{"id":6084,"date":"2023-09-18T04:31:15","date_gmt":"2023-09-18T04:31:15","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6084"},"modified":"2023-09-18T04:31:15","modified_gmt":"2023-09-18T04:31:15","slug":"ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 11 (Constructions)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 11 Constructions <\/strong>Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 11 Constructions<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 11.1<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.2<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">In each of the following, give also the justification of the construction:<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. \u00a0Draw 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><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Steps of construction:<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6602\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"382\" height=\"293\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png 425w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1-300x230.png 300w\" sizes=\"auto, (max-width: 382px) 100vw, 382px\" \/> <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step I &#8211;<\/strong> Take a point O as the centre and 6 cm radius. Draw a circle. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Take a point P in such a manner that OP = 10cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>With O and P as centres and radius more than half of OP draw arcs above and below OP to intersect at X and Y. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>Draw line segment XY to intersect OP at M. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>With M as the centre and OM as radius draw a circle to intersect the given circle at Q and R. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VI &#8211;\u00a0 <\/strong>Join PQ and PR. PQ and PR are the tangents, where PQ and PR = 8cm.\u00a0 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof<br \/>\n<\/strong>Since PQ \u22a5 OQ (Angle\u00a0in a\u00a0semicircle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220PQO = 90\u00ba <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In right \u0394PQO, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OP = 10 cm, OQ = 6cm (radius) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PQ\u00b2 = OP\u00b2 &#8211; OQ\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (10)\u00b2 &#8211; (6)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 100 &#8211; 36 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 64 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PQ = \u221a64 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 8 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly, we can prove that PR = 8 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\nSolution &#8211; <\/strong>Steps of construction:<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6603\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"386\" height=\"333\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans2.png 386w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans2-300x259.png 300w\" sizes=\"auto, (max-width: 386px) 100vw, 386px\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step I &#8211;\u00a0 <\/strong>Take \u2018O\u2019 as centre and\u00a0radius\u00a04 cm and 6 cm respectively to\u00a0draw two\u00a0circles.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Take a point \u2018P\u2019 on the bigger circle and join OP.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>With \u2018O\u2019 and \u2018P\u2019 as centre and radius more than half of OP draw arcs above and below OP to intersect at X and Y.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>Join XY to intersect OP at M.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>With M as centre and OM as radius draw a circle to cut the smaller circle at Q and R.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VI &#8211; <\/strong>Join PQ and PR.\u00a0PQ and PR are the required\u00a0tangents, where PQ = PR = 4.5 cm (approx.)<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof<br \/>\n<\/strong>\u2220PQO = 90\u00ba (Angle\u00a0in a\u00a0semi-circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 PQ \u22a5 OQ <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OQ is the radius of the smaller circle and PQ is the tangent at Q. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In the right \u0394PQO, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OP = 6 cm (radius of the bigger circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OQ = 4 cm (radius of the smaller circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PQ\u00b2 = (OP)\u00b2 &#8211; (OQ)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (6)\u00b2 &#8211; (4)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 36 &#8211; 16 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 20 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PQ = \u221a20 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 4.5 (approx) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly, we can prove PR = 4.5 (approx.)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\nSolution &#8211; <\/strong>Steps of construction:<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6604\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"400\" height=\"317\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans3.png 400w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans3-300x238.png 300w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step I &#8211; <\/strong>Draw a circle with O as centre and radius as 3 cm.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Draw a\u00a0diameter\u00a0of it and extend both the sides and mark the points as P, Q such that OP = OQ = 7 cm.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>Draw the\u00a0perpendicular\u00a0bisectors of OP and OQ to intersect PQ at M and N respectively.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>With M as centre and OM as radius draw a circle to cut the given circle at A and C. With N as the centre and ON as radius draw a circle to cut the given circle at B and D.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>Join PA, PC, QB, QD.\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PA, PC and QB, QD are the required tangents from P and Q respectively.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof<br \/>\n<\/strong>\u2220PAO = \u2220QBO = 90\u00b0 (Angle\u00a0in a\u00a0semi-circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 PA \u22a5 AO, QB \u22a5 BO <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In\u00a0right angle triangle PAO, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OP = OQ = 7cm (By\u00a0construction) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OA = OB = 3cm (radius\u00a0of the given\u00a0circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PA\u00b2 = (OP)\u00b2 &#8211; (OA)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (7)\u00b2 &#8211; (3)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 49 &#8211; 9 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 40 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PA = \u221a40 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 6.3 (approx.) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Also, in right triangle QBO, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">QB\u00b2 = (OQ)\u00b2 &#8211; (OB)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (7)\u00b2 &#8211; (3)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 49 &#8211; 9 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 40 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">QB = \u221a40 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 6.3 (approx)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60\u00b0.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Steps of construction:<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6605\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"393\" height=\"259\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans4.png 443w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans4-300x198.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans4-350x230.png 350w\" sizes=\"auto, (max-width: 393px) 100vw, 393px\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step I &#8211;\u00a0<\/strong>With O as the centre and 5cm as radius draw a circle.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Take a point A on the circumference of the\u00a0circle\u00a0and join OA.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>Draw AX perpendicular to OA.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>Construct \u2220AOB = 120\u00b0 where B lies on the circumference.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>Draw BY\u00a0perpendicular\u00a0to OB.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VI &#8211; <\/strong>Both AX and BY intersect at P.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VII &#8211; <\/strong>PA and PB are the required tangents inclined at 60\u00b0.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof<br \/>\n<\/strong>\u2220OAP = \u2220OBP = 90\u00b0 (By construction) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220AOB = 120\u00b0 (By construction) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In\u00a0quadrilateral\u00a0OAPB, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220APB = 360\u00b0 &#8211; [\u2220OAP + \u2220OBP + \u2220AOB] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 360\u00b0 &#8211; [90\u00b0 + 90\u00b0 + 120\u00b0] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 360\u00b0 &#8211; 300\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 60\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence PA and PB are the required\u00a0tangents\u00a0inclined at 60\u00b0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Steps of construction:\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6606\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"413\" height=\"387\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans5.png 413w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans5-300x281.png 300w\" sizes=\"auto, (max-width: 413px) 100vw, 413px\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step I &#8211; <\/strong>Draw AB = 8 cm. With A and B as centres, radii\u00a0as 4 cm and 3 cm respectively draw two circles.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Draw the\u00a0perpendicular\u00a0bisector of AB, intersecting AB at O.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>With O as the centre and OA as radius draw a circle that intersects the two\u00a0circles\u00a0at P, Q, R and S.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>Join BP, BQ, AR and AS.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>BP, BQ are the tangents from B to the circle with centre A.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VI &#8211; <\/strong>AR,\u00a0AS are the\u00a0tangents\u00a0from A to the circle with centre B.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof<br \/>\n<\/strong>\u2220APB = \u2220AQB = 90\u00b0 (Angle\u00a0in a\u00a0semi-circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 AP \u22a5 PB and AQ \u22a5 QB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, BP and BQ are the tangents to the circle with centre A. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly, AR and AS are the tangents to the circle with centre B.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \u2220B = 90\u00b0. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Steps of construction:\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6607\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"355\" height=\"300\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans6.png 355w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans6-300x254.png 300w\" sizes=\"auto, (max-width: 355px) 100vw, 355px\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step I &#8211; <\/strong>Draw BC = 8 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Draw the\u00a0perpendicular\u00a0at B and cut BA = 6 cm on it. Join AC and right \u0394ABC is obtained.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>Draw BD perpendicular to AC.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>Since \u2220BDC = 90\u00b0 and the circle has to pass through B, C and D, BC must be the diameter of this circle. So, take O as the midpoint of BC and with O as centre and OB as\u00a0radius\u00a0draw a circle that will pass through B, C and D.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>To draw tangents from A to the circle with centre O.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VI &#8211; <\/strong>Join OA, and draw its perpendicular bisector to intersect OA at point E.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VII &#8211; <\/strong>With E as centre and EA as radius draw a\u00a0circle\u00a0that intersects the previous circle at B and F.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step VIII &#8211; <\/strong>Join AF.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IX &#8211; <\/strong>Thus, AF and AB are the required tangents to the circle with centre O. <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof:<br \/>\n<\/strong>\u2220ABO = \u2220AFO = 90\u00b0 (Angle\u00a0in a\u00a0semi-circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 AB \u22a5 OB and AF \u22a5 OF (We know that the line joining the centre of a circle to the tangent is always perpendicular) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence AB and AF are the tangents from A to the circle with centre O.\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\nSolution &#8211; <\/strong>Steps of construction:\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6608\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"355\" height=\"411\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans7.png 355w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans7-259x300.png 259w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans7-300x347.png 300w\" sizes=\"auto, (max-width: 355px) 100vw, 355px\" \/><strong><br \/>\nStep I &#8211;\u00a0 <\/strong>Draw any circle using a bangle.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>To find its centre:<\/span><\/p>\n<ul style=\"text-align: justify;\">\n<li><span style=\"color: #000000; font-family: Georgia, Palatino;\">Draw two chords on\u00a0the circle say AB and CD.<\/span><\/li>\n<li><span style=\"color: #000000; font-family: Georgia, Palatino;\">Draw the perpendicular bisectors of AB and CD to intersect at O.<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>Now, \u2018O\u2019 is\u00a0the centre of the\u00a0circle (since the\u00a0perpendiculars\u00a0drawn from the centre of a circle to any\u00a0chord\u00a0bisect the chord and vice versa). <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>To draw the tangents from a point \u2018P\u2019 outside the circle:<\/span><\/p>\n<ul style=\"text-align: justify;\">\n<li><span style=\"color: #000000; font-family: Georgia, Palatino;\">Take a\u00a0point P outside the circle and draw the perpendicular bisector of OP which meets at OP at O\u2019.<\/span><\/li>\n<li><span style=\"color: #000000; font-family: Georgia, Palatino;\">With O\u2019 as the centre and OO\u2019 as\u00a0radius\u00a0draw a\u00a0circle\u00a0that cuts the given circle at Q and R.<\/span><\/li>\n<li><span style=\"color: #000000; font-family: Georgia, Palatino;\">Join PQ and PR.<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>Thus, PQ and PR are the required tangents.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof<br \/>\n<\/strong>\u2220QOP = \u2220ORP = 90\u00b0 (Angle\u00a0in a\u00a0semi-circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 OQ \u22a5 QP and OR \u22a5 RP. (We know that the line joining the centre of a circle to the tangent is always perpendicular) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, we have PQ and PR as\u00a0the\u00a0tangents\u00a0to the given circle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>\u00a0<\/strong><\/span><\/p>\n<p><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 11 (Constructions)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 11 Constructions NCERT Class 10 Maths Solution Ex &#8211; 11.1 Exercise &#8211; 11.2 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1102,1104,1044,1049,1048],"class_list":["post-6084","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-11-constructions-solutions","tag-ncert-class-10-mathematics-exercise-11-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 10 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.2\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions Class 10 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.2\" \/>\n<meta property=\"og:url\" content=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/\" \/>\n<meta property=\"og:site_name\" content=\"TheExamPillar NCERT\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-18T04:31:15+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png\" \/>\n<meta name=\"author\" content=\"Admin\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@exampillar\" \/>\n<meta name=\"twitter:site\" content=\"@exampillar\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Admin\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"9 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/\"},\"author\":{\"name\":\"Admin\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"headline\":\"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2\",\"datePublished\":\"2023-09-18T04:31:15+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/\"},\"wordCount\":1441,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex11.2-Ans1.png\",\"keywords\":[\"Class 10 NCERT Mathematics Solutions\",\"NCERT Class 10 Mathematics\u00a0Chapter 11 Constructions Solutions\",\"NCERT Class 10 Mathematics Exercise 11.2 Solutions\",\"NCERT Class 10 Mathematics Solutions Class 10 NCERT Solutions\",\"NCERT Solutions Class 10 Mathematics\",\"NCERT Solutions Class 10 Maths\"],\"articleSection\":[\"Class 10 Maths\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/\",\"name\":\"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2 | TheExamPillar NCERT\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#primaryimage\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex11.2-Ans1.png\",\"datePublished\":\"2023-09-18T04:31:15+00:00\",\"description\":\"NCERT Solutions Class 10 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.2\",\"breadcrumb\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#primaryimage\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex11.2-Ans1.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex11.2-Ans1.png\",\"width\":425,\"height\":326,\"caption\":\"NCERT Class 10 Maths Solution\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\\\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\",\"name\":\"TheExamPillar NCERT\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"alternateName\":\"NCERT Solution\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":[\"Person\",\"Organization\"],\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\",\"name\":\"Admin\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"width\":512,\"height\":512,\"caption\":\"Admin\"},\"logo\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\"},\"sameAs\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\"],\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/author\\\/ncert_eng_vikram\\\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2 | TheExamPillar NCERT","description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.2","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2","og_description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.2","og_url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/","og_site_name":"TheExamPillar NCERT","article_published_time":"2023-09-18T04:31:15+00:00","og_image":[{"url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png","type":"","width":"","height":""}],"author":"Admin","twitter_card":"summary_large_image","twitter_creator":"@exampillar","twitter_site":"@exampillar","twitter_misc":{"Written by":"Admin","Est. reading time":"9 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#article","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/"},"author":{"name":"Admin","@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"headline":"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2","datePublished":"2023-09-18T04:31:15+00:00","mainEntityOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/"},"wordCount":1441,"commentCount":0,"publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#primaryimage"},"thumbnailUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png","keywords":["Class 10 NCERT Mathematics Solutions","NCERT Class 10 Mathematics\u00a0Chapter 11 Constructions Solutions","NCERT Class 10 Mathematics Exercise 11.2 Solutions","NCERT Class 10 Mathematics Solutions Class 10 NCERT Solutions","NCERT Solutions Class 10 Mathematics","NCERT Solutions Class 10 Maths"],"articleSection":["Class 10 Maths"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/","url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/","name":"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2 | TheExamPillar NCERT","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/#website"},"primaryImageOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#primaryimage"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#primaryimage"},"thumbnailUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png","datePublished":"2023-09-18T04:31:15+00:00","description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.2","breadcrumb":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#primaryimage","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.2-Ans1.png","width":425,"height":326,"caption":"NCERT Class 10 Maths Solution"},{"@type":"BreadcrumbList","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/theexampillar.com\/ncert\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2"}]},{"@type":"WebSite","@id":"https:\/\/theexampillar.com\/ncert\/#website","url":"https:\/\/theexampillar.com\/ncert\/","name":"TheExamPillar NCERT","description":"","publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"alternateName":"NCERT Solution","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/theexampillar.com\/ncert\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":["Person","Organization"],"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1","name":"Admin","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","width":512,"height":512,"caption":"Admin"},"logo":{"@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png"},"sameAs":["https:\/\/theexampillar.com\/ncert"],"url":"https:\/\/theexampillar.com\/ncert\/author\/ncert_eng_vikram\/"}]}},"_links":{"self":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6084","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/comments?post=6084"}],"version-history":[{"count":4,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6084\/revisions"}],"predecessor-version":[{"id":6628,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6084\/revisions\/6628"}],"wp:attachment":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/media?parent=6084"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/categories?post=6084"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/tags?post=6084"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}