{"id":6080,"date":"2023-09-15T09:46:08","date_gmt":"2023-09-15T09:46:08","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6080"},"modified":"2023-09-15T09:46:08","modified_gmt":"2023-09-15T09:46:08","slug":"ncert-solutions-class-10-maths-chapter-10-circles-ex-10-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-10-circles-ex-10-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 10 Circles Ex 10.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 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 10 Mathematics <strong>Chapter &#8211; 10 Circles <\/strong>Exercise 10.2 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><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-10-circles-ex-10-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 10.1<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 10.2<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>In Q.1 to 3, choose the correct option and give a justification.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. From point Q, the length of the tangent to a circle is 24 cm, and the distance of Q from the centre is 25 cm. The radius of the circle is<br \/>\n<\/strong>(A) 7 cm<br \/>\n(B) 12 cm<br \/>\n(C) 15 cm<br \/>\n(D) 24.5 cm<br \/>\n<strong>Solution &#8211;\u00a0 <\/strong>Let O be the centre of the circle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6565\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"195\" height=\"131\" \/><br \/>\nSo, OP is perpendicular to PQ, i.e., OP \u22a5 PQ<br \/>\nIt is given that<br \/>\nOQ = 25 cm and PQ = 24 cm<br \/>\nBy using <strong>Pythagoras\u2019 theorem<\/strong> in \u25b3OPQ,<br \/>\nOQ<sup>2<\/sup>\u00a0= OP<sup>2<\/sup>\u00a0+PQ<sup>2<br \/>\n<\/sup>(25)<sup>2\u00a0<\/sup>= OP<sup>2<\/sup>+(24)<sup>2<br \/>\n<\/sup>OP<sup>2<\/sup> = 625 &#8211; 576<br \/>\nOP<sup>2<\/sup>\u00a0= 49<br \/>\nOP = 7 cm<br \/>\nSo, option A, i.e., 7 cm, is the radius of the given circle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that \u2220POQ = 110\u00b0, then \u2220PTQ is equal to<br \/>\n<\/strong>(A) 60\u00b0<br \/>\n(B) 70\u00b0<br \/>\n(C) 80\u00b0<br \/>\n(D) 90\u00b0<br \/>\n<strong>Solution &#8211; <\/strong>It is given that TP and TQ are tangents.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6566\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"195\" height=\"123\" \/><br \/>\nSo, OP \u22a5 PT and TQ \u22a5 OQ<br \/>\n\u2234 \u2220OPT = \u2220OQT = 90\u00b0<br \/>\nIn quadrilateral POQT,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sum of all interior angles = 360<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" width=\"11\" height=\"17\" name=\"Object11\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OPT + \u2220POQ + \u2220OQT + \u2220PTQ = 360<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" width=\"11\" height=\"17\" name=\"Object12\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 90<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" width=\"11\" height=\"17\" name=\"Object13\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" \/>+ 110\u00ba + 90<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" width=\"11\" height=\"17\" name=\"Object14\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" \/>\u00a0+<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_68d40597.gif\" width=\"17\" height=\"16\" name=\"Object15\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_68d40597.gif\" \/>PTQ = 360<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" width=\"11\" height=\"17\" name=\"Object16\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2105\/Chapter%2010_html_2245040b.gif\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220PTQ = 70\u00b0<br \/>\nSo, \u2220PTQ is 70\u00b0 which is option B.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80\u00b0, then \u2220 POA is equal to<br \/>\n<\/strong>(A) 50\u00b0<br \/>\n(B) 60\u00b0<br \/>\n(C) 70\u00b0<br \/>\n(D) 80\u00b0<br \/>\n<strong>Solution &#8211;\u00a0 <\/strong>It is given that PA and PB are tangents.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6567 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans3-300x203.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"229\" height=\"155\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans3-300x203.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans3.png 738w\" sizes=\"auto, (max-width: 229px) 100vw, 229px\" \/><br \/>\nNow, in the above diagram, OA is the radius to tangent PA, and OB is the radius to tangent PB.<br \/>\nSo, OA is perpendicular to PA, and OB is perpendicular to PB, i.e., OA \u22a5 PA and OB \u22a5 PB.<br \/>\nSo, \u2220OBP = \u2220OAP = 90\u00b0<br \/>\nNow, in the quadrilateral AOBP,<br \/>\nThe sum of all the interior angles will be 360\u00b0.<br \/>\nSo, \u2220AOB + \u2220OAP + \u2220OBP + \u2220APB = 360\u00b0<br \/>\nPutting their values, we get<br \/>\n\u2220AOB + 90\u00b0 + 90\u00b0 + 80\u00b0= 360\u00b0<br \/>\n\u2220AOB + 260\u00b0 = 360\u00b0<br \/>\n\u2220AOB = 100\u00b0<br \/>\nNow, consider the triangles \u25b3OPB and \u25b3OPA. Here,<br \/>\nAP = BP (Since the tangents from a point are always equal)<br \/>\nOA = OB (Which are the radii of the circle)<br \/>\nOP = OP (It is the common side)<br \/>\nNow, we can say that triangles OPB and OPA are similar using SSS congruency.<br \/>\n\u2234 \u25b3OPB \u2245 \u25b3OPA<br \/>\nSo, \u2220POB = \u2220POA<br \/>\n\u2220AOB = \u2220POA+\u2220POB<br \/>\n2 (\u2220POA) = \u2220AOB<br \/>\nBy putting the respective values, we get<br \/>\n\u21d2 \u2220POA = 100\u00b0\/2 = 50\u00b0<br \/>\nAs the angle \u2220POA is 50\u00b0, option A is the correct option.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6568\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"130\" height=\"168\" \/><br \/>\nRadius drawn to these tangents will be perpendicular to the tangents.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Thus, OA \u22a5 RS and OB \u22a5 PQ<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OAR = 90\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OAS = 90\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OBP = 90\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OBQ = 90\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">It can be observed that<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OAR = \u2220OBQ (Alternate interior angles)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OAS = \u2220OBP (Alternate interior angles)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Since alternate interior angles are equal, lines PQ and RS will be parallel.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.<br \/>\nDraw QP \u22a5 RP at point P, such that point Q lies on the circle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6569 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans5-300x237.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"162\" height=\"128\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans5-300x237.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans5.png 394w\" sizes=\"auto, (max-width: 162px) 100vw, 162px\" \/><br \/>\n\u2220OPR = 90\u00b0 (radius \u22a5 tangent)<br \/>\nAlso, \u2220QPR = 90\u00b0 (Given)<br \/>\n\u2234 \u2220OPR = \u2220QPR<br \/>\nNow, the above case is possible only when centre O lies on the line QP.<br \/>\nHence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. The length of a tangent from point A at a distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let us consider a circle centered at point O.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6570\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"195\" height=\"131\" \/><br \/>\nAB is a tangent drawn on this circle from point A.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Given that,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OA = 5cm and AB = 4 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>In \u0394ABO,<br \/>\n<\/strong>OB \u22a5 AB (Radius \u22a5 tangent at the point of contact)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Applying Pythagoras theorem in \u0394ABO, we obtain<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB<sup>2<\/sup>\u00a0+ BO<sup>2<\/sup>\u00a0= OA<sup>2<br \/>\n<\/sup>4<sup>2\u00a0<\/sup>+ BO<sup>2<\/sup>\u00a0= 5<sup>2<br \/>\n<\/sup>16 + BO<sup>2<\/sup> = 25<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BO<sup>2<\/sup> = 9<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BO = 3<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, the radius of the circle is 3 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6571\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"159\" height=\"144\" \/><br \/>\nOA \u22a5 PQ (As OA is the radius of the circle)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Applying Pythagoras theorem in \u0394OAP, we obtain<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OA<sup>2\u00a0<\/sup>+ AP<sup>2\u00a0<\/sup>= OP<sup>2<br \/>\n<\/sup>3<sup>2\u00a0<\/sup>+ AP<sup>2\u00a0<\/sup>= 5<sup>2<br \/>\n<\/sup>9 + AP<sup>2\u00a0<\/sup>= 25<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AP<sup>2\u00a0<\/sup>= 16<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AP = 4<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>In \u0394OPQ,<br \/>\n<\/strong>Since OA \u22a5 PQ,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AP = AQ (Perpendicular from the center of the circle bisects the chord)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2124\/Chapter%2010_html_4dd19828.gif\" width=\"15\" height=\"13\" name=\"Object32\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2124\/Chapter%2010_html_4dd19828.gif\" \/>PQ = 2AP = 2 \u00d7 4 = 8<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, the length of the chord of the larger circle is 8 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>The figure given is:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6572\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Que8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"217\" height=\"218\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Que8.png 217w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Que8-150x150.png 150w\" sizes=\"auto, (max-width: 217px) 100vw, 217px\" \/><br \/>\nIt can be observed that<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">DR = DS (Tangents on the circle from point D) &#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">CR = CQ (Tangents on the circle from point C) &#8212;&#8212;&#8212; (2)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BP = BQ (Tangents on the circle from point B) &#8212;&#8212;&#8212; (3)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AP = AS (Tangents on the circle from point A) &#8212;&#8212;&#8212; (4)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Adding all these equations, we obtain<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">DR + CR + BP + AP = DS + CQ + BQ + AS<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">CD + AB = AD + BC<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>9. In Fig. 10.13, XY and X\u2032Y\u2032 are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X\u2032Y\u2032 at B. Prove that \u2220 AOB = 90\u00b0.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6573\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Que9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"236\" height=\"182\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Let us join point O to C.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6574\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"236\" height=\"167\" \/><br \/>\nIn \u0394OPA and \u0394OCA,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OP = OC (Radii of the same circle)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AP = AC (Tangents from point A)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AO = AO (Common side)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u0394OPA\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2131\/Chapter%2010_html_m3437cc6c.gif\" width=\"15\" height=\"13\" name=\"Object34\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2131\/Chapter%2010_html_m3437cc6c.gif\" \/>\u0394OCA (SSS congruence criterion)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, P \u2194 C, A \u2194 A, O \u2194 O<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220POA = \u2220COA\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (i)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly, \u0394OQB\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"progressive\" src=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2131\/Chapter%2010_html_m3437cc6c.gif\" width=\"15\" height=\"13\" name=\"Object35\" align=\"ABSMIDDLE\" data-href=\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/curr\/1\/10\/9\/137\/2131\/Chapter%2010_html_m3437cc6c.gif\" \/>\u0394OCB<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220QOB = \u2220COB &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Since POQ is a diameter of the circle, it is a straight line.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, \u2220POA + \u2220COA + \u2220COB + \u2220QOB = 180\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">From equations (i) and (ii), it can be observed that<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">2\u2220COA + 2 \u2220COB = 180\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220COA + \u2220COB = 90\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220AOB = 90\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends \u2220AOB at center O of the circle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6575\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"182\" height=\"126\" \/><br \/>\nIt can be observed that<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OA (radius) \u22a5 PA (tangent)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, \u2220OAP = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly, OB (radius) \u22a5 PB (tangent)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OBP = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In quadrilateral OAPB,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sum of all interior angles = 360\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OAP +\u2220APB+\u2220PBO +\u2220BOA = 360\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">90\u00ba + \u2220APB + 90\u00ba + \u2220BOA = 360\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220APB + \u2220BOA = 180\u00ba<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>11. Prove that the parallelogram circumscribing a circle is a rhombus.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Since ABCD is a parallelogram,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = CD &#8212;&#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BC = AD &#8212;&#8212;&#8212;&#8212;&#8211; (ii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6576\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans11.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"136\" height=\"113\" \/><br \/>\n<\/strong>It can be observed that<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">DR = DS (Tangents on the circle from point D)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">CR = CQ (Tangents on the circle from point C)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BP = BQ (Tangents on the circle from point B)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AP = AS (Tangents on the circle from point A)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Adding all these equations, we obtain<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">DR + CR + BP + AP = DS + CQ + BQ + AS<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">CD + AB = AD + BC<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">On putting the values of equations (i) and (ii) in this equation, we obtain<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">2AB = 2BC<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = BC &#8212;&#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Comparing equations (i), (ii), and (iii), we obtain<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = BC = CD = DA<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, ABCD is a rhombus.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm, respectively (see Fig. 10.14). Find the sides AB and AC.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be\u00a0<i>x.<br \/>\n<\/i><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6577\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans12.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"194\" height=\"163\" \/><br \/>\nFrom the diagram, BD = 8 cm, CD = 6 cm<br \/>\nLet, AE = AF = x (The lengths of tangents drawn from an external point to a circle are equal)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">CD = CE = 6 cm (Tangents\u00a0from point C)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BD = BF = 8 cm (Tangents from point B)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Using\u00a0Heron&#8217;s formula, area of the triangle = \u221as (s &#8211; a)(s &#8211; b)(s &#8211; c)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">where <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">s = 1\/2 (a + b + c) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a, b and c are sides of a triangle <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = AB = x + 8 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">b = BC = 8 + 6 = 14 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">c = CA = 6 + x <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">s = 1\/2 ( x + 8 + 14 + 6 + x)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">s = 1\/2 (2x + 28) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">s = x + 14 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of \u0394ABC = \u221a(x + 14) ( x + 14 &#8211; x &#8211; 8)( x + 14 &#8211; 14)( x + 14 &#8211; x &#8211; 6)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u221a( x + 14) \u00d7 (6) \u00d7 (x) \u00d7 (8) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u221a48x ( x + 14) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u221a48 (x\u00b2 + 14x) square units &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (i) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of \u0394ABC = Area of \u0394AOC + Area of \u0394AOB + Area of \u0394BOC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 ( x + 6) \u00d7 4 + 1\/2 ( x + 8) \u00d7 4 + 1\/2 (14 \u00d7 4)\u00a0 \u00a0 \u00a0 [Since area of a triangle = 1\/2 \u00d7 l \u00d7 b \u00d7 h] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2x + 12 + 2x + 16 + 28 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 4x + 56 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 4( x + 14) square units &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Equating (i) and (ii) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u221a48 (x\u00b2 + 14x) = 4(x + 14) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Squaring both sides <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">48(x\u00b2 + 14x) = 4\u00b2 (x + 14)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">48(x\u00b2 + 14x) = 16 (x\u00b2 + 28x + 196)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Using (a + b)\u00b2 = a\u00b2 + 2ab + b\u00b2] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">48\/16 (x\u00b2 + 14x) = x\u00b2 + 28x + 196 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">3x\u00b2 + 42x = x\u00b2 + 28x + 196<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">3x\u00b2 &#8211; x\u00b2 + 42x &#8211; 28x &#8211; 196 = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">2x\u00b2 + 14x &#8211; 196 = 0 (divide this equation by 2) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">x\u00b2 + 7x &#8211; 98 = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Solving by\u00a0factorization\u00a0method,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">x\u00b2 + 14x &#8211; 7x &#8211; 98 = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">x ( x + 14) &#8211; 7 ( x + 14) = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">( x + 14)( x &#8211; 7) = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">x + 14 = 0 and x &#8211; 7 = 0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">x = &#8211; 14 and x = 7 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Since x represents length, it cannot be negative. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 x = 7 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = a = x + 8 = 7 + 8 = 15 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AC = c = 6 + x = 6 + 7 = 13 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Thus, AB = 15 cm and AC = 13 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>We know that\u00a0tangents\u00a0drawn from a\u00a0point\u00a0outside\u00a0the circle, subtend equal angles at the centre.<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6579\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans13.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"275\" height=\"256\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans13.png 385w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex10.2-Ans13-300x279.png 300w\" sizes=\"auto, (max-width: 275px) 100vw, 275px\" \/><br \/>\nIn the above figure, P, Q, R, S are points of contact <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AS = AP \u00a0(The tangents drawn from an external point to a circle are equal.) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220SOA = \u2220POA = \u22201 = \u22202 (Tangents drawn from a point outside of the circle, subtend equal angles at the centre) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u22203 = \u22204, \u22205 = \u22206, \u22207 = \u22208 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Since complete angle is 360\u00b0 at the centre, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">We have, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u22201 + \u22202 + \u22203 + \u22204 + \u22205 + \u22206 + \u22207 + \u22208 = 360\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">2 (\u22201 + \u22208 + \u22204 + \u22205) = 360\u00b0 (or) 2 (\u22202 + \u22203 + \u22206 + \u22207) = 360\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u22201 + \u22208 + \u22204 + \u22205 = 180\u00b0 (or) \u22202 + \u22203 + \u22206 + \u22207 = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">From above figure, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u22201 + \u22208 = \u2220AOD, \u22204 + \u22205 = \u2220BOC and \u22202 + \u22203 = \u2220AOB, \u22206 + \u22207 = \u2220COD <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Thus we have,\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220AOD + \u2220BOC = 180\u00b0 (or) \u2220AOB + \u2220COD = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220AOD and \u2220BOC are angles subtended by opposite sides of\u00a0quadrilateral\u00a0circumscribing a circle and the sum of the two is 180\u00b0. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence proved.<\/span><\/p>\n<p><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 10 (Circles)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 10 Circles Exercise 10.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 10 Circles NCERT Class 10 Maths Solution Ex &#8211; 10.1 Exercise &#8211; 10.2 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1099,1101,1044,1049,1048],"class_list":["post-6080","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-10-circles-solutions","tag-ncert-class-10-mathematics-exercise-10-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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