{"id":4520,"date":"2023-02-15T05:08:28","date_gmt":"2023-02-15T05:08:28","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4520"},"modified":"2023-02-15T05:13:01","modified_gmt":"2023-02-15T05:13:01","slug":"ncert-solutions-class-9-maths-chapter-10-circles-ex-10-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-10-circles-ex-10-4\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 10 Circles Ex 10.4"},"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 10.4 has been provided here to help the students in solving the questions from this exercise.<\/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-10-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 10.5<\/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; 10.4<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Two circles of radii 5 cm and 3 cm intersect at two points, and the distance between their centres is 4 cm. Find the length of the common chord.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>The\u00a0perpendicular bisector\u00a0of the common chord passes through the centers of both circles.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4537\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"342\" height=\"293\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans1.png 462w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans1-300x257.png 300w\" sizes=\"auto, (max-width: 342px) 100vw, 342px\" \/><br \/>\nGiven that the\u00a0circles intersect at two\u00a0points, so we can draw the above figure. Let AB be the common chord. Let O and O\u2019 be the centers of the circles, respectively. <\/span><br \/>\n<span style=\"color: #000000;\">O\u2019A = 5 cm, OA = 3 cm<\/span><br \/>\n<span style=\"color: #000000;\">OO\u2019 = 4 cm\u00a0 \u00a0 \u00a0 \u00a0(Given distance between the centres is 4cm)<\/span><br \/>\n<span style=\"color: #000000;\">Since the radius of the bigger circle is more than the distance between the 2 centers, we can say that the center of the smaller circle lies inside, the bigger circle itself.<\/span><br \/>\n<span style=\"color: #000000;\">OO\u2019 is the perpendicular bisector of AB.<\/span><br \/>\n<span style=\"color: #000000;\">So, OA = OB = 3 cm<\/span><br \/>\n<span style=\"color: #000000;\">AB = 3 cm + 3 cm = 6 cm (Since, O is the mid point of AB)<\/span><br \/>\n<span style=\"color: #000000;\">The length of the common chord is 6 cm.<\/span><br \/>\n<span style=\"color: #000000;\">It is also evident that the common chord is the diameter of the smaller circle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Let AB and CD be the 2 equal chords. AB = CD. Let the chords intersect at point E. Join OE.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4538\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"330\" height=\"302\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans2.png 330w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans2-300x275.png 300w\" sizes=\"auto, (max-width: 330px) 100vw, 330px\" \/><br \/>\nTo prove AE = CE and BE = DE. <\/span><br \/>\n<span style=\"color: #000000;\">Draw perpendiculars from the center O to the chords. This\u00a0Perpendicular bisects\u00a0the chord AB at M and CD at N.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, AM = MB = CN = DN\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000;\">In \u2206OME and \u2206ONE<\/span><br \/>\n<span style=\"color: #000000;\">\u2220M = \u2220N = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">OE = OE<\/span><br \/>\n<span style=\"color: #000000;\">OM = ON (Equal chords are\u00a0equidistant\u00a0from the center.)<\/span><br \/>\n<span style=\"color: #000000;\">By\u00a0RHS criteria, \u2206OME and \u2206ONE are congruent.<\/span><br \/>\n<span style=\"color: #000000;\">So by CPCT, ME = NE\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<\/span><br \/>\n<span style=\"color: #000000;\">We know that: CE = CN + NE and AE = AM + ME<\/span><br \/>\n<span style=\"color: #000000;\">From (i) and (ii), it is evident CE = AE<\/span><br \/>\n<span style=\"color: #000000;\">DE = CD &#8211; CE and BE = AB &#8211; AE<\/span><br \/>\n<span style=\"color: #000000;\">AB and CD are equal, CE and AE are equal. So, DE and BE are also equal. It is proved corresponding segments of equal chords are equal.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Let AB and CD be the two\u00a0equal chords. AB = CD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4538\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"319\" height=\"292\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans2.png 330w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans2-300x275.png 300w\" sizes=\"auto, (max-width: 319px) 100vw, 319px\" \/><br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000;\">Let the chords intersect at point E. Join OE. <\/span><br \/>\n<span style=\"color: #000000;\">Draw perpendiculars from the center O\u00a0to the chords. The\u00a0Perpendicular bisects the chord AB at M and CD at N. <\/span><br \/>\n<span style=\"color: #000000;\"><strong>To prove:<\/strong> \u2220OEN = \u2220OEM.<\/span><br \/>\n<span style=\"color: #000000;\">In \u2206OME and \u2206ONE,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220M = \u2220N = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">OE = OE<\/span><br \/>\n<span style=\"color: #000000;\">OM = ON (Equal chords are equidistant from the center.)<\/span><br \/>\n<span style=\"color: #000000;\">By RHS criteria, \u2206OME and \u2206ONE are\u00a0congruent. So, by CPCT, \u2220OEN = \u2220OEM<\/span><br \/>\n<span style=\"color: #000000;\">Hence proved that the line joining the point of intersection of two\u00a0equal chords to the center makes equal angles with the chords.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4542\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"386\" height=\"272\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Q4.png 386w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Q4-300x211.png 300w\" sizes=\"auto, (max-width: 386px) 100vw, 386px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Draw a perpendicular from the center of the circle OM to the line AD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4539\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"357\" height=\"315\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans4.png 441w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans4-300x265.png 300w\" sizes=\"auto, (max-width: 357px) 100vw, 357px\" \/><br \/>\nWe can see that BC is the chord of the smaller circle, and AD is the chord of the bigger circle. <\/span><br \/>\n<span style=\"color: #000000;\">We know that\u00a0perpendicular\u00a0drawn from the center of the circle\u00a0bisects\u00a0the chord.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BM = MC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000;\">and, <\/span><br \/>\n<span style=\"color: #000000;\">AM = MD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;- (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Subtracting (ii) from (i), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">AM \u2212 BM = DM\u00a0\u2212 CM<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB = CD<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Three girls, Reshma, Salma and Mandip, are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, and Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Let O be the center of the circle, and R, M and S denote Reshma, Mandip, and Salma respectively.<br \/>\nDraw a perpendicular OA to RS from O. Then RA = AS = 3 m.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4540\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"311\" height=\"286\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans5.png 311w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans5-300x276.png 300w\" sizes=\"auto, (max-width: 311px) 100vw, 311px\" \/><br \/>\nUsing the\u00a0Pythagoras theorem, we get OA = 4 m. <\/span><br \/>\n<span style=\"color: #000000;\">We can see that quadrilateral ORSM takes the shape of a kite. (Because OR = OM and RS = SM).<\/span><br \/>\n<span style=\"color: #000000;\">We know that the diagonals of a kite are perpendicular, and the main diagonal bisects the other diagonal.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220RNS will be 90\u00b0 and RN = NM<\/span><br \/>\n<span style=\"color: #000000;\">Area of \u2206ORS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 RS \u00d7 OA<\/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\" \/> \u00d7 6 \u00d7 4 <\/span><br \/>\n<span style=\"color: #000000;\">= 12\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">Also <\/span><br \/>\n<span style=\"color: #000000;\">Area of \u2206ORS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 OS \u00d7 RN<\/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\" \/> \u00d7 5 \u00d7 RN\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (2)<\/span><br \/>\n<span style=\"color: #000000;\">From equation (1) and (2)<\/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\" \/> \u00d7 5 \u00d7 RN = 12 <\/span><br \/>\n<span style=\"color: #000000;\">RN = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{24}{5}\" alt=\"\\frac{24}{5}\" align=\"absmiddle\" \/> = 4.8m<\/span><br \/>\n<span style=\"color: #000000;\">RM = 2 \u00d7 RN = 2 \u00d7 4.8 = 9.6 m<\/span><br \/>\n<span style=\"color: #000000;\">Distance between Reshma and Salma is 9.6 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. A circular park of radius 20m is situated in a colony. Three boys, Ankur, Syed and David, are sitting at equal distances on its boundary, each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Centre and Centroid are the same for an equilateral triangle, and it divides the median in the ratio 2 : 1.<br \/>\nLet A, D, S denote the positions of Ankur, David, and Syed, respectively.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4541\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"438\" height=\"323\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans6.png 489w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans6-300x221.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-10.4-Ans6-480x354.png 480w\" sizes=\"auto, (max-width: 438px) 100vw, 438px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">\u2206ADS is an\u00a0equilateral triangle\u00a0since all the 3 boys are equidistant from one another.<\/span><br \/>\n<span style=\"color: #000000;\">Let B denote the mid-point of DS, and hence AB is the median and perpendicular bisector of DS.<\/span><br \/>\n<span style=\"color: #000000;\">Hence \u2206ABS is a right-angled triangle with \u2220ABS = 90\u00ba.<\/span><br \/>\n<span style=\"color: #000000;\">O (centroid) divides the line AB in the ratio 2 : 1. So OA : OB = 2 : 1.<\/span><br \/>\n<span style=\"color: #000000;\">OA\/OB = 2\/1 <\/span><br \/>\n<span style=\"color: #000000;\">Since OA = 20\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">thus, OB = 10m<\/span><br \/>\n<span style=\"color: #000000;\">AB = OA + OB = 20 + 10 = 30m\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;(1)<\/span><br \/>\n<span style=\"color: #000000;\">Let the side of the equilateral triangle \u2206ADS be 2x.<\/span><br \/>\n<span style=\"color: #000000;\">AD = DS = SA = 2x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;(2)<\/span><br \/>\n<span style=\"color: #000000;\">Since B is the mid-point of DS, we get BS = BD = x\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;(3)<\/span><br \/>\n<span style=\"color: #000000;\">Applying Pythagoras theorem to \u2206ABD, we get:<\/span><br \/>\n<span style=\"color: #000000;\">AD<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ BD<sup>2<br \/>\n<\/sup>(2x)<sup>2<\/sup>\u00a0= 30<sup>2<\/sup>\u00a0+ x<sup>2<br \/>\n<\/sup>4x<sup>2<\/sup>\u00a0= 900 + x<sup>2<br \/>\n<\/sup>3x<sup>2<\/sup>\u00a0= 900<\/span><br \/>\n<span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0= 300<\/span><br \/>\n<span style=\"color: #000000;\">x = 10\u221a3<\/span><br \/>\n<span style=\"color: #000000;\">x = 17.32 <\/span><br \/>\n<span style=\"color: #000000;\">AD = DS = SA = 2x = 2 \u00d7 10\u221a3 = 20\u221a3<br \/>\nAD = DS = SA = 2x = 2 \u00d7 17.32 = 34.64 m<br \/>\nLength of the string = Distance between them = AD or DS or SA = 34.64 m. 0r 20\u221a3m.\u00a0<\/span><\/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 10.4 has been provided here to help the students in solving the questions from this exercise. 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-4520","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) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 9 Maths Chapter 10 Circles Ex 10.4 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 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