSolution:
\n(i) Angle subtended in the circumference, \u2220BAD =?
\n(ii) Angle subtended in the circumference, \u2220BCD =?
\nIn this figure BD is a chord, OB is the radius, it is equal to OD.
\n\u2234 OB = OD = BD
\n\u2234 \u2206OBD is equilateral triangle.
\n\u2234 each angle is equal to 60\u00b0.
\n\u2234 angle subtended at the centre \u2220BOD = 60\u00b0.
\n(i) Angle subtended in the circumference
\n\u2220BAD= \\(\\frac{1}{2}\\) \u00d7 angle subtended at centre \u2220BOD
\n= \\(\\frac{1}{2}\\) \u00d7 60\u00b0
\n\u2234 \u2220BAD = 30\u00b0.<\/p>\n

(ii) The sum of either pair of opposite angles of a cyclic quadrilateral is 180\u00b0.
\n\u2234 In cyclic quadrilateral ABCD,
\n\u2220BAD + \u2220ACD = 180
\n30 + \u2220ACD = 180 (\u2235 \u2220BAD = 30\u00b0)
\n\u2234 \u2220ADC = 180 – 30
\n\u2234 \u2220ACD = 150\u00b0.<\/p>\n

Question 3.
\nIn Fig., \u2220PQR = 100\u00b0. where P, Q and R are points on a circle with centre O. Find \u2220OPR.
\n\"KSSEB
\nSolution:
\nIf \u2220PQR = 100\u00b0, then \u2220OPR = ?
\n\u2220PQR = 100\u00b0.
\n\u2234 \u2220POR = ?
\n\u2220POR = 2 \u00d7 \u2220PQR = 2 \u00d7 100\u00b0
\n\u2220POR = 200\u00b0
\n\u2234 \u2220POR – Reflex angle \u2220POR = 360\u00b0
\n\u2234 \u2220POR = 360 – 200
\n\u2234 \u2220POR = 160\u00b0
\nIn \u2206POR, OP = OR radii.
\n\u2234 \u2220OPR = \u2220ORB
\n\u2234 \u2220OPR + \u2220ORP + \u2220POR = 180\u00b0
\n\u2220OPR + \u2220OPR + 160 = 180
\n2\u2220OPR+ 160 = 180
\n2\u2220OPR = 180 – 160
\n2\u2220OPR = 20
\n\u2220OPR = \\(\\frac{20}{2}\\)
\n\u2234 \u2220OPR= 10\u00b0.<\/p>\n

Question 4.
\nIn Fig., \u2220ABC = 69\u00b0, \u2220ACB = 31\u00b0, find \u2220BDC.
\n\"KSSEB
\nSolution:
\nIf \u2220ABC = 69\u00b0, \u2220ACB = 31\u00b0, then \u2220BDC = ?
\nIn \u2206ABC,
\n\u2220ABC = \u2220ACB + \u2220BAC = 180\u00b0
\n69 + 31 + \u2220BAC = 180
\n100 + \u2220BAC = 180
\n\u2220BAC = 180 – 100
\n\u2220BAC = 80\u00b0
\n\u2220BAC and \u2220BDC are angles in same segment. These are equal.
\n\u2220BDC = \u2220BAC = 80\u00b0
\n\u2234 \u2220BDC = 80\u00b0.<\/p>\n

Question 5.
\nIn Fig., A, B, C, and D are four points on a circle. AC and BD intersect at a point E such that \u2220BEC = 130\u00b0 and \u2220ECD = 20\u00b0. Find \u2220BAC.
\n\"KSSEB
\nSolution:
\n\u2220BEC = 130\u00b0, \u2220ECD = 20\u00b0, \u2220BAC = ?
\nAngles formed by arc AD are \u2220ABD, \u2220ACD is equal.
\n\u2234 \u2220ABD = \u2220ACB = 20\u00b0
\n\u2220ABD = 20\u00b0
\n\u2220AEB + \u2220BEC = 180 (adjacent angles)
\n\u2220AEB + 130\u00b0 = 180
\n\u2220AEB = 180 – 130
\n\u2234 \u2220AEB = 50\u00b0
\nNow, in \u2206BAE,
\n\u2220BAE = \u2220ABE + \u2220AEB = 180\u00b0
\n\u2220BAE + 20 + 50 = 180
\n\u2220BAE + 70 = 180
\n\u2220BAE = 180 – 70
\n\u2234 \u2220BAE = 110\u00b0
\nBut \u2220BAE and \u2220BAC are the same.
\n\u2234 \u2220BAC = 110\u00b0<\/p>\n

Question 6.
\nABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \u2220DBC = 70\u00b0, \u2220BAC = 30\u00b0, find \u2220BCD. Further, if AB = BC, find \u2220ECD.
\n\"KSSEB
\nSolution:
\n\u2220DBC = 70\u00b0, \u2220BAC = 30\u00b0, then \u2220BCD =?
\nAB = BC, then \u2220ECD = ?
\n\u2220DAC and \u2220DBC are angles in the same segment.
\n\u2234 \u2220DAC = \u2220DBC = 70\u00b0
\n\u2234 \u2220DAC = 70\u00b0
\nABCD is a cyclic quadrilateral.
\n\u2234 Sum of opposite angles is 180\u00b0.
\n\u2220DAB + \u2220DCB = 180
\n100 + \u2220DCB = 180
\n[\u2235 \u2220DAC + \u2220BAC = \u2220DAB 70 + 30 = 100]
\n\u2220DCB = 180 – 100
\n\u2234 \u2220DCB = 80
\n\u2220DCB = \u2220BCD = 80
\n\u2234 \u2220BCD = 80
\nIn \u2206ABC, AB = AC,
\n\u2234 \u2220BAC = \u2220BCA = 30\u00b0
\n\u2220BCA = 30\u00b0
\n\u2220ECD = \u2220BCD – \u2220BCA = 80 – 30
\n\u2234 \u2220ECD = 50\u00b0.<\/p>\n

Question 7.
\nIf diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
\n\"KSSEB
\nSolution:
\nData: In cyclic quadrilateral ABCD, AC and BD are diameters of circle.
\nTo Prove: ABCD is a rectangle.
\nProof: AC is a diameter. \u2220ABC is angle in semicircle. Angle in semicircle is a right angle.
\n\u2234 \u2220ABC = 90\u00b0 \u2220ADC = 90\u00b0
\nSimilarly, BD is a diamgers, \u2220DAB, \u2220DCB are angles in semicircle.
\n\u2220DAB = 90\u00b0 \u2220DCB = 90\u00b0
\nNow, four angles of quadrilateral ABCD are right angles.
\n\u2234 \u2220A = \u2220B = \u2220C = \u2220D = 90\u00b0
\n\u2234 ABCD is a rectangle.<\/p>\n

Question 8.
\nIf the non-parallel sides of a trapezium are equal, prove that it is cyclic.
\n\"KSSEB
\nSolution:
\nData: ABCD is a trapezium, DC || AB and AD = BC which are non-parallel sides.
\nTo Prove: ABCD is a cyclic quadrilateral.
\nProof: ABCD is a trapezium.
\nAB || CD and AD = BC.
\n\u2234 \u2220DAB + \u2220CDA = 180\u00b0 …………. (i)
\n(sum of interior angles)
\nSimilarly, \u2220DCB + \u2220ABC = 180\u00b0 …………. (ii)
\nAs we know, AD = BC,
\n\u2234\u2220DAB = \u2220CBA
\nSubstituting Eqn. (i) in Eqn. (ii),
\n\u2220CBA + \u2220CDA = 180\u00b0
\n\u2220DAB + \u2220DCB = 180\u00b0
\nIf sum of angles of a quadrilateral is 180\u00b0, then it is cyclic quadrilateral.
\n\u2234 ABCD is a cyclic quadrilateral.<\/p>\n

Question 9.
\nTwo circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that \u2220ACP = \u2220QCD.
\n\"KSSEB
\nSolution:
\nTwo circles are drawn taking PQ and PR of a triangle as diameter. Let these intersect at P and S.
\nTo Prove: \u2220ACP = \u2220QCD
\nProof: \u2220ABP = \u2220QBD ………….. (i) (vertically opposite angles)
\n\u2220ABP = \u2220ACP ……….. (ii) (angles in the same segment)
\nSimilarly, \u2220QCD = \u2220QBD …………. (iii)(angles in the same segment)
\nFrom (i), (ii), and (iii),
\n\u2220ACP = \u2220QCD.<\/p>\n

Question 10.
\nIf circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
\n\"KSSEB
\nSolution:
\nData: Two circles are drawn taking PQ and PR of a triangle as diameter. Let these intersect at P and S.
\nTo Prove: The point of intersection ‘S’ is on the third side QR of \u2206PQR.
\nConstruction: Join PS.
\nProof: QAP is a diameter.
\n\u2234 \u2220QSP = 90\u00b0 (angle in the semi-circle) Similarly, ABR is a diameter.
\n\u2220PSR – 90\u00b0 (angle in the semicircle)
\n\u2220QSR = \u2220QSP + \u2220RSP = 90 + 90
\n\u2220QSR = 180\u00b0
\n\u2234 \u2220QSR is a straight angle.
\n\u2234 QSR is a straight line.
\n\u2234 Point ‘S’ is on the third side QR of \u2206PQR.<\/p>\n

Question 11.
\n\u2206ABC and \u2206ADC are two right triangles with common hypotenuse AC. Prove that \u2220CAD = \u2220CBD.
\n\"KSSEB
\nSolution:
\nData: \u2206ABC and \u2206ADC are right-angled triangles having common hypotenuse AC.
\nTo Prove: \u2220CAD = \u2220CBD
\nProof: In \u2206ABC, \u2220ABC = 90\u00b0
\n\u2234 \u2220BAC + \u2220BCA = 90\u00b0 …………. (i)
\nIn \u2206ADC, \u2220ADC = 90\u00b0
\n\u2234 \u2220DAC + \u2220DCA = 90\u00b0 …………… (ii)
\nAdding (i) and (ii),
\n\u2220BAC + \u2220BCA + \u2220DAC + \u2220DCA = 90 + 90
\n(\u2220BAC + \u2220DAC) + (\u2220BCA + \u2220DCA) = 180\u00b0
\n\u2220BAD + \u2220BCD = 180\u00b0
\n\u2234 \u2220ABC + \u2220ADC = 180\u00b0
\nIf opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral.
\n\u2234 \u2220CAD = \u2220CBD (\u2235 Angles in the same segment).<\/p>\n

Question 12.
\nProve that a cyclic parallelogram is a rectangle.
\n\"KSSEB
\nSolution:
\nABCD is a cyclic parallelogram in the circle with ‘O’ centre.
\nTo Prove: ABCD is a rectangle.
\nProof: ABCD is a cyclic parallelogram.
\n\u2234 AB || DC and AD || BC
\n\u2220A = \u2220C (Opposite angles of parallelogram)
\nBut, \u2220A + \u2220C = 180\u00b0 (Opposites angles of cyclic quadrilateral)
\n\u2220A + \u2220A = 180\u00b0
\n2\u2220A = 180\u00b0
\n\u2220A = \\(\\frac{180}{2}\\)
\n\u2234 \u2220A = 90\u00b0
\nIf each angle of the parallelogram is a right angle, it is a rectangle.
\n\u2234 ABCD is a rectangle.<\/p>\n

KSEEB Solutions for Class 9 Maths<\/a><\/h5>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.5 Question 1. In Fig., A, B, and C are three points on a circle with centre O such that \u2220BOC = 30\u00b0 and \u2220AOB = 60\u00b0. If D is a point on the circle other than the arc ABC, find \u2220ADC. Solution: \u2220AOB + \u2220BOC …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2502"}],"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\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2502"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2502\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2502"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2502"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2502"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}