{"id":127887,"date":"2021-09-15T14:46:13","date_gmt":"2021-09-15T09:16:13","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=127887"},"modified":"2021-09-15T14:46:17","modified_gmt":"2021-09-15T09:16:17","slug":"rd-sharma-class-10-solutions-chapter-10-exercise-10-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-10-exercise-10-2\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 10 Circles Exercise 10.2 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127914\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-10-Exercise-10.2-1.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-10-Exercise-10.2-1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-10-Exercise-10.2-1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2:\u00a0<\/strong>This exercise has exciting and difficult questions based on the concepts of tangent to a circle, tangent from a point on a circle, and tangent length. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> was prepared by our expert team at Kopykitab to assist students in clearing conceptual doubts and performing well in their exams. Students should comprehend the proper methods for solving questions in this exercise, which are available as <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-10-circles\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 10<\/strong><\/a> Exercise 10.2 PDF in the link provided below.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d7a1d4ea791\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" 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id=\"download-rd-sharma-class-10-solutions-chapter-10-exercise-102-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-10-Exercise-10.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-10-Exercise-10.2.pdf\">RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-class-10-solutions-chapter-10-exercise-102-important-question-with-answers\"><\/span>Access answers to RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 1.<br \/>If PT is a tangent at T to a circle whose center is O and OP = 17 cm, OT = 8 cm. Find the length of the tangent segment PT.<br \/>Solution:<br \/>PT is the tangent to the circle with center O, at T<br \/>Radius OT = 8 cm, OP = 17 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/41891576255_8122088a64_o.png\" alt=\"RD Sharma Class 10 Chapter 10 Circles \" width=\"286\" height=\"178\" \/><br \/>PT is the tangent segment<br \/>Now in right \u2206OPT,<br \/>OP\u00b2 = OT\u00b2 + PT\u00b2 (Pythagoras Theorem)<br \/>=&gt; (17)\u00b2 = (8)\u00b2 + PT\u00b2<br \/>=&gt; 289 = 64 + PT\u00b2<br \/>=&gt; PT\u00b2 = 289 \u2013 64 = 225 = (15)\u00b2<br \/>PT = 15 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 2.<br \/>Find the length of a tangent drawn to a circle with a radius of 5 cm, from a point 13 cm from the center of the circle.<br \/>Solution:<br \/>From a point P outside the circle with centre O, PT is the tangent to the circle and radius<br \/>OT = 5 cm, OP = 15 cm<br \/><img src=\"https:\/\/farm1.staticflickr.com\/876\/41891576365_32d3284b20_o.png\" alt=\"Circles Class 10 RD Sharma\" width=\"291\" height=\"178\" \/><br \/>OT \u22a5 PT<br \/>Now in right \u2206OPT,<br \/>OP\u00b2 = OT\u00b2 + PT\u00b2 (Pythagoras Theorem)<br \/>(13)\u00b2 = (5)\u00b2 + PT\u00b2<br \/>=&gt; 169 = 25 + PT\u00b2<br \/>=&gt; PT\u00b2 = 169 \u2013 25 = 144 = (12)\u00b2<br \/>PT = 12 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 3.<br \/>A point P is 26 cm away from the center O of a circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.<br \/>Solution:<br \/>From a point P outside the circle of center 0 and radius OT, PT is the tangent to the circle<br \/>OP = 26 cm, PT = 10 cm<br \/><img src=\"https:\/\/farm1.staticflickr.com\/879\/41891576525_4d9c69dd90_o.png\" alt=\"RD Sharma Class 10 Solutions Circles \" width=\"281\" height=\"173\" \/><br \/>Now in right \u2206OPT<br \/>Let r be the radius<br \/>OP\u00b2 = OT\u00b2 + PT\u00b2 (Pythagoras Theorem)<br \/>=&gt; (26)\u00b2 = r\u00b2 + (10)\u00b2<br \/>=&gt; 676 = r\u00b2 + 100<br \/>=&gt; 676 \u2013 100 = r\u00b2<br \/>=&gt; r\u00b2 = 576 = (24)\u00b2<br \/>r = 24<br \/>Hence radius of the circle = 24 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 4.<br \/>If from any point on the common chord of two intersecting circles, tangents are drawn to the circles, prove that they are equal.<br \/>Solution:<br \/>Given: QR is the common chord of two circles intersecting each other at Q and R<br \/>P is a point on RQ when produced From PT and RS are the tangents drawn to tire circles with centers O and C respectively<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/41891576735_5f1a833abb_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 10 Circles \" width=\"314\" height=\"223\" \/><br \/>To prove : PT = PS<br \/>Proof: PT is the tangent and PQR is the secant to the circle with center O<br \/>PT\u00b2 = PQ x PR \u2026.(i)<br \/>Similarly, PS is the tangent and PQR is the secant to the circle with center C<br \/>PS\u00b2 = PQ x PR \u2026.(ii)<br \/>From (i) and (ii)<br \/>PT\u00b2 = PS\u00b2<br \/>PT = PS<br \/>Hence proved.<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 5.<br \/>If the sides of a quadrilateral touch a circle, prove that the sum of a pair of opposite sides is equal to the sum of the other pair.<br \/>Solution:<br \/>Given: The sides of a quadrilateral ABCD touch the circle at P, Q, R, and S respectively<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/41891577025_bea8cc8be6_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 10 Circles \" width=\"209\" height=\"214\" \/><br \/>To prove : AB + CD = AP + BC<br \/>Proof : AP and AS are the tangents to the circle from A<br \/>AP = AS \u2026.(i)<br \/>Similarly BP = BQ \u2026\u2026(ii)<br \/>CR = CQ \u2026.(iii)<br \/>and DR = DS \u2026.(iv)<br \/>Adding, we get<br \/>AP + BP + CR + DR = AS + BQ + CQ + DS<br \/>=&gt; (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)<br \/>=&gt; AB + CD = AD + BC<br \/>Hence proved.<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 6.<br \/>Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle. [NCERT Exemplar]<br \/>Solution:<br \/>Let C<sub>1<\/sub>\u00a0and C<sub>2<\/sub> be the two circles having the same center O. AC is a chord that touches the C<sub>1<\/sub>\u00a0at point D<br \/><img src=\"https:\/\/farm1.staticflickr.com\/899\/41891577155_8d293a5360_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 10 Circles \" width=\"199\" height=\"203\" \/><br \/>Join OD.<br \/>Also, OD \u22a5 AC<br \/>AD = DC = 4 cm<br \/>[perpendicular line OD bisects the chord]<br \/>In right angled \u2206AOD,<br \/>OA\u00b2 = AD\u00b2 + DO\u00b2<br \/>[by Pythagoras theorem, i.e.,<br \/>(hypotenuse)\u00b2 = (base)\u00b2 + (perpendicular)\u00b2]<br \/>=&gt; DO\u00b2 = 5\u00b2 \u2013 4\u00b2 = 25 \u2013 16 = 9<br \/>=&gt; DO = 3 cm<br \/>Radius of the inner circle OD = 3 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 7.<br \/>A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. [NCERT Exemplar]<br \/>Solution:<br \/>Given: Chord PQ is parallel tangent at R.<br \/>To prove: R bisects the arc PRQ.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/887\/41891577405_b4d46809df_o.png\" alt=\"Learncbse.In Class 10 Chapter 10 Circles \" width=\"211\" height=\"183\" \/><br \/>Proof: \u22201 = \u22202 [alternate interior angles]<br \/>\u22201 = \u22203<br \/>[angle between tangent and chord is equal to the angle made by a chord in alternate segment]<br \/>\u22202 = \u22203<br \/>=&gt; PR = QR<br \/>[sides opposite to equal angles are equal]<br \/>=&gt; PR = QR<br \/>So, R bisects PQ.<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 8.<br \/>Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A. [NCERT Exemplar]<br \/>Solution:<br \/>Given, AB is the diameter of the circle.<br \/>A tangent is drawn from point A.<br \/>Draw a chord CD parallel to the tangent MAN.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1757\/41891577565_09856e71db_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 10 Circles \" width=\"194\" height=\"200\" \/><br \/>So, the CD is a chord of the circle and OA is a radius of the circle.<br \/>\u2220MAO = 90\u00b0<br \/>[Tangent at any point of a circle is perpendicular to the radius through the point of contact]<br \/>\u2220CEO = \u2220MAO [corresponding angles]<br \/>\u2220CEO = 90\u00b0<br \/>Thus, OE bisects CD<br \/>[perpendicular from the centre of the circle to a chord bisects the chord]<br \/>Similarly, the diameter AB bisects all. A chord that is parallel to the tangent at point A.<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 9.<br \/>If AB, AC, PQ are the tangents in the figure, and AB = 5 cm, find the perimeter of \u2206APQ.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/41891577725_cd833134ce_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 10 Circles \" width=\"265\" height=\"286\" \/><br \/>Solution:<br \/>Given : AB, AC and PQ are the tangents to the circle as shown in the figure above and AB = 5 cm<br \/>To find : The perimeter of \u2206APQ<br \/>Proof: PB and PX are the tangents to the circle<br \/>PB = PX<br \/>Similarly QC and QX are the tangents from<br \/>QC = QX<br \/>and AB and AC are the tangents from A<br \/>AB = AC<br \/>Now perimeter of \u2206APQ<br \/>= AP + PQ + AQ<br \/>= AP + PX + QX + AQ<br \/>= AP + PB + QC + AQ { PB = PX and QC = QX}<br \/>= AB + AC<br \/>= AB + AB (AB=AC)<br \/>= 2 AB = 2 x 5 = 10 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 10.<br \/>Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the center.<br \/>Solution:<br \/>Given: PQ and RS are parallel tangents of a circle<br \/>RMP is the intercept of the tangent between PQ and RS<br \/>RO and PQ are joined<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/41891577875_e3695ac413_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 10 Circles \" width=\"351\" height=\"278\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1740\/41891577825_7e5c300bb7_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 10 Circles \" width=\"352\" height=\"494\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 11.<br \/>In the figure, PQ is tangent at a point R of the circle with center O. If \u2220TRQ = 30\u00b0, find m \u2220PRS<br \/>Solution:<br \/>In the figure,<br \/>PRQ is tangent to the circle with center O at R<br \/><img src=\"https:\/\/farm1.staticflickr.com\/888\/42792512631_dd2d7d4f4d_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 10 Circles \" width=\"216\" height=\"191\" \/><br \/>RT and RS are joined such that \u2220TRQ = 30\u00b0<br \/>Let \u2220PRS = x\u00b0<br \/>Now \u2220SRX = 90\u00b0 (angle in a semicircle)<br \/>But \u2220TRQ + \u2220SRT + \u2220PRS = 180\u00b0 (Angles of a line)<br \/>=&gt; 30\u00b0 + 90\u00b0 + x\u00b0 = 180\u00b0<br \/>=&gt; 120\u00b0 + x\u00b0 = 180\u00b0<br \/>=&gt; x\u00b0 = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<br \/>\u2220PRS = 60\u00b0<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 12.<br \/>If PA and PB are tangents from an outside point P, such that PA = 10 cm and \u2220APB = 60\u00b0. Find the length of chord AB.<br \/>Solution:<br \/>PA and PB are the tangents from a point PQ outside the circle with center O<br \/>PA = 10 cm and \u2220APB = 60\u00b0<br \/><img src=\"https:\/\/farm1.staticflickr.com\/883\/41891578045_5cec550378_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 10 Circles \" width=\"291\" height=\"199\" \/><br \/>Tangents drawn from a point outside the circle are equal<br \/>PA = PB = 10 cm \u2220PAB = \u2220PBA<br \/>(Angles opposite to equal sides)<br \/>But in \u2206APB,<br \/>\u2220APB + \u2220PAB + \u2220PBA = 180\u00b0 (Angles of a triangle)<br \/>=&gt; 60\u00b0 + \u2220PAB + \u2220PAB = 180\u00b0<br \/>=&gt; 2 \u2220PAB = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<br \/>\u2220PAB = 60\u00b0<br \/>\u2220PBA = \u2220PAB = 60\u00b0<br \/>PA = PB = AB = 10 cm<br \/>Hence length of chord AB = 10 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 13.<br \/>In a right triangle ABC in which \u2220B = 90\u00b0, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC. [NCERT Exemplar]<br \/>Solution:<br \/>Let O be the center of the given circle. Suppose, the tangent at P meets BC at Q.<br \/>Join BP.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/899\/42792512811_f2229c9d48_o.png\" alt=\"Class 10 RD Sharma Chapter 10 Circles \" width=\"239\" height=\"204\" \/><br \/>To prove: BQ = QC<br \/>[angles in alternate segment]<br \/>Proof : \u2220ABC = 90\u00b0<br \/>[tangent at any point of a circle is perpendicular to the radius through the point of contact]<br \/>In \u2206ABC, \u22201 + \u22205 = 90\u00b0<br \/>[angle sum property, \u2220ABC = 90\u00b0]<br \/>\u22203 = \u22201<br \/>[angle between tangent and the chord equals angle made by the chord in alternate segment]<br \/>\u22203 + \u22205 = 90\u00b0 \u2026\u2026..(i)<br \/>Also, \u2220APB = 90\u00b0 [angle in semi-circle]<br \/>\u22203 + \u22204 = 90\u00b0 \u2026\u2026.(ii)<br \/>[\u2220APB + \u2220BPC = 180\u00b0, linear pair]<br \/>From Eqs. (i) and (ii), we get<br \/>\u22203 + \u22205 = \u22203 + \u22204<br \/>\u22205 = \u22204<br \/>=&gt; PQ = QC<br \/>[sides opposite to equal angles are equal]<br \/>Also, QP = QB<br \/>[tangents drawn from an internal point to a circle are equal]<br \/>=&gt; QB = QC<br \/>Hence proved.<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 14.<br \/>From an external point P, tangents PA and PB are drawn to a circle with center O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of \u2206PCD.<br \/>Solution:<br \/>PA and PB are the tangents drawn from a point P outside the circle with center O<br \/>CD is another tangent to the circle at point E which intersects PA and PB at C and D respectively<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1725\/42792513031_70b9745f02_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 10 Circles \" width=\"301\" height=\"214\" \/><br \/>PA = 14 cm<br \/>PA and PB are the tangents to the circle from P<br \/>PA = PB = 14 cm<br \/>Now CA and CE are the tangents from C<br \/>CA = CE \u2026.(i)<br \/>Similarly DB and DE are the tangents from D<br \/>DB = DE \u2026.(ii)<br \/>Now perimeter of \u2206PCD<br \/>= PC + PD + CD<br \/>= PC + PD + CE + DE<br \/>= PC + CE + PD + DE<br \/>= PC + CA + PD = DB {From (i) and (ii)}<br \/>= PA + PB<br \/>= 14 + 14<br \/>= 28 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 15.<br \/>In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle. [CBSE 2002]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/41891578575_5390556250_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 10 Circles \" width=\"208\" height=\"274\" \/><br \/>Solution:<br \/>In right \u2206ABC, \u2220B = 90\u00b0<br \/>BC = 6 cm, AB = 8 cm<br \/>Let r be the radius of incircle whose centre is O and touches the sides A B, BC and CA at P, Q and R respectively<br \/>AP and AR are the tangents to the circle AP = AR<br \/>Similarly CR = CQ and BQ = BP<br \/>OP and OQ are radii of the circle<br \/>OP \u22a5 AB and OQ \u22a5 BC and \u2220B = 90\u00b0 (given)<br \/>BPOQ is a square<br \/>BP = BQ = r<br \/>AR = AP = AB \u2013 BD = 8 \u2013 r<br \/>and CR = CQ = BC \u2013 BQ = 6 \u2013 r<br \/>But AC\u00b2 = AB\u00b2 + BC\u00b2 (Pythagoras Theorem)<br \/>= (8)\u00b2 + (6)\u00b2 = 64 + 36 = 100 = (10)\u00b2<br \/>AC = 10 cm<br \/>=&gt; AR + CR = 10<br \/>=&gt; 8 \u2013 r + 6 \u2013 r = 10<br \/>=&gt; 14 \u2013 2r = 10<br \/>=&gt; 2r = 14 \u2013 10 = 4<br \/>=&gt; r = 2<br \/>Radius of the incircle = 2 cm<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 16.<br \/>Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the endpoints of the arc. [NCERT Exemplar]<br \/>Solution:<br \/>Let mid-point of an arc AMB be M and TMT\u2019 be the tangent to the circle.<br \/>Join AB, AM, and MB.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/41891578745_c42bf50cd4_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 10 Circles \" width=\"199\" height=\"180\" \/><br \/>Since, arc AM = arc MB<br \/>=&gt; Chord AM = Chord MB<br \/>In \u2206AMB, AM = MB<br \/>=&gt; \u2220MAB = \u2220MBA \u2026\u2026(i)<br \/>[equal sides corresponding to the equal angle]<br \/>Since, TMT\u2019 is a tangent line.<br \/>\u2220AMT = \u2220MBA<br \/>[angle in alternate segment are equal]<br \/>\u2220AMT = \u2220MAB [from Eq. (i)]<br \/>But \u2220AMT and \u2220MAB are alternate angles, which is possible only when AB || TMT\u2019<br \/>Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the endpoints of the arc.<br \/>Hence proved<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Question 17.<br \/>From a point P, two tangents PA and PB are drawn to a circle with center O. If OP = diameter of the circle shows that \u2206APB is equilateral.<br \/>Solution:<br \/>Given: From a point P outside the circle with center O, PA and PB are the tangents to the circle such that OP is a diameter.<br \/>AB is joined.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/42792513391_a1eee3592f_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 10 Circles \" width=\"271\" height=\"195\" \/><br \/>To prove: APB is an equilateral triangle<br \/>Const : Join OP, AQ, OA<br \/>Proof : OP = 2r<br \/>=&gt; OQ + QP = 2r<br \/>=&gt; OQ = QP = r (OQ = r)<br \/>Now in right \u2206OAP,<br \/>OP is its hypotenuse and Q is its mid point<br \/>OA = AQ = OQ<br \/>(mid-point of hypotenuse of a right triangle is equidistances from its vertices)<br \/>\u2206OAQ is equilateral triangle \u2220AOQ = 60\u00b0<br \/>Now in right \u2206OAP,<br \/>\u2220APO = 90\u00b0 \u2013 60\u00b0 = 30\u00b0<br \/>=&gt; \u2220APB = 2 \u2220APO = 2 x 30\u00b0 = 60\u00b0<br \/>But PA = PB (Tangents from P to the circle)<br \/>=&gt; \u2220PAB = \u2220PBA = 60\u00b0<br \/>Hence \u2206APB is an equilateral triangle.<\/p>\n<p>Question 18.<br \/>Two tangents segments PA and PB are drawn to a circle with center O such that \u2220APB = 120\u00b0. Prove that OP = 2 AP. [CBSE 2014]<br \/>Solution:<br \/>Given: From a point P. Outside the circle with center O, PA and PB are tangents drawn and \u2220APB = 120\u00b0<br \/>OP is joined To prove: OP = 2 AP<br \/>Const: Take midpoint M of OP and join AM, join also OA and OB.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/888\/42792513521_2442efcb3c_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 10 Circles \" width=\"204\" height=\"182\" \/><br \/>Proof : In right \u2206OAP,<br \/>\u2220OPA =\u00a012\u00a0\u2220APB =\u00a012\u00a0x 120\u00b0 = 60\u00b0<br \/>\u2220AOP = 90\u00b0 \u2013 60\u00b0 = 30\u00b0<br \/>M is mid point of hypotenuse OP of \u2206OAP<br \/>MO = MA = MP<br \/>\u2220OAM = \u2220AOM = 30\u00b0 and \u2220PAM = 90\u00b0 \u2013 30\u00b0 = 60\u00b0<br \/>\u2206AMP is an equilateral triangle<br \/>MA = MP = AP<br \/>But M is mid point of OP<br \/>OP = 2 MP = 2 AP<br \/>Hence proved.<\/p>\n<p>Question 19.<br \/>If \u2206ABC is isosceles with AB = AC and C (0, r) is the incircle of the \u2206ABC touching BC at L. Prove that L bisects BC.<br \/>Solution:<br \/>Given: In \u2206ABC, AB = AC, and a circle with center O and radius r touches the side BC of \u2206ABC at L.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/896\/42792513731_5d63f78ffb_o.png\" alt=\"RD Sharma 10 Chapter 10 Circles \" width=\"176\" height=\"251\" \/><br \/>To prove: L is the midpoint of BC.<br \/>Proof: AM and AN are the tangents to the circle from A<br \/>AM = AN<br \/>But AB = AC (given)<br \/>AB \u2013 AN = AC \u2013 AM<br \/>BN = CM<br \/>Now BL and BN are the tangents from B<br \/>BL = BN<br \/>Similarly, CL and CM are tangents<br \/>CL = CM<br \/>But BM = CM (proved)<br \/>BL = CL<br \/>L is the midpoint of BC.<\/p>\n<p>Question 20.<br \/>AB is diameter and AC is a chord of a circle with center O such that \u2220BAC = 30\u00b0. The tangent at C intersects AB at a point D. Prove that BC = BD. [NCERT Exemplar]<br \/>Solution:<br \/>To prove, BC = BD<br \/>Join BC and OC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/42792513911_92758b601a_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 10 Circles \" width=\"296\" height=\"215\" \/><br \/>Given, \u2220BAC = 30\u00b0<br \/>=&gt; \u2220BCD = 30\u00b0<br \/>[angle between tangent and chord is equal to angle made by chord in the alternate segment]<br \/>\u2220ACD = \u2220ACO + \u2220OCD<br \/>\u2220ACD = 30\u00b0 + 90\u00b0 = 120\u00b0<br \/>[OC \u22a5 CD and OA = OC = radius =&gt; \u2220OAC = \u2220OCA = 30\u00b0]<br \/>In \u2206ACD,<br \/>\u2220CAD + \u2220ACD + \u2220ADC = 180\u00b0<br \/>[since, sum of all interior angles of a triangle is 180\u00b0]<br \/>=&gt; 30\u00b0 + 120\u00b0 + \u2220ADC = 180\u00b0<br \/>=&gt; \u2220ADC = 180\u00b0 \u2013 (30\u00b0 + 120\u00b0) = 30\u00b0<br \/>Now, in \u2206BCD,<br \/>\u2220BCD = \u2220BDC = 30\u00b0<br \/>=&gt; BC = BD<br \/>[since, sides opposite to equal angles are equal]<\/p>\n<p>Question 21.<br \/>In the figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm, and CD = 4 cm. Find AD. [CBSE 2002]<br \/>Solution:<br \/>A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively.<br \/>AB = 6 cm, BC = 7 cm, CD = 4cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1735\/42792514071_36dc0107c6_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 10 Circles \" width=\"200\" height=\"184\" \/><br \/>Let AD = x<br \/>AP and AS are the tangents to the circle<br \/>AP = AS<br \/>Similarly,<br \/>BP = BQ<br \/>CQ = CR<br \/>and OR = DS<br \/>AB + CD = AD + BC<br \/>=&gt; 6 + 4 = 7 + x<br \/>=&gt; 10 = 7 + x<br \/>=&gt;x = 10 \u2013 7 = 3<br \/>AD = 3 cm<\/p>\n<p>Question 22.<br \/>Prove that the perpendicular at the point contact to the tangent to a circle passes through the center of the circle.<br \/>Solution:<br \/>Given: TS is a tangent to the circle with center O at P, OP is joined<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/42792514171_6193bd13ca_o.png\" alt=\"RD Sharma 10 Solutions Chapter 10 Circles \" width=\"238\" height=\"184\" \/><br \/>To prove: OP is perpendicular to TS which passes through the center of the circle<br \/>Construction: Draw a line OR which intersect the circle at Q and meets the tangent TS at R<br \/>Proof: OP = OQ<br \/>(radii of the same circle) and OQ &lt; OR =&gt; OP &lt; OR<br \/>Similarly, we can prove that OP is less than all lines which can be drawn from O to TS<br \/>OP is the shortest<br \/>OP is perpendicular to TS<br \/>Perpendicular through P will pass through the centre of the circle<br \/>Hence proved.<\/p>\n<p>Question 23.<br \/>Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/42792514511_469dff1d85_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 10 Circles \" width=\"254\" height=\"216\" \/><br \/>Solution:<br \/>Given: Two circles with centres O and C touch each other externally at P. PT is its common tangent<br \/>From a point T on PT, TR and TQ are the tangents drawn to the circles<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/42792514331_abf1d3e115_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 10 Circles \" width=\"273\" height=\"217\" \/><br \/>To prove : TQ = TR<br \/>Proof: From T, TR and TP are two tangents to the circle with center O<br \/>TR = TP \u2026.(i)<br \/>Similarly, from T,<br \/>TQ and TP are two tangents to the circle with center C<br \/>TQ = TP \u2026.(ii)<br \/>From (i) and (ii)<br \/>TQ = TR<br \/>Hence proved.<\/p>\n<p>Question 24.<br \/>A is a point at a distance of 13 cm from the center O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the \u2206ABC. [NCERT Exemplar]<br \/>Solution:<br \/>Given: Two tangents are drawn from an external point A to the circle with centre O, Tangent BC is drawn at a point R, the radius of the circle equ to 5 cm.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/888\/42792514661_948c358ee9_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 10 Circles \" width=\"273\" height=\"157\" \/><br \/>To find the Perimeter of \u2206ABC.<br \/>Proof : \u2220OPA = 90\u00b0<br \/>[Tangent at any point of a circle is perpendicular to the radius through the point of contact]<br \/>OA\u00b2 = OP\u00b2 + PA\u00b2 [by Pythagoras Theorem]<br \/>(13)\u00b2 = 5\u00b2 + PA\u00b2<br \/>=&gt; PA\u00b2 = 144 = 12\u00b2<br \/>=&gt; PA = 12 cm<br \/>Now, perimeter of \u2206ABC = AB + BC + CA = (AB + BR) + (RC + CA)<br \/>= AB + BP + CQ + CA [BR = BP, RC = CQ tangents from internal point to a circle are equal]<br \/>= AP + AQ = 2AP = 2 x (12) = 24 cm<br \/>[AP = AQ tangent from internal point to a circle are equal]<br \/>Hence, the perimeter of \u2206ABC = 24 cm.<\/p>\n<p>Question 25.<br \/>In the figure, a circle is inscribed in a quadrilateral ABCD in which \u2220B = 90\u00b0. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius r of the circle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/42792515241_7e17c62929_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 10 Circles \" width=\"193\" height=\"183\" \/><br \/>Solution:<br \/>In the figure, O is the centre of the circle inscribed in a quadrilateral ABCD and \u2220B = 90\u00b0<br \/>AD = 23 cm, AB = 29 cm, DS = 5 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1754\/42792514981_34ab82702a_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 10 Circles \" width=\"191\" height=\"186\" \/><br \/>OP = OQ (radii of the same circle)<br \/>AB and BC are tangents to the circle and OP and OQ are radii<br \/>OP \u22a5 BC and OQ \u22a5 AB<br \/>\u2220OPB = \u2220OQB = 90\u00b0<br \/>PBQO is a square<br \/>DS and DR are tangents to the circle<br \/>DR = DS = 5 cm<br \/>AR = AD \u2013 DR = 23 \u2013 5 = 18 cm<br \/>AR and AQ are the tangents to the circle<br \/>AQ = AR = 18 cm But AB = 29 cm<br \/>BQ = AB \u2013 AQ = 29 \u2013 18 = 11 cm<br \/>Side of square PBQO is 11 cm<br \/>OP = 11 cm<br \/>Hence radius of the circle = 11 cm<\/p>\n<p>Question 26.<br \/>In the figure, there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point of P, tangents PA and PB are drawn to these circles. If AP =12 cm, find the length of BP. [CBSE 2010]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/42792515931_b68cbd73b5_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 10 Circles \" width=\"247\" height=\"160\" \/><br \/>Solution:<br \/>Two concentric circles with centre O with radii 5 cm and 3 cm respectively from a<br \/>point P, PA, and BP are tangents drawn to their circles<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/42792515671_d9a781fd7f_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 10 Circles \" width=\"250\" height=\"163\" \/><br \/>AP = 12 cm<br \/>To find BP<br \/>In right \u2206OAP,<br \/>OP\u00b2 = OA\u00b2 + AP\u00b2 (Pythagoras Theorem)<br \/>= (5)\u00b2 + (12)\u00b2 = 25 + 144<br \/>= 169 = (13)\u00b2<br \/>OP = 13 cm<br \/>Now in right \u2206OBP,<br \/>OP\u00b2 = OB\u00b2 + BP\u00b2<br \/>=&gt; (13)\u00b2 = (3)\u00b2 + BP\u00b2<br \/>=&gt; 169 = 9 + BP\u00b2<br \/>=&gt; BP\u00b2 = 169 \u2013 9 = 160 = 16 x 10<br \/>BP = \u221a(16 x 10) = 4\u221a10 cm<\/p>\n<p>Question 27.<br \/>In the figure, AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA. [CBSE 2010]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/42742789502_1ce9cc8933_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 10 Circles \" width=\"362\" height=\"561\" \/><br \/>Solution:<br \/>In the figure, AB is the chord of the circle with centre O and radius of 10 cm.<br \/>Two tangents from P are drawn to the circle touching it at A and B respectively<br \/>AB is joined with intersects OP at L<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1740\/42792516131_9d66490efb_o.png\" alt=\"RD Sharma 10 Solutions Chapter 10 Circles \" width=\"354\" height=\"405\" \/><br \/><img src=\"https:\/\/farm1.staticflickr.com\/895\/42792516381_84e2055eca_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 10 Circles \" width=\"220\" height=\"110\" \/><\/p>\n<p>Question 28.<br \/>In the figure, PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN. [CBSE 2010]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/42792516981_fcd48129c9_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 10 Circles \" width=\"246\" height=\"180\" \/><br \/>Solution:<br \/>Given: In the figure, PA and PB are the tangents to the circle with centre O from a point P outside it LN touches it at M<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/42792516861_d029e3ef30_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 10 Circles \" width=\"243\" height=\"176\" \/><br \/>To prove : PL + LM = PN + MN<br \/>Prove : PA and PB are tangents to the circle from P<br \/>PA = PB<br \/>Similarly from L, LA and LM are tangents<br \/>LA = LM<br \/>Similarly NB = NM<br \/>Now PA = PB =&gt; PL + LA = PN + NB<br \/>PL + LM = PN + NM<br \/>Hence proved.<\/p>\n<p>Question 29.<br \/>In the figure, BDC is a tangent to the given circle at point D such that BD = 30 to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/891\/40982330330_3c9db340c3_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 10 Circles \" width=\"219\" height=\"209\" \/><br \/>Solution:<br \/>In the figure, BDC is a tangent to the given circle with centre O and D is a point such that<br \/>BD = 30 cm and CD = 7 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/27923477947_2436fb8ea0_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 10 Circles \" width=\"190\" height=\"208\" \/><br \/>BE and CF are other two tangents drawn from B and C respectively which meet at A on producing this and \u2206BAC is a right angle so formed<br \/>To find : (i) AF and (ii) radius of the circle<br \/>Join OE and OF<br \/>OE = OF radii of the circle<br \/>OE \u22a5 AB and OF \u22a5 AC<br \/>OEAF is a square<br \/>BD and BE are the tangents from B<br \/>BE = BD = 30 cm and similarly<br \/>CF = CD = 7 cm<br \/>Let r be the radius of the circle<br \/>OF = AF = AE = r<br \/>AB = 30 + r and AC = 7 + r and BC = 30 + 7 = 37 cm<br \/>Now in right \u2206ABC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/27923477757_c52b199efe_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 10 Circles \" width=\"351\" height=\"303\" \/><\/p>\n<p>Question 30.<br \/>If d<sub>1<\/sub>, d<sub>2<\/sub>\u00a0(d<sub>2<\/sub>\u00a0&gt; d<sub>1<\/sub>) be the diameters of two concentric circles and c be the length of a chord of a circle that is tangent to the other circle, prove that d22=c2+d21. [NCERT Exemplar]<br \/>Solution:<br \/>Let AB be a chord of a circle that touches the other circle at C. Then \u2206OCB is the right triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/27923478077_629d2b17a0_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 10 Circles \" width=\"321\" height=\"322\" \/><\/p>\n<p>Question 31.<br \/>In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that \u2220RPQ = 30\u00b0. A chord RS is drawn parallel to the tangent PQ. Find \u2220RQS. [CBSE 2015, NCERT Exemplar]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/27923478387_62be5b92a3_o.png\" alt=\"Learncbse.In Class 10 Chapter 10 Circles \" width=\"291\" height=\"155\" \/><br \/>Solution:<br \/>In the given figure,<br \/>PQ and PR are tangents to the circle with centre O drawn from P<br \/>\u2220RPQ = 30\u00b0<br \/>Chord RS || PQ is drawn<br \/>To find \u2220RQS<br \/>PQ = PR (tangents to the circle)<br \/>\u2220PRQ = \u2220PQR But \u2220RPQ = 30\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/27923478247_8b6d71cba2_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 10 Circles \" width=\"349\" height=\"192\" \/><\/p>\n<p>Question 32.<br \/>From an external point P, tangents PA = PB are drawn to a circle with centre O. If \u2220PAB = 50\u00b0, then find \u2220AOB. [CBSE 2016]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1727\/27923478667_a590ed0ce2_o.png\" alt=\"RD Sharma Class 10 Solutions Circles \" width=\"274\" height=\"196\" \/><br \/>PA = PB [tangents drawn from external point are equal]<br \/>\u2220PBA = \u2220PAB = 50\u00b0 [angles equal to opposite sides]<br \/>\u2220APB = 180\u00b0 \u2013 50\u00b0 \u2013 50\u00b0 = 80\u00b0 [angle-sum property of a A]<br \/>In cyclic quad. OAPB<br \/>\u2220AOB + \u2220APB = 180\u00b0 [sum of opposite angles of a cyclic quadrilateral is 180\u00b0]<br \/>\u2220AOB + 80\u00b0 = 180\u00b0<br \/>\u2220AOB = 180\u00b0- 80\u00b0 = 100\u00b0<\/p>\n<p>Question 33.<br \/>In the figure, two tangents AB and AC are drawn to a circle with centre O such that \u2220BAC = 120\u00b0. Prove that OA = 2AB.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/876\/42742790372_83b032e604_o.png\" alt=\"RD Sharma Class 10 Chapter 10 Circles \" width=\"164\" height=\"188\" \/><br \/>Solution:<br \/>Given: In the figure, O is the centre of the circle.<br \/>AB and AC are the tangents to the circle from A such that<br \/>\u2220BAC = 120\u00b0 .<br \/>To prove : OA = 2AB<br \/>Proof: In \u2206OAB and \u2206OAC<br \/>\u2220OBA = \u2220OCA \u2013 90\u00b0 (OB and OC are radii)<br \/>OA = OA (common)<br \/>OB = OC (radii of the circle)<br \/>\u2206OAB ~ \u2206OAC<br \/>\u2220OAB = \u2220OAC = 60\u00b0<br \/>Now in right \u2206OAB,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/42742790312_a2a58673e5_o.png\" alt=\"Circles Class 10 RD Sharma\" width=\"359\" height=\"147\" \/><\/p>\n<p>Question 34.<br \/>The lengths of three consecutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side.<br \/>Solution:<br \/>In quadrilateral ABCD which is circumerscribing it<br \/>BC = 4 cm, CD = 5 cm and DA = 7 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/42742790502_c1854abbc1_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 10 Circles \" width=\"249\" height=\"229\" \/><br \/>We know that if a quad, is circumscribed in a circle, then<br \/>AB + CD = AD + BC<br \/>=&gt; AB + 5 = 4 + 7<br \/>=&gt; AB + 5 = 11<br \/>AB = 11 \u2013 5 = 6<br \/>AB = 6 cm<\/p>\n<p>Question 35.<br \/>The common tangents AB and CD to two circles with centres O and O\u2019 intersect at E between their centres. Prove that the points O, E, and O\u2019 are collinear. [NCERT Exemplar]<br \/>Solution:<br \/>Joint AO, OC, and O\u2019D, O\u2019B<br \/>Now, in \u2206EO\u2019D and \u2206EO\u2019B<br \/>O\u2019D = O\u2019B [radius]<br \/>O\u2019E = O\u2019E [common side]<br \/>ED = EB<br \/><img src=\"https:\/\/farm1.staticflickr.com\/875\/27923479587_51930a5a0d_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 10 Circles \" width=\"354\" height=\"564\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/27923479217_69ce25f978_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 10 Circles \" width=\"355\" height=\"425\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/42742790782_9b2d7f0ca3_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 10 Circles \" width=\"347\" height=\"546\" \/><\/p>\n<p>Question 36.<br \/>In the figure, common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/27923479827_859d1c6961_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 10 Circles \" width=\"362\" height=\"192\" \/><br \/>Solution:<br \/>Given: Two common tangents PQ and RS intersect each other at A.<br \/>To prove : PQ = RS<br \/>Proof: From A, AQ and AR are two tangents are drawn to the circle with centre O.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/42742791302_354aaa4e61_o.png\" alt=\"Class 10 RD Sharma Chapter 10 Circles \" width=\"358\" height=\"201\" \/><br \/>AP = AR \u2026.(i)<br \/>Similarly AQ and AS are the tangents to the circle with centre C<br \/>AQ = AS \u2026.(ii)<br \/>Adding (i) and (ii)<br \/>AP + AQ = AR + AS<br \/>=&gt; PQ = RS<br \/>Hence proved.<\/p>\n<p>Question 37.<br \/>Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE 2014]<br \/>Solution:<br \/>Let R be the radius of the outer circle and r be the radius if a small circle of two concentric circle<br \/>AB is the chord of the outer circle and touches the smaller circle at P<br \/>Join OP, OA<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/27923480107_ee51c3d590_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 10 Circles \" width=\"316\" height=\"329\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1757\/27923479977_25355173a7_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 10 Circles \" width=\"282\" height=\"141\" \/><\/p>\n<p>Question 38.<br \/>AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. [NCERT Exemplar]<br \/>Solution:<br \/>Given: AB and CD are tangents to two circles of equal radii.<br \/>To prove :<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1756\/42742791702_a0d92e1055_o.png\" alt=\"RD Sharma 10 Chapter 10 Circles \" width=\"366\" height=\"161\" \/><br \/>Construction: Join OA, OC, O\u2019B, and O\u2019D<br \/>Proof: Now, \u2220OAB = 90\u00b0<br \/>[tangent at any point of a circle is perpendicular to the radius through the point of contact]<br \/>Thus, AC is a straight line.<br \/>Also, \u2220OAB + \u2220OCD = 180\u00b0<br \/>AB || CD<br \/>Similarly, BD is a straight line and \u2220O\u2019BA = \u2220O\u2019DC = 90\u00b0<br \/>Also, AC = BD<br \/>[radii of two circles are equal] In quadrilateral ABCD,<br \/>\u2220A = \u2220B = \u2220C = \u2220D = 90\u00b0<br \/>andAC = BD<br \/>ABCD is a rectangle<br \/>Hence, AB = CD<br \/>[opposite sides of rectangle are equal]<\/p>\n<p>Question 39.<br \/>A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If an area of \u2206PQR is 336 cm\u00b2, find the sides PQ and PR. [CBSE 2014]<br \/>Solution:<br \/>\u2206PQR is circumscribed by a circle with centre O and a radius of 8 cm<br \/>T is a point of contact that divides the line segment OT into two parts such that<br \/>QT = 14 cm and TR = 16 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/27923480307_cc7cca52c7_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 10 Circles \" width=\"261\" height=\"172\" \/><br \/>Area of \u2206PQR = 336 cm\u00b2<br \/>Let PS = x cm<br \/>QT and QS are tangents to the circle from Q<br \/>QS = QT = 14 cm<br \/>Similarly RU and RT are tangents to the circle<br \/>RT = RU = 16 cm<br \/>Similarly PS and PU are tangents from P<br \/>PS = PU = x<br \/>Now PQ = x + 14 and PR = x + 16 and QR = 14 + 16 = 30 cm<br \/>Now area of \u2206PQR = Area of \u2206POQ + area of \u2206QOR + area of \u2206POR<br \/><img src=\"https:\/\/farm1.staticflickr.com\/880\/42742791872_22396eacaf_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 10 Circles \" width=\"351\" height=\"333\" \/><\/p>\n<p>Question 40.<br \/>In the figure, the tangent at a point C of a circle and a diameter AB when extended intersect at P. If \u2220PCA = 110\u00b0, find \u2220CBA. [NCERT Exemplar]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/42742792142_3a2f813c50_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 10 Circles \" width=\"305\" height=\"202\" \/><br \/>Solution:<br \/>Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.<br \/>Join OC. Here, OC is a radius.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/42743081162_d454908582_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 10 Circles \" width=\"304\" height=\"210\" \/><br \/>Since, tangent at any point of a circle is perpendicular to the radius through point of contact circle.<br \/>OC \u22a5 PC<br \/>Now, \u2220PCA = 110\u00b0 [given]<br \/>=&gt; \u2220PCO + \u2220OCA = 110\u00b0<br \/>=&gt; 90\u00b0 + \u2220OCA = 110\u00b0<br \/>=&gt; \u2220OCA = 20\u00b0<br \/>OC = OA = Radius of circle<br \/>\u2220OCA = \u2220OAC = 20\u00b0<br \/>[since, two sides are equal, then their opposite angles are equal]<br \/>Since, PC is a tangent, so<br \/>\u2220BCP = \u2220CAB = 20\u00b0<br \/>[angles in a alternate segment are equal]<br \/>In \u2206PBC, \u2220P + \u2220C + \u2220A= 180\u00b0<br \/>\u2220P = 180\u00b0 \u2013 (\u2220C + \u2220A)<br \/>\u2220P = 180\u00b0 \u2013 (110\u00b0 + 20\u00b0)<br \/>\u2220P = 180\u00b0 \u2013 130\u00b0 = 50\u00b0<br \/>In \u2206PBC,<br \/>\u2220BPC + \u2220PCB + \u2220PBC = 180\u00b0<br \/>[sum of all interior angles of any triangle is 180\u00b0]<br \/>=&gt; 50\u00b0 + 20\u00b0 + \u2220PBC = 180\u00b0<br \/>=&gt; \u2220PBC = 180\u00b0 \u2013 70\u00b0<br \/>\u2220PBC = 110\u00b0<br \/>Since, \u2206PB is a straight line.<br \/>\u2220PBC + \u2220CBA = 180\u00b0<br \/>\u2220CBA = 180\u00b0 \u2013 110\u00b0 = 70\u00b0<\/p>\n<p>Question 41.<br \/>AB is a chord of a circle with centre O, AOC is diameter, and AT is the tangent at A as shown in the figure. Prove that \u2220BAT = \u2220ACB. [NCERT Exemplar]<br \/><img src=\"https:\/\/farm1.staticflickr.com\/884\/42742792302_2c6583076f_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 10 Circles \" width=\"249\" height=\"208\" \/><br \/>Solution:<br \/>Since, AC is a diameter line, so angle in semicircle makes an angle 90\u00b0.<br \/>\u2220ABC = 90\u00b0 [by property]<br \/>In \u2206ABC,<br \/>\u2220CAB + \u2220ABC + \u2220ACB = 180\u00b0<br \/>[ sum of all interior angles of any triangle is 180\u00b0]<br \/>=&gt; \u2220CAB + \u2220ACB = 180\u00b0 \u2013 90\u00b0 = 90\u00b0 \u2026\u2026\u2026.(i)<br \/>Since the diameter of a circle is perpendicular to the tangent.<br \/>i.e. CA \u22a5 AT<br \/>\u2220CAT = 90\u00b0<br \/>=&gt; \u2220CAB + \u2220BAT = 90\u00b0 \u2026\u2026.(ii)<br \/>From Eqs. (i) and (ii),<br \/>\u2220CAB + \u2220ACB = \u2220CAB + \u2220BAT<br \/>=&gt; \u2220ACB = \u2220BAT<br \/>Hence proved.<\/p>\n<p>Question 42.<br \/>In the given figure, an \u2206ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when an area of \u2206ABC is 84 cm\u00b2. [CBSE 2015]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/42742792742_7f830033b2_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 10 Circles \" width=\"166\" height=\"174\" \/><br \/>Solution:<br \/>In the given figure,<br \/>In \u2206ABC is circle is inscribed touching it at D, E, and F respectively.<br \/>Radius of the circle (r) = 4cm<br \/>OD \u22a5 BC, then<br \/>OD = 4 cm, BD = 8 cm, DC = 6 cm<br \/>Join OE and OF<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/42742792492_8660620e05_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 10 Circles \" width=\"261\" height=\"323\" \/><br \/><img src=\"https:\/\/farm1.staticflickr.com\/876\/42742792652_74613d46f0_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 10 Circles \" width=\"353\" height=\"355\" \/><\/p>\n<p>Question 43.<br \/>In the given figure, AB is a diameter of a circle with centre O, and AT is a tangent. If \u2220AOQ = 58\u00b0, find \u2220ATQ. [CBSE 2015]<br \/><img src=\"https:\/\/farm1.staticflickr.com\/899\/42742792952_cdd280ca52_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 10 Circles \" width=\"192\" height=\"192\" \/><br \/>Solution:<br \/>In the given figure,<br \/>AB is the diameter, AT is the tangent<br \/>and \u2220AOQ = 58\u00b0<br \/>To find \u2220ATQ<br \/>Arc AQ subtends \u2220AOQ at the centre and \u2220ABQ at the remaining part of the circle<br \/>\u2220ABQ =\u00a012\u00a0\u2220AOQ =\u00a012\u00a0x 58\u00b0 = 29\u00b0<br \/>Now in \u2206ABT,<br \/>\u2220BAT = 90\u00b0 ( OA \u22a5 AT)<br \/>\u2220ABT + \u2220ATB = 90\u00b0<br \/>=&gt; \u2220ABT + \u2220ATQ = 90\u00b0<br \/>=&gt; 29\u00b0 + \u2220ATQ = 90\u00b0<br \/>=&gt; \u2220ATQ = 90\u00b0- 29\u00b0 = 61\u00b0<\/p>\n<p>Question 44.<br \/>In the figure, OQ : PQ = 3:4 and perimeter of \u2206POQ = 60 cm. Determine PQ, QR, and OP.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/42742793242_120de1b4d9_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 10 Circles \" width=\"262\" height=\"193\" \/><br \/>Solution:<br \/>In the figure, OQ : PQ = 3:4<br \/><img src=\"https:\/\/farm1.staticflickr.com\/900\/42742793072_2e1a979afa_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 10 Circles \" width=\"268\" height=\"193\" \/><br \/>Perimeter of \u2206POQ = 60 cm<br \/>To find PQ, QR and OP<br \/>OQ : PQ = 3 : 4<br \/>Let OQ = 3x and PQ = 4x<br \/>Now in right \u2206OPQ,<br \/>OP\u00b2 = OQ\u00b2 + PQ\u00b2 = (3x)\u00b2 + (4x)\u00b2 = 9x\u00b2 + 16x\u00b2 = 25x\u00b2 = (5x)\u00b2<br \/>OP = 5x<br \/>But OQ + QP + OP = 60 cm<br \/>3x + 4x + 5x = 60<br \/>=&gt; 12x = 60<br \/>x = 5<br \/>PQ = 4x = 4 x 5 = 20 cm<br \/>QR = 2 OQ = 2 x 3x = 6 x 5 = 30 cm<br \/>OP = 5x = 5 x 5 = 25 cm<\/p>\n<p>Question 45.<br \/>Equal circles with centre O and O\u2019 touch each other at X. OO\u2019 produced to meet a circle with centre O\u2019, at A. AC is a tangent to the circle whose centre is O. O\u2019 D is perpendicular to AC. Find the value of\u00a0DO\u2032CO<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/42742793742_de4ab23909_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 10 Circles \" width=\"323\" height=\"177\" \/><br \/>Solution:<br \/>Two equal circles with centre O and O\u2019 touch each other externally at X<br \/>OO\u2019 produced to meet at A<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/42742793402_8ffdc71a4c_o.png\" alt=\"Learncbse.In Class 10 Chapter 10 Circles \" width=\"321\" height=\"180\" \/><br \/>AC is the tangent of a circle with centre O,<br \/>O\u2019D \u22a5 AC is drawn OC is joined<br \/>AC is tangent and OC is the radius<br \/>OC \u22a5 AC<br \/>O\u2019D \u22a5 AC<br \/>OC || O\u2019D<br \/>Now O\u2019A =\u00a012\u00a0A x or\u00a012\u00a0AO<br \/>Now in O\u2019AD and AOAC<br \/>\u2220A = \u2220A (common)<br \/>\u2220AO\u2019D = \u2220AOC (corresponding angles)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/42742793582_042460ba4d_o.png\" alt=\"Learncbse.In Class 10 Chapter 10 Circles \" width=\"355\" height=\"112\" \/><\/p>\n<p>Question 46.<br \/>In the figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that \u2206AEO ~ \u2206ABC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/42742794012_053a1a5174_o.png\" alt=\"Learncbse.In Class 10 Chapter 10 Circles \" width=\"213\" height=\"193\" \/><br \/>Solution:<br \/>Given: In the figure, BC is a tangent to the circle with centre O at B.<br \/>AB is diameter AC is joined which intersects the circle at P<br \/>OE bisects AP<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/42742793862_527dafc26e_o.png\" alt=\"RD Sharma Class 10 Solutions Circles \" width=\"215\" height=\"191\" \/><br \/>To prove: \u2206AEO ~ \u2206ABC<br \/>Proof: In \u2206OAE and \u2206OPE<br \/>OE = OE (common)<br \/>OA = OP \u2018 (radii of the same circle)<br \/>EA = EP (given)<br \/>\u2206OAE = \u2206OPE (SSS axiom)<br \/>\u2220OEA = \u2220OEP<br \/>But \u2220OEA + \u2220OEP = 180\u00b0<br \/>\u2220OEA = 90\u00b0<br \/>Now in \u2206AEO and \u2206ABC<br \/>\u2220OEA = \u2220ABC (each 90\u00b0)<br \/>\u2220A = \u2220A (common)<br \/>\u2206AEO ~ \u2206ABC (AA axiom)<br \/>Hence proved.<\/p>\n<p>Question 47.<br \/>In the figure, PO \u22a5 QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/42742794522_42fc110ef8_o.png\" alt=\"RD Sharma Class 10 Solutions Circles \" width=\"188\" height=\"164\" \/><br \/>Solution:<br \/>Given: In the figure, O is the centre of the circle<br \/>PO \u22a5 QO<br \/>The tangents at P and Q intersect each other at T<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/42742794172_ff91178013_o.png\" alt=\"RD Sharma Class 10 Solutions Circles \" width=\"180\" height=\"168\" \/><br \/>To prove: PQ and OT are right bisectors of each other<br \/>Proof: PT and QT are tangents to the circle<br \/>PT = QT<br \/>OP and OQ are radii of the circle and \u2220POQ = 90\u00b0 ( PO \u22a5 QO)<br \/>OQTP is a square Where PQ and OT are diagonals<br \/>Diagonals of a square bisect each other at right angles<br \/>PQ and OT bisect each other at right angles<br \/>Hence PQ and QT are right bisectors of each other.<\/p>\n<p>Question 48.<br \/>In the figure, O is the centre of the circle and BCD is tangent to it at C. Prove that \u2220BAC + \u2220ACD = 90\u00b0.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/879\/41891587255_e1a4d095f2_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 10 Circles \" width=\"291\" height=\"153\" \/><br \/>Solution:<br \/>Given: In the figure, O is the centre of the circle BCD is a tangent, CP is a chord<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/42742794722_0883154bab_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 10 Circles \" width=\"275\" height=\"148\" \/><br \/>prove : \u2220BAC + \u2220ACD = 90\u00b0<br \/>Proof: \u2220ACD = \u2220CPA (Angles in the alternate segment)<br \/>But in \u2206ACP,<br \/>\u2220ACP = 90\u00b0 (Angle in a semicircle)<br \/>\u2220PAC + \u2220CPA = 90\u00b0<br \/>=&gt; \u2220BAC + \u2220ACD = 90\u00b0<br \/>(\u2220ACD = \u2220CPA proved)<br \/>Hence proved.<\/p>\n<p>Question 49.<br \/>Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines. [NCERT Exemplar]<br \/>Solution:<br \/>Given: Two tangents PQ and PR are drawn from an external point P to a circle with centre O.<br \/>To prove: Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.<br \/><img src=\"https:\/\/farm1.staticflickr.com\/880\/42742795192_a051fd8bdc_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 10 Circles \" width=\"303\" height=\"174\" \/><br \/>Construction : Join OR, and OQ.<br \/>In \u2206POR and \u2206POQ<br \/>\u2220PRO = \u2220PQO = 90\u00b0<br \/>[tangent at any point of a circle is perpendicular to the radius through the point of contact]<br \/>OR = OQ [radii of same circle]<br \/>Since OP is common.<br \/>\u2206PRO = \u2206PQO [RHS]<br \/>Hence, \u2220RPO = \u2220QPO [by CPCT]<br \/>Thus, O lies on the angle bisector of PR and PQ.<br \/>Hence proved.<\/p>\n<p>Question 50.<br \/>In the figure, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the lengths of PS. [CBSE 2017]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1732\/42742795512_4ea3fc60f5_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 10 Circles \" width=\"267\" height=\"257\" \/><br \/>Solution:<br \/>Construction: Join OS and OP.<br \/>Consider \u2206POS<br \/>We have,<br \/>PO = OS<br \/><img src=\"https:\/\/farm1.staticflickr.com\/876\/42742795392_ccc3059fec_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 10 Circles \" width=\"218\" height=\"215\" \/><br \/>\u2206POS is an isosceles triangle.<br \/>We know that in an isosceles triangle, if a line is drawn perpendicular to the base of the triangle from the common vertex of the equal sides, then that line will bisect the base (unequal side).<br \/>And PQ = PR = 5 cm<br \/>[PRT and PQS are tangents to the inner circle to the inner circle from a point P lying on the outer circle]<br \/>We have, PQ = QS<br \/>It is given that, PQ = 5 cm<br \/>QS = 5 cm<br \/>From the figure, we have<br \/>PS = PQ + QS<br \/>=&gt; PS = 5 + 5<br \/>=&gt; PS = 10 cm<\/p>\n<p>Question 51.<br \/>In the figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOP is a diameter. If \u2220POR = 130\u00b0 and S is a point on the circle, find \u22201 + \u22202. [CBSE 2017]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/42742795862_46a6735c57_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 10 Circles \" width=\"314\" height=\"222\" \/><br \/>Solution:<br \/>Construction: Join RT.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/42075085334_ab8020c541_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 10 Circles \" width=\"320\" height=\"213\" \/><br \/>Given, \u2220POR = 130\u00b0<br \/>\u2220POQ = 180\u00b0- (\u2220POR) = 180\u00b0 \u2013 130\u00b0 = 50\u00b0<br \/>Since, PQ is a tangent<br \/>\u2220PQO = 90\u00b0<br \/>Now, In \u2206POQ,<br \/>\u2220POQ + \u2220PQO + \u2220QPO = 180\u00b0<br \/>=&gt; 50\u00b0 + 90\u00b0 + \u22201 = 180\u00b0<br \/>=&gt; \u22201 = 180\u00b0 \u2013 140\u00b0<br \/>=&gt; \u22201 = 40\u00b0<br \/>Now, In \u2206RST<br \/>\u2220RST =\u00a012\u00a0\u2220ROT<br \/>[Angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference]<br \/>=&gt; \u22202 =\u00a012\u00a0x 130\u00b0 = 65\u00b0<br \/>Now \u22201 + \u22202 = 40\u00b0 + 65\u00b0 = 105\u00b0<\/p>\n<p>Question 52.<br \/>In the figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC = PD. [CBSE 2017]<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1731\/42742796022_9eeaee96f7_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 10 Circles \" width=\"321\" height=\"212\" \/><br \/>Solution:<br \/>PA = PB = 12 cm \u2026(i)<br \/>QC = AC = 3cm \u2026(ii)<br \/>QD = BD = 3 cm \u2026(iii)<br \/>[Tangents drawn from an external point are equal]<br \/>To find : PC + PD<br \/>= (PA \u2013 AC) + (PB \u2013 BD)<br \/>= (12 \u2013 3) + (12 \u2013 3) [From (i), (ii), and (iii)]<br \/>= 9 + 9 = 18 cm<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 10\u00a0 Exercise 10.2. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-10-exercise-102\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631695893187\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-10-exercise-102-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631695988421\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-key-benefits-of-learning-rd-sharma-class-10-solutions-chapter-10-exercise-102\"><\/span>What are the key benefits of learning RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 answers were quite helpful. This makes it simple to clear any questions about arithmetic progression. Easily answer all of the questions that the Class 10 students have given for exercise.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631696060794\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-10-exercise-102-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 10 Exercise 10.2:\u00a0This exercise has exciting and difficult questions based on the concepts of tangent to a circle, tangent from a point on a circle, and tangent length. The RD Sharma Class 10 Solutions was prepared by our expert team at Kopykitab to assist students in clearing conceptual doubts &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 10 Circles Exercise 10.2 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-10-exercise-10-2\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 10 Circles Exercise 10.2 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":127914,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127887"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=127887"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127887\/revisions"}],"predecessor-version":[{"id":127959,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127887\/revisions\/127959"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127914"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=127887"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=127887"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=127887"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}