{"id":4950,"date":"2016-05-19T09:49:00","date_gmt":"2016-05-19T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/"},"modified":"2023-03-22T15:18:14","modified_gmt":"2023-03-22T09:48:14","slug":"ncert-solutions-class-10-maths-exercise-11-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Exercise 11.2"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 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\" style=\"cursor:inherit\">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=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#NCERT_Solutions_for_Class_10_Maths_Constructions\" >NCERT Solutions for Class 10 Maths\u00a0Constructions<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#1_Draw_a_circle_of_radius_6_cm_From_a_point_10_cm_away_from_its_centre_construct_the_pair_of_tangents_to_the_circle_and_measure_their_lengths\" >1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#2_Construct_a_tangent_to_a_circle_of_radius_4_cm_from_a_point_on_the_concentric_circle_of_radius_6_cm_and_measure_its_length_Also_verify_the_measurement_by_actual_calculation\" >2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#3_Draw_a_circle_of_radius_3_cm_Take_two_points_P_and_Q_on_one_of_its_extended_diameter_each_at_a_distance_of_7_cm_from_its_centre_Draw_tangents_to_the_circle_from_these_two_points_P_and_Q\" >3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#4_Draw_a_pair_of_tangents_to_a_circle_of_radius_5_cm_which_are_inclined_to_each_other_at_an_angle_of\" >4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#5_Draw_a_line_segment_AB_of_length_8_cm_Taking_A_as_centre_draw_a_circle_of_radius_4_cm_and_taking_B_as_centre_draw_another_circle_of_radius_3_cm_Construct_tangents_to_each_circle_from_the_centre_of_the_other_circle\" >5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#6_Let_ABC_be_a_right_triangle_in_which_AB_6_cm_BC_8_cm_and_B_BD_is_the_perpendicular_from_B_on_AC_The_circle_through_B_C_D_is_drawn_Construct_the_tangents_from_A_to_this_circle\" >6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B =  BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#7_Draw_a_circle_with_the_help_of_a_bangle_Take_a_point_outside_the_circle_Construct_the_pair_of_tangents_from_this_point_to_the_circle\" >7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#NCERT_Solutions_for_Class_10_Maths_Exercise_112\" >NCERT Solutions for Class 10 Maths Exercise 11.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-11-2\/#CBSE_app_for_Class_10\" >CBSE app for Class 10<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 10 Maths Exercise 11.2 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Maths Constructions<\/strong><strong>\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics-constructions\/1211\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10%20Maths%20Book.jpg\" alt=\"NCERT Solutions for Class 10 Maths Exercise 11.2\" width=\"180\" height=\"222\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Constructions\"><\/span>NCERT Solutions for Class 10 Maths\u00a0Constructions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><strong>Exercise 11.2<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>In each of the following, give the justification of the construction also:<\/strong><\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_Draw_a_circle_of_radius_6_cm_From_a_point_10_cm_away_from_its_centre_construct_the_pair_of_tangents_to_the_circle_and_measure_their_lengths\"><\/span><strong>1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. Given<\/strong>: A circle whose centre is O and radius is 6 cm and a point P is 10 cm away from its centre.<\/p>\n<p style=\"text-align: justify;\"><strong>To construct<\/strong>: To construct the pair of tangents to the circle and measure their lengths.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of Construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 9\" style=\"height: 164px; width: 171px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image001.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Join PO and bisect it. Let M be the mid-point of PO.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Join PQ and PR.<\/p>\n<p style=\"text-align: justify;\">Then PQ and PR are the required two tangents.<\/p>\n<p style=\"text-align: justify;\">By measurement, PQ = PR = 8 cm<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>: Join OQ and OR.<\/p>\n<p style=\"text-align: justify;\">Since <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>OQP and <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>ORP are the angles in semicircles.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image003.png\" \/><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>OQP <img decoding=\"async\" style=\"height: 21px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image004.png\" \/> = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>ORP<\/p>\n<p style=\"text-align: justify;\">Also, since OQ, OR are radii of the circle, PQ and PR will be the tangents to the circle at Q and R respectively.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image003.png\" \/> We may see that the circle with OP as diameter increases the given circle in two points. Therefore, only two tangents can be draw.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_Construct_a_tangent_to_a_circle_of_radius_4_cm_from_a_point_on_the_concentric_circle_of_radius_6_cm_and_measure_its_length_Also_verify_the_measurement_by_actual_calculation\"><\/span><strong>2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. To construct<\/strong>: To construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its lengths. Also to verify the measurements by actual calculation.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of Construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 10\" style=\"height: 159px; width: 154px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image005.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Join PO and bisect it. Let M be the mid-point of PO.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the point Q and R.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Join PQ.<\/p>\n<p style=\"text-align: justify;\">Then PQ is the required tangent.<\/p>\n<p style=\"text-align: justify;\">By measurement, PQ = 4.5 cm<\/p>\n<p style=\"text-align: justify;\">By actual calculation,<\/p>\n<p style=\"text-align: justify;\">PQ = <img decoding=\"async\" style=\"height: 35px; width: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image006.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 27px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image007.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image008.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 33px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image009.png\" \/> = 4.47 cm<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>: Join OQ. Then <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>PQO is an angle in the semicircle and therefore,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>PQO = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/>PQ <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image012.png\" \/> OQ<\/p>\n<p style=\"text-align: justify;\">Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_Draw_a_circle_of_radius_3_cm_Take_two_points_P_and_Q_on_one_of_its_extended_diameter_each_at_a_distance_of_7_cm_from_its_centre_Draw_tangents_to_the_circle_from_these_two_points_P_and_Q\"><\/span><strong>3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. To construct<\/strong>: A circle of radius 3 cm and take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre and then draw tangents to the circle from these two points P and Q.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of Construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 12\" style=\"height: 198px; width: 266px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image013.jpg\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Bisect PO. Let M be the mid-point of PO.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points A and B.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Join PA and PB.<\/p>\n<p style=\"text-align: justify;\">Then PA and PB are the required two tangents.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> Bisect QO. Let N be the mid-point of QO.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Taking N as centre and NO as radius, draw a circle. Let it intersects the given circle at the points C and D.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Join QC and QD.<\/p>\n<p style=\"text-align: justify;\">Then QC and QD are the required two tangents.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>: Join OA and OB.<\/p>\n<p style=\"text-align: justify;\">Then <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>PAO is an angle in the semicircle and therefore <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>PAO = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/>.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/>PA <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image012.png\" \/> OA<\/p>\n<p style=\"text-align: justify;\">Since OA is a radius of the given circle, PA has to be a tangent to the circle. Similarly, PB is also a tangent to the circle.<\/p>\n<p style=\"text-align: justify;\">Again join OC and OD.<\/p>\n<p style=\"text-align: justify;\">Then <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>QCO is an angle in the semicircle and therefore <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>QCO = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Since OC is a radius of the given circle, QC has to be a tangent to the circle. Similarly, QD is also a tangent to the circle.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_Draw_a_pair_of_tangents_to_a_circle_of_radius_5_cm_which_are_inclined_to_each_other_at_an_angle_of\"><\/span><strong>4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of <\/strong><img decoding=\"async\" style=\"height: 21px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image014.png\" \/><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. To construct<\/strong>: A pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of <img decoding=\"async\" style=\"height: 21px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Steps of Construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 13\" style=\"height: 141px; width: 225px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image015.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a circle of radius 5 cm with centre O.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Draw an angle AOB of <img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image016.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> At A and B, draw <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/> angles which meet at C.<\/p>\n<p style=\"text-align: justify;\">Then AC and BC are the required tangents which are inclined to each other at an angle of <img decoding=\"async\" style=\"height: 21px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image017.png\" \/><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>OAC = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/>and OA is a radius.<\/p>\n<p style=\"text-align: justify;\">[By construction]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image003.png\" \/>AC is a tangent to the circle.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image017.png\" \/><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>OBC = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/>and OB is a radius.<\/p>\n<p style=\"text-align: justify;\">[By construction]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image003.png\" \/>BC is a tangent to the circle.<\/p>\n<p style=\"text-align: justify;\">Now, in quadrilateral OACB,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>AOB + <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>OAC + <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>OBC + <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>ACB = <img decoding=\"async\" style=\"height: 19px; width: 35px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image018.png\" \/><\/p>\n<p style=\"text-align: justify;\">[Angle sum property of a quadrilateral]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image019.png\" \/> + <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>ACB = <img decoding=\"async\" style=\"height: 19px; width: 35px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image018.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 35px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image020.png\" \/> + <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>ACB = <img decoding=\"async\" style=\"height: 19px; width: 35px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image018.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/> <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>ACB = <img decoding=\"async\" style=\"height: 19px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image021.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_Draw_a_line_segment_AB_of_length_8_cm_Taking_A_as_centre_draw_a_circle_of_radius_4_cm_and_taking_B_as_centre_draw_another_circle_of_radius_3_cm_Construct_tangents_to_each_circle_from_the_centre_of_the_other_circle\"><\/span><strong>5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. To construct<\/strong>: A line segment of length 8 cm and taking A as centre, to draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Also, to construct tangents to each circle from the centre to the other circle.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of Construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 14\" style=\"height: 212px; width: 236px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image022.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Bisect BA. Let M be the mid-point of BA.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points P and Q.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Join BP and BQ.<\/p>\n<p style=\"text-align: justify;\">Then, BP and BQ are the required two tangents from B to the circle with centre A.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> Again, Let M be the mid-point of AB.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Taking M as centre and MB as radius, draw a circle. Let it intersects the given circle at the points R and S.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Join AR and AS.<\/p>\n<p style=\"text-align: justify;\">Then, AR and AS are the required two tangents from A to the circle with centre B.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>: Join BP and BQ.<\/p>\n<p style=\"text-align: justify;\">Then <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>APB being an angle in the semicircle is <img decoding=\"async\" style=\"height: 21px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image023.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/>BP <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image012.png\" \/> AP<\/p>\n<p style=\"text-align: justify;\">Since AP is a radius of the circle with centre A, BP has to be a tangent to a circle with centre A. Similarly, BQ is also a tangent to the circle with centre A.<\/p>\n<p style=\"text-align: justify;\">Again join AR and AS.<\/p>\n<p style=\"text-align: justify;\">Then <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>ARB being an angle in the semicircle is <img decoding=\"async\" style=\"height: 21px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image023.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/>AR <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image012.png\" \/> BR<\/p>\n<p style=\"text-align: justify;\">Since BR is a radius of the circle with centre B, AR has to be a tangent to a circle with centre B. Similarly, AS is also a tangent to the circle with centre B.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_Let_ABC_be_a_right_triangle_in_which_AB_6_cm_BC_8_cm_and_B_BD_is_the_perpendicular_from_B_on_AC_The_circle_through_B_C_D_is_drawn_Construct_the_tangents_from_A_to_this_circle\"><\/span><strong>6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and <\/strong><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/><strong>B = <\/strong><img decoding=\"async\" style=\"height: 21px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image023.png\" \/><strong> BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. To construct<\/strong>: A right triangle ABC with AB = 6 cm, BC = 8 cm and <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>B = <img decoding=\"async\" style=\"height: 21px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image023.png\" \/> BD is the perpendicular from B on AC and the tangents from A to this circle.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of Construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 15\" style=\"height: 161px; width: 156px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image024.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a right triangle ABC with AB = 6 cm, BC = 8 cm and <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>B = <img decoding=\"async\" style=\"height: 21px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image023.png\" \/> Also, draw perpendicular BD on AC.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Join AO and bisect it at M (here O is the centre of circle through B, C, D).<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points B and E.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> Join AB and AE.<\/p>\n<p style=\"text-align: justify;\">Then AB and AE are the required two tangents.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>: Join OE.<\/p>\n<p style=\"text-align: justify;\">Then, <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>AEO is an angle in the semicircle.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>AEO = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/> AE <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image012.png\" \/> OE<\/p>\n<p style=\"text-align: justify;\">Since OE is a radius of the given circle, AE has to be a tangent to the circle. Similarly, AB is also a tangent to the circle.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7_Draw_a_circle_with_the_help_of_a_bangle_Take_a_point_outside_the_circle_Construct_the_pair_of_tangents_from_this_point_to_the_circle\"><\/span><strong>7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. To construct<\/strong>: A circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of Construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 16\" style=\"height: 191px; width: 241px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image025.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a circle with the help of a bangle.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Take two non-parallel chords AB and CD of this circle.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Draw the perpendicular bisectors of AB and CD. Let these intersect at O. Then O is the centre of the circle draw.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> Take a point P outside the circle.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Join PO and bisect it. Let M be the mid-point of PO.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.<\/p>\n<p style=\"text-align: justify;\"><strong>(g)<\/strong> Join PQ and PR.<\/p>\n<p style=\"text-align: justify;\">Then PQ and PR are the required two tangents.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>: Join OQ and OR.<\/p>\n<p style=\"text-align: justify;\">Then, <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>PQO is an angle in the semicircle.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image002.png\" \/>PQO = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image010.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image011.png\" \/> PQ <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch11\/Ex11.2\/image012.png\" \/> OQ<\/p>\n<p style=\"text-align: justify;\">Since OQ is a radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Exercise_112\"><\/span>NCERT Solutions for Class 10 Maths Exercise 11.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions for Class 10 Maths Exercise 11.2 PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_10\"><\/span>CBSE app for Class 10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 10 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\">the best app for CBSE\u00a0<\/a><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 10 Maths Exercise 11.2 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. 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