{"id":25811,"date":"2021-06-24T11:38:51","date_gmt":"2021-06-24T06:08:51","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25811"},"modified":"2021-06-24T11:40:05","modified_gmt":"2021-06-24T06:10:05","slug":"ncert-solutions-for-class-9-maths-chapter-10-ex-10-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-10-ex-10-5\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 10 Circles Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.5<\/h2>\n

Question 1.
\nIn Fig. 10.36, A, B and C are three points on a circle with centre O such that \u2220BOC = 30\u00b0 and \u2220AOB = 60\u00b0. If D In a point on the circle other than the arc ABC, find \u2220ADC.
\n\"NCERT
\nSolution:
\nWe have given,
\n\u2220BOC = 30\u00b0 and \u2220AOB = 60\u00b0
\n\u2220AQC = 30\u00b0 + 60\u00b0
\n\u2234 \u2220AOC = 90\u00b0 …….(i)
\nAgain, \u2220AOC and \u2220ADC subtended by same arc AC.
\nWe know that the 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\u2220AOC = 2\u2220ADC
\n\u21d2 90\u00b0 = 2\u2220ADC (From (i))
\n\u21d2 \u2220ADC = 45\u00b0<\/p>\n

\"NCERT<\/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\"NCERT
\nSolution:
\nIn circle whose centre is O. AB is a chord equal to its radius.
\n\u2234 OAB is a an equilateral triangle.
\nWe know that each angle of an equilateral triangle is 60\u00b0.
\n\u2234 AOB = 60\u00b0
\nNow, \u2220AOB = 2\u2220AQB
\n(The angle subtended by an arc at the centre is double the angle subtanded by it at any point on the remaining part of the circle)
\n\u21d2 60\u00b0 = 2\u2220AQB
\n\u21d2 \u2220AQB = 30\u00b0
\nAgain, QAPB is a cyclic quadrilateral.
\n\u2234 \u2220AQB + \u2220APB = 180\u00b0 (Sum of opposite angles, of a cyclic quadrilateral is 180\u00b0)
\n\u21d2 30 + \u2220APB = 180\u00b0
\n\u21d2 \u2220APB = 180\u00b0 – 30\u00b0 = 150\u00b0
\nTherefore, angle subtended by the chord at a point on the minor arc is 150\u00b0 and at a point on major arc is 30\u00b0.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn Fig. 10.37, \u2220PQR = 100\u00b0, where P, Q and R are points on a circle with centre O. Find \u2220OPR.
\n\"NCERT
\nSolution:
\nWe have given \u2220PQR = 100\u00b0.
\nSince, angle subtended by an arc at the centre is double tire angle subtended by it at any point on the remaining part of the circle.
\n\u2234 \u22201 = 2\u2220PQR
\n\u21d2 \u22201 = 2 \u00d7 100\u00b0
\n\u21d2 \u22201 = 200\u00b0
\nAgain, \u22201 + \u22202 = 360\u00b0
\n\u21d2 200\u00b0 + \u22202 = 360\u00b0 (\u2235 \u22201 = 200\u00b0)
\n\u21d2 \u22202 = 160\u00b0
\nNow, In \u2206POR
\nOP = OR (Radii of circle)
\n\u2234 \u2220OPR = \u2220ORP (Angle opposite to equal sides of \u2206 are equal)
\nTherefore,
\n\u2220OPR + \u2220ORP + \u22202 = 180 (Angle sum property of triangle)
\n\u21d2 \u2220OPR + \u2220OPR + 160 = 180
\n\u21d2 2\u2220OPR = 20
\n\u21d2 \u2220OPR = 10\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nIn Fig. 10.38, \u2220ABC = 69\u00b0, \u2220ACB = 31\u00b0, find \u2220BDC.
\n\"NCERT
\nSolution:
\nIn \u2206ABC,
\n\u2220ABC = 69\u00b0 and \u2220ACB = 31\u00b0
\n\u2234 \u2220ABC + \u2220ACB + \u2220BAC = 180\u00b0 (Angle sum property of \u2206)
\n\u21d2 69\u00b0+ 31\u00b0 + \u2220BAC = 180\u00b0
\n\u21d2 \u2220BAC = 80\u00b0
\nWe know that angles in the same segment of a circle are equal.
\n\u2234 \u2220BAC = \u2220BDC
\n\u2220BDC = 80\u00b0 (\u2235 \u2220BAC = 80\u00b0)<\/p>\n

Question 5.
\nIn Fig. 10.39, 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\"NCERT
\nSolution:
\nWe have, \u2220BEC = 130\u00b0 and \u2220ECD = 20\u00b0
\nAgain, \u2220BEC + \u2220CED = 180\u00b0 (linear pair)
\n\u21d2 130\u00b0 + \u2220CED = 180\u00b0
\n\u21d2 \u2220CED = 50\u00b0
\nNow, In triangle CDE
\n\u2220CED + \u2220ECD + \u2220CDE = 180\u00b0 (Sum of all angles of a \u2206)
\n\u21d2 50\u00b0 + 20\u00b0 + \u2220CDE = 180
\n\u21d2 \u2220CDE = 110\u00b0
\nWe know that angles on the same segments of a circle are equal.
\n\u2234 \u2220CDB = \u2220BAC
\nor, \u2220BAC = 110\u00b0 (\u2235 \u2220CDB = 110\u00b0)<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \u2220DBC = 70\u00b0, \u2220BAC is 30\u00b0, find \u2220BCD. Further, if AB = BC, find \u2220ECD.
\nSolution:
\nWe know that angles on the same segment of a circle are equal.
\n\"NCERT
\n\u2234 \u2220DAC = \u2220DBC
\nor, \u2220DAC = 70\u00b0 (\u2235 \u2220DBC = 70\u00b0)
\nAgain, ABCD is a cyclic quadrilateral.
\n\u2234 \u2220DAB + \u2220BCD = 180\u00b0
\n(Sum of opposite angles of a cyclic quadrilateral is 180\u00b0)
\n100\u00b0 + \u2220BCD = 180\u00b0 (\u2235\u2220DAB = \u2220DAC + \u2220BAC)
\n\u21d2 \u2220BCD = 80\u00b0
\nNow, in triangle ABC,
\nAB = BC (Given)
\n\u2234 \u2220BAC = \u2220BCA (Angle opposite to equal sides are equal)
\nor, \u2220BCA = 30\u00b0
\nAgain, \u2220BCD = \u2220BCA + \u2220ACD
\n\u21d2 80\u00b0 = 30\u00b0 + \u2220ACD
\n\u21d2 \u2220ACD = 50\u00b0
\n\u2234 \u2220ECD = 50\u00b0<\/p>\n

Question 7.
\nIf diagonals of a cyclic quadrilateral are diameter of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
\n\"NCERT
\nSolution:
\nGiven: ABCD is a cyclic quadrilateral in which diagonal AC and BD are diameter of circle.
\nTo prove that: ABCD is a rectangle.
\nProof: In ABCD,
\nAC = BD (Diameter of circle are equal)
\nAgain, \u2220ABC = 90\u00b0 (Angle in a semicircle is a right angle)
\nSimilarly, \u2220BAD = \u2220ADC = \u2220BCD = 90\u00b0
\nNow, we know that, if a quadrilateral has both diagonals are equal and each angle is 90\u00b0 then the quadrilateral is a rectangle.
\n\u2234 ABCD is a rectangle.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nIf the non parallel sides of a trapezium are equal, prove that it is cyclic.
\nSolution:
\nGiven: A trapezium ABCD in which AB || CD and AD = BC.
\nTo prove that: ABCD is a cyclic trapezium.
\nConstruction: Draw DE \u22a5 AB and CF \u22a5 AB.
\nProof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that \u2220B + \u2220D = 180\u00b0.
\n\"NCERT
\nIn triangles DEA and CFB, we have
\nAD = BC (Given)
\n\u2220DEA = \u2220CFB (Each equal to 90\u00b0)
\nDE = CF (Distance between two parallel lines is always same)
\nSo, by S- A-S congruency condition
\n\u2206DEA \u2245 \u2206CFB
\n\u2234 \u2220A = \u2220B and \u2220ADE = \u2220BCF (By CPCT)
\n\u21d2 90\u00b0 + \u2220ADE = 90\u00b0 + \u2220BCF (Add 90\u00b0 both side)
\n\u21d2 \u2220EDC + \u2220ADE = \u2220FCD + \u2220BCF (\u2235 \u2220EDC = 90\u00b0 and \u2220CD = 90\u00b0)
\n\u21d2 \u2220ADC = \u2220BCD
\n\u21d2 \u2220D = \u2220C
\nThus, \u2220A = \u2220B and \u2220C = \u2220D
\n\u2234 \u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0 (Sum of the angles of a equal is 360\u00b0)
\n\u21d2 2\u2220B + 2\u2220D = 360\u00b0
\n\u21d2 \u2220B + \u2220D = 180\u00b0
\nHence ABCD is a cyclic quadrilateral.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nTwo circles intersect at two points B and C. Through B, two line segment ABD and PBQ are drawn to intersect the circle at A, D and P, Q respectively (see Fig. 10.40). Prove that \u2220ACP = \u2220QCD.
\n\"NCERT
\nSolution:
\nWe know that angles in the same segment of a circle are equal.
\n\u2220D = \u2220Q and \u2220P = \u2220A
\nAgain, In \u2206PQC,
\n\u2220P + \u2220Q + \u2220PCQ = 180\u00b0 ……(i)
\n(Sum of all three angles of the triangle is 180\u00b0)
\nNow, In \u2206ADC,
\n\u2220A + \u2220D + \u2220ACD = 180\u00b0 …….(ii)
\n(Sum of all three angles of a triangle is 180\u00b0)
\nFrom equation (i) and (ii)
\n\u2220P + \u2220Q + \u2220PCQ = \u2220A + \u2220D + \u2220ACD
\n\u21d2 \u2220P + \u2220Q + \u2220PCQ = \u2220P + \u2220Q + \u2220ACD (\u2235 \u2220A = \u2220P and \u2220D = \u2220Q prove above)
\n\u2234 \u2220PCQ = \u2220ACD
\n\u21d2 \u2220PCQ – \u2220PCD = \u2220ACD – \u2220PCD (Subtract \u2220PCD both side)
\n\u21d2 \u2220QCD = \u2220ACP
\n\u21d2 \u2220ACP = \u2220QCD<\/p>\n

\"NCERT<\/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.
\nSolution:
\nJoin AD
\nSince angle in a semicircle is right angle.
\nTherefore,
\n\"NCERT
\n\u2220ADB = 90\u00b0 and \u2220ADC = 90\u00b0
\n\u21d2 \u2220ADB + \u2220ADC = 90\u00b0 + 90\u00b0
\n\u21d2 \u2220ADB + \u2220ADC = 180\u00b0
\n\u21d2 \u2220BDC = 180\u00b0
\nTherefore, BDC is a straight line or, D must be lie on BC.<\/p>\n

Question 11.
\nABC and ADC are two right triangles with common hypotenuse AC. Prove that \u2220CAD = \u2220CBD.
\nSolution:
\nGiven: ABC and DBC are two right angle triangles on same base AC.
\n\"NCERT
\nTo prove that: \u2220CAD = \u2220CBD
\nConstruction: Take AC as a diameter and draw a circle and join BD.
\nProof: We know that angle in a semicircle is a right angle.
\nTherefore, vertex B of \u2206ABC and vertex D of \u2206ADC must be lie on the circumference of the circle.
\nAgain we know that angles in the same segment of a circle are equal.
\n\u2234 \u2220CAD = \u2220CBD.<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nProve that a cyclic parallelogram is a rectangle.
\nSolution:
\nLet ABCD be acyclic parallelogram. In order to prove that it is a rectangle, it is sufficient to show that one of the angles of parallelogram ABCD is a right angle.
\n\"NCERT
\nNow ABCD is a parallelogram.
\n\u2220B = \u2220D ……(i) [\u2235 opposite angle of a || gm are equal]
\nAlso, ABCD is a cyclic quadrilateral
\n\u2234 \u2220B + \u2220D = 180\u00b0 ……(ii) (Sum of opposite angles of cyclic quadrilateral)
\n\u21d2 \u2220B + \u2220B = 180\u00b0 [\u2235 \u2220B = \u2220D from (i)]
\n\u21d2 2\u2220B = 180\u00b0
\n\u21d2 \u2220B = 90\u00b0
\nHence ABCD is a rectangle.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.5 Question 1. In Fig. 10.36, A, B and C are three points on a circle with centre O such that \u2220BOC …<\/p>\n

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.5 Question 1. 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