{"id":4521,"date":"2023-02-15T05:08:34","date_gmt":"2023-02-15T05:08:34","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4521"},"modified":"2023-02-15T05:12:49","modified_gmt":"2023-02-15T05:12:49","slug":"ncert-solutions-class-9-maths-chapter-10-circles-ex-10-5","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-5\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 10 Circles Ex 10.5"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 10 (Circles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 10 Circles <\/strong>Exercise 10.5 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 10: Circles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 10.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 10.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 10.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 10.4<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-6\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 10.6<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 10.5<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In Fig. 10.36, A, B and C are three points on a circle with centre O, such that\u00a0<b>\u2220<\/b>BOC = 30\u00b0 and\u00a0<b>\u2220<\/b>AOB = 60\u00b0. If D is a point on the circle other than the arc ABC, find\u00a0<b>\u2220<\/b>ADC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4548\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"196\" height=\"231\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong>The\u00a0angle subtended by an arc\u00a0at the center is double the angle subtended by it at any point on the remaining part of the\u00a0circle.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOC = \u2220AOB + \u2220BOC = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOC = 2 \u2220ADC (By Theorem 10.8)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADC = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220AOC <\/span><\/p>\n<p><span style=\"color: #000000;\">\u2220ADC = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 90 = 45\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ADC = 45\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. A 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.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.<br \/>\nA quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle and the sum of either pair of opposite angles of a cyclic\u00a0quadrilateral\u00a0is 180\u00b0.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4549\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"257\" height=\"259\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans2.png 298w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans2-150x150.png 150w\" sizes=\"auto, (max-width: 257px) 100vw, 257px\" \/><br \/>\nHere OA = OB = AB <\/span><br \/>\n<span style=\"color: #000000;\">Hence \u2206ABO becomes an\u00a0equilateral triangle.<\/span><br \/>\n<span style=\"color: #000000;\">Draw 2 points C and D on the circle such that they lie on the\u00a0major arc and minor arc, respectively.<\/span><br \/>\n<span style=\"color: #000000;\">Since \u2206ABO is an equilateral triangle, we get \u2220AOB = 60\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">For the arc AB, \u2220AOB = 2\u2220ACB as we know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ACB = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220AOB = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 60 = 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">As you can notice the points A, B, C, and D lie on the circle. Hence ABCD is a cyclic quadrilateral.<\/span><br \/>\n<span style=\"color: #000000;\">We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220ACB + \u2220ADB = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">30\u00b0 + \u2220ADB = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADB = 150\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">So, when the chord of a circle is equal to the\u00a0radius of the circle, the angle subtended by the chord at a point on the minor arc is 150\u00b0 and also at a point on the major arc is 30\u00b0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In Fig. 10.37,\u00a0<b>\u2220<\/b>PQR = 100\u00b0, where P, Q and R are points on a circle with centre O. Find\u00a0<b>\u2220<\/b>OPR.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4546\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"210\" height=\"243\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Since the angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.<br \/>\nSo, the reflex\u00a0<b>\u2220<\/b>POR = 2 \u00d7 <b>\u2220<\/b>PQR<br \/>\nWe know the values of angle PQR as 100\u00b0.<br \/>\nSo,\u00a0<b>\u2220<\/b>POR = 2\u00d7100\u00b0 = 200\u00b0<br \/>\n\u2234\u00a0<b>\u2220<\/b>POR = 360\u00b0 &#8211; 200\u00b0 = 160\u00b0<br \/>\nNow, in \u0394OPR,<br \/>\nOP and OR are the radii of the circle.<br \/>\nSo, OP = OR<br \/>\nAlso,\u00a0<b>\u2220<\/b>OPR =\u00a0<b>\u2220<\/b>ORP<br \/>\nNow, we know the sum of the angles in a triangle is equal to 180 degrees.<br \/>\nSo,<br \/>\n<b>\u2220<\/b>POR + <b>\u2220<\/b>OPR + <b>\u2220<\/b>ORP = 180\u00b0<br \/>\n<b>\u2220<\/b>OPR + <b>\u2220<\/b>OPR = 180\u00b0 &#8211; 160\u00b0<br \/>\nAs\u00a0<b>\u2220<\/b>OPR =\u00a0<b>\u2220<\/b>ORP<br \/>\n2 <b>\u2220<\/b>OPR = 20\u00b0<br \/>\nThus,\u00a0<b>\u2220<\/b>OPR = 10\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Fig. 10.38,\u00a0<b>\u2220<\/b>ABC = 69\u00b0,\u00a0<b>\u2220<\/b>ACB = 31\u00b0, find\u00a0<b>\u2220<\/b>BDC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4550\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"269\" height=\"281\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q4.png 535w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q4-288x300.png 288w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q4-480x501.png 480w\" sizes=\"auto, (max-width: 269px) 100vw, 269px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>We know that angles in the segment of the circle are equal, so,<br \/>\n\u2220BAC = \u2220BDC<br \/>\nNow, in the \u0394ABC, the sum of all the interior angles will be 180\u00b0.<br \/>\nSo, \u2220ABC + \u2220BAC + \u2220ACB = 180\u00b0<br \/>\nNow, by putting the values,<br \/>\n\u2220BAC = 180\u00b0 &#8211; 69\u00b0 &#8211; 31\u00b0<br \/>\n\u2234 \u2220BDC = 80\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E, such that \u2220 BEC = 130\u00b0 and \u2220 ECD = 20\u00b0. Find BAC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4547\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"217\" height=\"238\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Consider the straight-line BD. As the line AC intersects with the line BD, the sum of two adjacent angles so formed is 180\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, \u2220BEC + \u2220DEC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">130\u00b0 + \u2220DEC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DEC = 180\u00b0 &#8211; 130\u00b0 = 50\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Consider the \u2206DEC, the sum of all angles will be 180\u00ba.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DEC + \u2220EDC + \u2220ECD = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">50\u00b0 + \u2220EDC + 20\u00b0 = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220EDC = 180\u00b0 &#8211; 70\u00b0 = 110\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220BDC = \u2220EDC = 110\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">We know that angles in the same segment of a\u00a0circle\u00a0are equal.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220BAC = \u2220BDC = 110\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. ABCD is a cyclic quadrilateral whose diagonals intersect at point E. If \u2220 DBC = 70\u00b0, \u2220 BAC is 30\u00b0, find \u2220 BCD. Further, if AB = BC, find \u2220 ECD.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the following diagram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4551\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"273\" height=\"283\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans6.png 313w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans6-290x300.png 290w\" sizes=\"auto, (max-width: 273px) 100vw, 273px\" \/><br \/>\nIn the triangles, ABD and\u00a0BCD, \u2220CAD = \u2220CBD = 70\u00b0. (Angles in the same segment are equal)<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u2220BAD = \u2220CAB +\u00a0\u2220DAC<\/span><br \/>\n<span style=\"color: #000000;\">= 30\u00b0 +\u00a070\u00b0\u00a0 = 100\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220BAD =\u00a0100\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Since ABCD is a cyclic\u00a0quadrilateral, the sum of either pair of opposite angles of a cyclic quadrilateral is 180\u00ba.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD + \u2220BCD = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BCD = 180\u00b0 &#8211; 100\u00b0 = 80\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u2220BCD =\u00a080\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Also given AB = BC.<\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220BCA = \u2220BAC = 30\u00b0 (Base angles of\u00a0isosceles triangle\u00a0are equal)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ECD = \u2220BCD &#8211; \u2220BCA <\/span><br \/>\n<span style=\"color: #000000;\">\u2220ECD = 80\u00b0 &#8211; 30\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2220ECD = 50\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u2220ECD =\u00a050\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Draw a cyclic quadrilateral ABCD inside a circle with centre O, such that its diagonal AC and BD are two diameters of the circle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4552\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"297\" height=\"282\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans7.png 319w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans7-300x285.png 300w\" sizes=\"auto, (max-width: 297px) 100vw, 297px\" \/><br \/>\nWe know that the angles in the semi-circle are equal.<br \/>\nSo, \u2220 ABC = \u2220 BCD = \u2220 CDA = \u2220 DAB = 90\u00b0<br \/>\nSo, as each internal angle is 90\u00b0, it can be said that the quadrilateral ABCD is a rectangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>We know that, if the sum of a pair of opposite angles of a\u00a0quadrilateral\u00a0is 180\u00b0, the quadrilateral is cyclic.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Draw a trapezium ABCD with AB || CD<\/span><br \/>\n<span style=\"color: #000000;\">AD and BC are the non-parallel sides that are equal. AD = BC. Draw AM \u22a5 CD and BN \u22a5 CD.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4553\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"390\" height=\"257\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans8.png 390w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans8-300x198.png 300w\" sizes=\"auto, (max-width: 390px) 100vw, 390px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394AMD and \u0394BNC<\/span><br \/>\n<span style=\"color: #000000;\">AD = BC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AMD = \u2220BNC (90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">AM = BN (Perpendicular distance between two\u00a0parallel lines\u00a0is same)<\/span><br \/>\n<span style=\"color: #000000;\">By\u00a0RHS\u00a0congruence, \u0394AMD \u2245 \u0394BNC.<\/span><br \/>\n<span style=\"color: #000000;\">Using CPCT, \u2220ADC = \u2220BCD\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (1)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD and \u2220ADC are on the same side of\u00a0transversal\u00a0AD.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD + \u2220ADC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD + \u2220BCD = 180\u00b0 [From equation(1)] <\/span><br \/>\n<span style=\"color: #000000;\">This equation proves that the sum of opposite angles is\u00a0supplementary. Hence, ABCD is a cyclic quadrilateral.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Two 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 (see Fig. 10.40). Prove that \u2220 ACP = \u2220 QCD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4545\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Q9.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"235\" height=\"206\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>\u2220ACP and \u2220ABP lie in the same segment. Similarly, \u2220DCQ and \u2220DBQ lie in the same segment.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">We know that angles in the same segment of a\u00a0circle\u00a0are equal.<\/span><br \/>\n<span style=\"color: #000000;\">So, we get \u2220ACP = \u2220ABP and \u2220QCD = \u2220QBD<\/span><br \/>\n<span style=\"color: #000000;\">Also, \u2220QBD = \u2220ABP (Vertically opposite angles)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, \u2220ACP = \u2220QCD.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>We know that an angle in a semicircle is a\u00a0right angle. By using this fact, we can show that BDC is a line that will lead to proving that the point of intersection lies on the third side.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4554\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans10.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"387\" height=\"225\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans10.png 387w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans10-300x174.png 300w\" sizes=\"auto, (max-width: 387px) 100vw, 387px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Since angle in a\u00a0semicircle\u00a0is a right angle, we get:<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADB = 90\u00b0 and \u2220ADC = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADB + \u2220ADC = 90\u00b0 + 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220ADB + \u2220ADC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 BDC is a straight line.<\/span><br \/>\n<span style=\"color: #000000;\">D lies on BC<\/span><br \/>\n<span style=\"color: #000000;\">Hence, the point of intersection of circles lies on the third side BC.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that \u2220 CAD = \u2220CBD.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong>We know that, the sum of all angles in a triangle is 180\u00b0. <\/span><br \/>\n<span style=\"color: #000000;\">If the sum of pair of opposite angles in a quadrilateral is 180\u00b0, then it is a cyclic\u00a0quadrilateral.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4555\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans11.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"307\" height=\"318\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans11.png 355w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans11-289x300.png 289w\" sizes=\"auto, (max-width: 307px) 100vw, 307px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394ABC,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC + \u2220BCA + \u2220CAB = 180\u00b0 (Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">90\u00b0 + \u2220BCA + \u2220CAB = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BCA + \u2220CAB = 90\u00b0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394ADC,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220CDA + \u2220ACD + \u2220DAC = 180\u00b0 (Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">90\u00b0 + \u2220ACD + \u2220DAC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ACD + \u2220DAC = 90\u00b0\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (2)<\/span><br \/>\n<span style=\"color: #000000;\">Adding Equations (1) and (2), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BCA + \u2220CAB + \u2220ACD + \u2220DAC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">(\u2220BCA +\u2220ACD) + (\u2220CAB + \u2220DAC) = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BCD + \u2220DAB = 180\u00b0\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (3)<\/span><br \/>\n<span style=\"color: #000000;\">However, it is given that<\/span><br \/>\n<span style=\"color: #000000;\">\u2220B + \u2220D = 90\u00b0 + 90\u00b0 = 180\u00b0\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (4)<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180\u00b0. Therefore, it is a cyclic quadrilateral. <\/span><br \/>\n<span style=\"color: #000000;\">Since it is a cylclic quadrilateral the below figure can be drawn.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4556\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans11ii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"332\" height=\"286\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans11ii.png 372w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans11ii-300x258.png 300w\" sizes=\"auto, (max-width: 332px) 100vw, 332px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Consider chord CD. \u2220CAD and \u2220CBD are formed on the same segment CD. <\/span><br \/>\n<span style=\"color: #000000;\">\u2220CAD = \u2220CBD (Angles in the same segment are equal)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>12. Prove that a cyclic parallelogram is a rectangle.<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>The sum of either pair of opposite angles of a cyclic quadrilateral is 180\u00b0. Using this fact, we can show each angle of a cyclic\u00a0parallelogram as 90\u00b0, proving the statement it is a rectangle.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4557\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.5-Ans12.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"267\" height=\"258\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Let ABCD be the cyclic parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">We know that opposite angles of a parallelogram are equal.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A = \u2220C and \u2220B = \u2220D\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">We know that the sum of either pair of opposite angles of a cyclic\u00a0quadrilateral\u00a0is 180\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220C = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220A = 180\u00b0 (From equation (1))<\/span><br \/>\n<span style=\"color: #000000;\">2\u2220A = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">We know that if one of the\u00a0interior angles\u00a0of a parallelogram is 90\u00b0, all the other angles will also be equal to 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">Since all the angles in the parallelogram are 90\u00b0, we can say that parallelogram ABCD is a\u00a0rectangle.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 10 (Circles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 10 Circles Exercise 10.5 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 10: Circles NCERT Solution Class 9 Maths Ex &#8211; 10.1 NCERT Solution Class 9 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[719,702,186,720,701,5],"class_list":["post-4521","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-10-circles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-10-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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