{"id":4558,"date":"2023-02-15T05:08:39","date_gmt":"2023-02-15T05:08:39","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4558"},"modified":"2023-02-15T05:12:41","modified_gmt":"2023-02-15T05:12:41","slug":"ncert-solutions-class-9-maths-chapter-10-circles-ex-10-6","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-6\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 10 Circles Ex 10.6"},"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 10.6 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-10-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 10.5<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 10.6<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. \u00a0Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the following diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4561\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"369\" height=\"235\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans1.png 488w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans1-300x191.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans1-480x306.png 480w\" sizes=\"auto, (max-width: 369px) 100vw, 369px\" \/><br \/>\nWe need to prove that \u2220OAO&#8217;= \u2220OBO&#8217;<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394OAO\u2019 and \u0394OBO\u2019<\/span><br \/>\n<span style=\"color: #000000;\">OA = OB (Radii of a circle with center O)<\/span><br \/>\n<span style=\"color: #000000;\">O\u2019A = O\u2019B (Radii of a circle with center O\u2019)<\/span><br \/>\n<span style=\"color: #000000;\">OO\u2019= OO\u2019 (Common)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, by\u00a0SSS\u00a0criteria, \u0394OAO\u2019 and \u0394OBO\u2019 are congruent to each other.<\/span><br \/>\n<span style=\"color: #000000;\">By\u00a0CPCT, \u2220OAO&#8217;= \u2220OBO&#8217;<\/span><br \/>\n<span style=\"color: #000000;\">Hence it is proved that the line of centers of two intersecting circles subtends equal angles at the two points of intersection.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 , find the radius of the circle.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Draw two\u00a0parallel chords AB and CD of lengths 5 cm and 11 cm. Let the center of the circle be O. Join one end of each chord to the center.<br \/>\n<strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4562\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"250\" height=\"263\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans2.png 297w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans2-285x300.png 285w\" sizes=\"auto, (max-width: 250px) 100vw, 250px\" \/><br \/>\n<\/strong>Thus, MB = 2.5 cm and ND = 5.5 cm [The perpendicular drawn from the center of the circle to the chords bisects it.] <\/span><br \/>\n<span style=\"color: #000000;\">Let OM = x and ON = 6 &#8211; x<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394OMB <\/span><br \/>\n<span style=\"color: #000000;\">By\u00a0Pythagoras theorem,<\/span><br \/>\n<span style=\"color: #000000;\">OM<sup>2<\/sup>\u00a0+ MB<sup>2<\/sup>\u00a0= OB<sup>2<br \/>\n<\/sup>x<sup>2<\/sup>\u00a0+ 2.5<sup>2<\/sup>\u00a0= OB<sup>2<br \/>\n<\/sup>x<sup>2<\/sup>\u00a0+ 6.25 = OB<sup>2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sup>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;(1)<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394OND<\/span><br \/>\n<span style=\"color: #000000;\">By Pythagoras theorem,<\/span><br \/>\n<span style=\"color: #000000;\">ON<sup>2<\/sup>\u00a0+ ND<sup>2<\/sup>\u00a0= OD<sup>2<br \/>\n<\/sup>(6 &#8211; x)\u00b2 + 5.5<sup>2<\/sup>\u00a0= OD<sup>2<br \/>\n<\/sup>36 + x<sup>2<\/sup>\u00a0&#8211; 12x + 30.25 = OD<sup>2<br \/>\n<\/sup>x<sup>2<\/sup>\u00a0&#8211; 12x + 66.25 = OD<sup>2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/sup>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;(2)<\/span><br \/>\n<span style=\"color: #000000;\">OB and OD are the radii of the circle. Therefore OB = OD.<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0OB<sup>2<\/sup>\u00a0=\u00a0OD<sup>2<br \/>\n<\/sup>Equating (1) and (2) we get,<\/span><br \/>\n<span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0+ 6.25 = x<sup>2<\/sup>\u00a0&#8211; 12x + 66.25<\/span><br \/>\n<span style=\"color: #000000;\">12x = 60<\/span><br \/>\n<span style=\"color: #000000;\">x = 5<\/span><br \/>\n<span style=\"color: #000000;\">Substituting the value of x in (1),<\/span><br \/>\n<span style=\"color: #000000;\">OB<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ 6.25<\/span><br \/>\n<span style=\"color: #000000;\">OB<sup>2<\/sup>\u00a0= 5<sup>2\u00a0<\/sup>+ 6.25<\/span><br \/>\n<span style=\"color: #000000;\">OB<sup>2<\/sup>\u00a0= 31.25<\/span><br \/>\n<span style=\"color: #000000;\">OB = 5.59 (approx.)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, we get the radius of the circle = 5.59 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance 4 cm from the centre, what is the distance of the other chord from the centre?<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Consider the following diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4563\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"251\" height=\"255\" \/><br \/>\nAB = 6 cm CD = 8 cm MB = 3 cm ND = 4 cm<\/span><br \/>\n<span style=\"color: #000000;\">Given OM = 4 cm and let ON = x cm Consider \u0394OMB<\/span><br \/>\n<span style=\"color: #000000;\">By Pythagoras theorem,<\/span><br \/>\n<span style=\"color: #000000;\">OM\u00b2 + MB\u00b2 = OB\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">4\u00b2 + 3\u00b2 = OB\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">OB\u00b2 = 25<\/span><br \/>\n<span style=\"color: #000000;\">OB = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">OB and OD are the radii of the circle.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore OD = OB = 5 cm.<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394OND<\/span><br \/>\n<span style=\"color: #000000;\">By\u00a0Pythagoras theorem,<\/span><br \/>\n<span style=\"color: #000000;\">ON\u00b2 + ND\u00b2 = OD\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">x\u00b2 + 4\u00b2 = 5\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">x\u00b2 = 25 &#8211; 16<\/span><br \/>\n<span style=\"color: #000000;\">x\u00b2 = 9<\/span><br \/>\n<span style=\"color: #000000;\">x = 3<\/span><br \/>\n<span style=\"color: #000000;\">The distance of the chord CD from the center is 3 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that \u2220ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4564\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"419\" height=\"290\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans4.png 508w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans4-300x208.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans4-480x333.png 480w\" sizes=\"auto, (max-width: 419px) 100vw, 419px\" \/><br \/>\n<strong>To prove:\u00a0<\/strong>\u2220ABC = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (\u2220DOE &#8211; \u2220AOC)<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394AOD and \u0394COE,<\/span><br \/>\n<span style=\"color: #000000;\">OA = OC (Radii of the circle)<\/span><br \/>\n<span style=\"color: #000000;\">OD = OE (Radii of the circle)<\/span><br \/>\n<span style=\"color: #000000;\">AD = CE (Given)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2206AOD \u2245 \u2206COE (SSS Congruence Rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220OAD = \u2220OCE (By CPCT) &#8212;&#8212;&#8212;&#8211; (1)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ODA = \u2220OEC (By CPCT) &#8212;&#8212;&#8212;&#8211; (2)<\/span><br \/>\n<span style=\"color: #000000;\">Also,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220OAD = \u2220ODA (As OA = OD) &#8212;&#8212;&#8212;&#8211; (3)<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (1), (2), and (3), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">\u2220OAD = \u2220OCE = \u2220ODA = \u2220OEC<\/span><br \/>\n<span style=\"color: #000000;\">Let \u2220OAD = \u2220OCE = \u2220ODA = \u2220OEC = x<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394OAC,<\/span><br \/>\n<span style=\"color: #000000;\">OA = OC<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220OCA = \u2220OAC (Angle a)<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ODE,<\/span><br \/>\n<span style=\"color: #000000;\">OD = OE<\/span><br \/>\n<span style=\"color: #000000;\">\u2220OED = \u2220ODE (Angle y)<\/span><br \/>\n<span style=\"color: #000000;\">ADEC is a cyclic quadrilateral.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220CAD + \u2220DEC = 180\u00b0 (Opposite angles are\u00a0supplementary)<\/span><br \/>\n<span style=\"color: #000000;\">x + a + x + y = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">2x + a + y = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">y = 180\u00b0 &#8211; 2x &#8211; a &#8212;&#8212;&#8212;&#8211; (4)<\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220DOE = 180\u00b0 &#8211; 2y and, \u2220AOC = 180\u00b0 \u2212 2a<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DOE &#8211; \u2220AOC = 2a &#8211; 2y<\/span><br \/>\n<span style=\"color: #000000;\">= 2a &#8211; 2 (180\u00b0 &#8211; 2x &#8211; a) [From equation (4)]<\/span><br \/>\n<span style=\"color: #000000;\">= 4a + 4x &#8211; 360\u00b0 &#8212;&#8212;&#8212;&#8211; (5)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAC + \u2220CAD = 180\u00b0 (Linear pair)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220BAC = 180\u00ba &#8211; \u2220CAD = 180\u00b0 &#8211; (a + x) &#8212;&#8212;&#8212;&#8211; (6)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, \u2220ACB = 180\u00b0 &#8211; (a + x) &#8212;&#8212;&#8212;&#8211; (7)<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABC,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC + \u2220BAC + \u2220ACB = 180\u00b0 (Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC = 180\u00b0 &#8211; \u2220BAC &#8211; \u2220ACB<\/span><br \/>\n<span style=\"color: #000000;\">= 180\u00b0 &#8211; (180\u00b0 &#8211; a &#8211; x) &#8211; (180\u00b0 &#8211; a &#8211; x) [From (6) and (7)]<\/span><br \/>\n<span style=\"color: #000000;\">= 2a + 2x -180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> [4a + 4x &#8211; 360\u00b0]<\/span><br \/>\n<span style=\"color: #000000;\">Using Equation (5)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (\u2220DOE &#8211; \u2220AOC)<\/span><br \/>\n<span style=\"color: #000000;\">Hence it is proved that \u2220ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4565\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"325\" height=\"288\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans5.png 405w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans5-300x266.png 300w\" sizes=\"auto, (max-width: 325px) 100vw, 325px\" \/><br \/>\nLet ABCD be a rhombus in which diagonals intersect at point O, and a circle is drawn by taking side CD as its diameter. We know that a diameter subtends 90\u00b0 on the arc. <\/span><br \/>\n<span style=\"color: #000000;\">Therefore,\u00a0\u2220COD = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Also, in the\u00a0rhombus, the diagonals intersect each other at 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOB = \u2220BOC = \u2220COD = \u2220DOA = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">But\u00a0\u2220COD is 90\u00b0 and this can only happen on a semicircle with\u00a0diameter\u00a0DC since the angle subtended by the diameter on a semicircle is 90\u00b0.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Clearly, point O has to lie on the circle.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the circle passes through the point of intersection of its diagonals O.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE, = AD.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4566\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"370\" height=\"294\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans6.png 424w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans6-300x238.png 300w\" sizes=\"auto, (max-width: 370px) 100vw, 370px\" \/><br \/>\nWe can see that ABCE is a cyclic quadrilateral. <\/span><br \/>\n<span style=\"color: #000000;\">We know that in a cyclic quadrilateral, the sum of the opposite angles is 180\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AEC + \u2220CBA = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AEC + \u2220AED = 180\u00b0 (Linear pair)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220AED = \u2220CBA\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8211; (1)<\/span><br \/>\n<span style=\"color: #000000;\">We know that in a parallelogram, opposite angles are equal.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADE = \u2220CBA\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (2)<\/span><br \/>\n<span style=\"color: #000000;\">From (1) and (2),<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AED = \u2220ADE<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AD = AE (sides opposite to equal angles in\u00a0a triangle are equal).<\/span><br \/>\n<span style=\"color: #000000;\">Hence proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. AC and BD are chords of a circle which bisect each other. Prove that<br \/>\n(i) <\/strong>AC and BD are diameters;<strong><br \/>\n(ii) <\/strong>ABCD is a rectangle.<br \/>\n<strong>Answer &#8211;\u00a0<\/strong>Consider the diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4567\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"293\" height=\"298\" \/><br \/>\nLet AC and BD be two chords intersecting at O.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AOB and \u0394COD,<\/span><br \/>\n<span style=\"color: #000000;\">OA = OC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">OB = OD (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOB = \u2220COD (Vertically opposite angles)<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0\u0394AOB \u2245 \u0394COD (SAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">AB = CD (By CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, it can be proved that \u0394AOD \u2245 \u0394COB<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0AD = CB (By CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Since in quadrilateral ABCD, opposite sides are equal in length, ABCD is a 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;\">Therefore,\u00a0\u2220A = \u2220C<\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220A + \u2220C = 180\u00b0 (ABCD is a cyclic\u00a0quadrilateral)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220A = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">2\u2220A = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220A = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">ABCD is a parallelogram and one of its interior angles is 90\u00b0, therefore, it is a rectangle.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A is the angle subtended by chord BD, \u2220A = 90\u00b0, therefore, BD should be the diameter of the circle [Since, angle in a semicircle is a right angle]<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, AC is the diameter of the circle.<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0(i) AC and BD are diameters, and (ii)\u00a0ABCD is a rectangle, proved<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90\u00b0\u2013(\u00bd)A, 90\u00b0\u2013(\u00bd)B and 90\u00b0\u2013(\u00bd)C.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the following diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4568\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"292\" height=\"322\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans8.png 362w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans8-272x300.png 272w\" sizes=\"auto, (max-width: 292px) 100vw, 292px\" \/><br \/>\nA diagram is constructed as per the given question.<\/span><br \/>\n<span style=\"color: #000000;\">It is given that BE is the bisector of \u2220B, AD is the bisector of \u2220A and CF is the bisector of \u2220C.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220ABE = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\angle&amp;space;B}{2}\" alt=\"\\frac{\\angle B}{2}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220ADE = \u2220ABE (Angles in the same segment for chord AE)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220ADE = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\angle&amp;space;B}{2}\" alt=\"\\frac{\\angle B}{2}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Similarly, \u2220ADF = \u2220ACF = \u2220C\/2 (Angle in the same segment for chord AF) <\/span><br \/>\n<span style=\"color: #000000;\">\u2220D = \u2220ADE + \u2220ADF <\/span><br \/>\n<span style=\"color: #000000;\">= <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\angle&amp;space;B}{2}\" alt=\"\\frac{\\angle B}{2}\" width=\"28\" height=\"38\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\angle&amp;space;C}{2}\" alt=\"\\frac{\\angle C}{2}\" align=\"absmiddle\" \/>\u00a0 [Since \u2220ADE = <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\angle&amp;space;B}{2}\" alt=\"\\frac{\\angle B}{2}\" width=\"28\" height=\"38\" align=\"absmiddle\" \/>\u00a0 and \u2220ADF = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\angle&amp;space;C}{2}\" alt=\"\\frac{\\angle C}{2}\" align=\"absmiddle\" \/> ]<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (\u2220B + \u2220C )<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (180\u00b0 &#8211; \u2220A) [Angle sum property of triangle\u00a0ABC]<\/span><br \/>\n<span style=\"color: #000000;\">= 90\u00b0 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> A<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, it can be proved for<\/span><br \/>\n<span style=\"color: #000000;\">\u2220E = 90\u00b0 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> B<\/span><\/p>\n<p><span style=\"color: #000000;\">\u2220F = 90\u00b0 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> C.<\/span><br \/>\n<span style=\"color: #000000;\">Thus we have proved that the angles of the triangle DEF are 90\u00b0 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> A, 90\u00b0 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> B, 90\u00b0 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> C.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4569\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans9.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"424\" height=\"227\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans9.png 1006w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans9-300x161.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans9-768x411.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans9-480x257.png 480w\" sizes=\"auto, (max-width: 424px) 100vw, 424px\" \/><br \/>\nAB is the common chord to both circles.<\/span><br \/>\n<span style=\"color: #000000;\">Since\u00a0the circles are congruent, their radii are equal.<\/span><br \/>\n<span style=\"color: #000000;\">In triangles ABX and ABY<\/span><br \/>\n<span style=\"color: #000000;\">AB = AB (Common)<\/span><br \/>\n<span style=\"color: #000000;\">AX = AY (equal radii)<\/span><br \/>\n<span style=\"color: #000000;\">BX = BY (equal radii)<\/span><br \/>\n<span style=\"color: #000000;\">By\u00a0SSS congruence criteria, triangles\u00a0ABX and ABY are congruent.<\/span><br \/>\n<span style=\"color: #000000;\">Hence \u2220X = \u2220Y [CPCT]\u00a0 &#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">Now, we know that the\u00a0angle 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;\">Thus, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220APB = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220X\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (2)<\/span><\/p>\n<p><span style=\"color: #000000;\">\u2220AQB = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220Y\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;- (3)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore,\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220APB = \u2220AQB [From equations (1), (2) and (3)]<\/span><br \/>\n<span style=\"color: #000000;\">Consider the \u0394BPQ,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220APB = \u2220AQB<\/span><br \/>\n<span style=\"color: #000000;\">This implies that \u0394BPQ is an\u00a0isosceles triangle\u00a0as base angles are equal.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, we get BP = BQ, sides opposite to equal sides in a triangle are equal.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. In any triangle ABC, if the angle bisector of \u2220A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Consider the diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4570\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans10.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"348\" height=\"349\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans10.png 375w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans10-300x300.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.6-Ans10-150x150.png 150w\" sizes=\"auto, (max-width: 348px) 100vw, 348px\" \/><br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000;\">Let AE be the angle bisector of\u00a0\u2220A.<\/span><br \/>\n<span style=\"color: #000000;\">We need to prove that ED is the\u00a0perpendicular bisector\u00a0of BC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAE = \u2220CAE\u00a0 &#8212;&#8212;&#8212;&#8212;- (1) [Since, AE is the angle bisector of \u2220A]<\/span><br \/>\n<span style=\"color: #000000;\">Now, \u2220EBC = \u2220CAE &#8212;&#8212;&#8212;&#8212;-\u00a0 (2) [Angles subtended by the same arc EC]\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Also, \u2220ECB = \u2220BAE &#8212;&#8212;&#8212;&#8212;-\u00a0 (3) [Angles subtended by the same arc BE]<\/span><br \/>\n<span style=\"color: #000000;\">But we know that,\u00a0\u2220BAE = \u2220CAE [From equation (1)]<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0\u2220EBC =\u00a0\u2220ECB [From equations (2) and (3)]<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, BE = EC [Sides opposite to equal angles are equal]<\/span><br \/>\n<span style=\"color: #000000;\">Thus, point E is\u00a0equidistant\u00a0from the points B and C. This is only possible when E lies on the perpendicular bisector of BC.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, ED is the perpendicular bisector of BC.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the perpendicular bisector of side BC and the angle bisector of \u2220A meet on the circumcircle of triangle ABC at point E.<\/span><\/p>\n<p style=\"text-align: justify;\">\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 10.6 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-4558","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 v27.5) - 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