{"id":6089,"date":"2023-09-20T10:00:18","date_gmt":"2023-09-20T10:00:18","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6089"},"modified":"2023-09-19T16:54:11","modified_gmt":"2023-09-19T16:54:11","slug":"ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2"},"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; 12 (Areas Related to Circles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 12 Areas Related to Circles <\/strong>Exercise 12.2 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 : 12 Areas Related to Circles<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 12.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 12.3<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 12.2<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Unless stated otherwise, use \u03c0 =22\/7.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Find the area of a sector of a circle with a radius 6 cm if the angle of the sector is 60\u00b0.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Area of the sector making angle \u03b8 = (\u03b8\/360\u00b0) \u00d7 \u03c0 r<sup>2<br \/>\n<\/sup>Area of the sector making angle 60\u00b0 = (60\u00b0\/360\u00b0) \u00d7 \u03c0 r<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= (1\/6) \u00d7 6<sup>2<\/sup>\u03c0\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 36\/6 \u03c0 cm<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 6 \u00d7 22\/7 cm<sup>2<br \/>\n<\/sup>= 132\/7 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Find the area of a quadrant of a circle whose circumference is 22 cm.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Quadrant of a circle means sector is making angle 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Circumference of the circle = 2\u03c0r = 22 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Radius of the circle = r = 22\/2\u03c0 cm = 7\/2 cm\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the sector making angle 90\u00b0 = (90\u00b0\/360\u00b0) \u00d7 \u03c0 r<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= (1\/4) \u00d7 (7\/2)<sup>2<\/sup>\u03c0\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (49\/16) \u03c0 cm<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">=\u00a0(49\/16)\u00a0\u00d7 (22\/7) cm<sup>2<br \/>\n<\/sup>= 77\/8 cm<sup>2<br \/>\n<\/sup>= 9.6 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Minute hand of clock acts as radius of the circle.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Radius of the circle (r) = 14 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Angle rotated by minute hand in 1 hour = 360\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Angle rotated by minute hand in 5 minutes = 360\u00b0 \u00d7 5\/60 = 30\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the sector making angle 30\u00b0 = (30\u00b0\/360\u00b0) \u00d7 \u03c0 r<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= (1\/12) \u00d7 14<sup>2<\/sup>\u03c0\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 196\/12 \u03c0 cm<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (49\/3) \u00d7 (22\/7) cm<sup>2<br \/>\n<\/sup>= 154\/3 cm<sup>2<br \/>\n<\/sup>Area swept by the minute hand in 5 minutes = 154\/3 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:<br \/>\n<\/strong><strong>(i)<\/strong> minor segment<br \/>\n<strong>(ii)<\/strong> major sector. (Use \u03c0 = 3.14)<strong><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>In a circle with radius r and the angle at the centre with degree measure \u03b8 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Area of the sector = \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<\/sup><br \/>\n<strong>(ii)<\/strong> Area of the segment = Area of the sector &#8211; Area of the corresponding triangle<br \/>\nArea of the right triangle = 1\/2 \u00d7 base \u00d7 height<br \/>\nLet&#8217;s draw a figure to visualize the area to be calculated.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6613\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"318\" height=\"298\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans4.png 318w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans4-300x281.png 300w\" sizes=\"auto, (max-width: 318px) 100vw, 318px\" \/><br \/>\nHere, Radius, r = 10 cm, \u03b8 = 90\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Visually it\u2019s clear from the figure that, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB is the\u00a0<a style=\"color: #000000;\" href=\"https:\/\/www.cuemath.com\/geometry\/Chords-of-a-circle\/\">chord<\/a>\u00a0that subtends a\u00a0<a style=\"color: #000000;\" href=\"https:\/\/www.cuemath.com\/geometry\/right-angle\/\">right angle<\/a> at the centre. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(i) Area of minor segment APB = Area of sector OAPB &#8211; Area of right triangle AOB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(ii) Area of major segment AQB = \u03c0r\u00b2 &#8211; Area of minor segment APB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the right triangle \u0394AOB = 1\/2 \u00d7 OA \u00d7 OB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)<\/strong> Area of minor segment APB = Area of sector OAPB &#8211; Area of right \u0394AOB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<\/sup> &#8211; 1\/2 \u00d7 OA \u00d7 OB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 90\u00b0\/360\u00b0 \u00d7 \u03c0r<sup>2<\/sup> &#8211; 1\/2 \u00d7 r \u00d7 r <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/4 \u03c0r<sup>2<\/sup>\u00a0-1\/2r<sup>2<br \/>\n<\/sup>= r<sup>2<\/sup> (1\/4 \u03c0 &#8211; 1\/2) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= r<sup>2\u00a0<\/sup>(3.14 &#8211; 2)\/4 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (r<sup>2<\/sup> \u00d7 1.14)\/4 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (10 \u00d7 10 \u00d7 1.14)\/4 cm\u00b2 (Since radius r is given as 10 cm) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 28.5 cm\u00b2<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Area of major sector AOBQ\u00a0= \u03c0r<sup>2<\/sup> &#8211; Area of minor sector OAPB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0r<sup>2<\/sup>\u00a0&#8211; \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>=\u00a0\u03c0r<sup>2<\/sup> (1 &#8211; 90\u00b0\/360\u00b0) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 3.14 \u00d7 (10 cm)<sup>2\u00a0<\/sup>\u00d7 3\/4 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 235.5 cm<sup>2 <\/sup><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. In a circle of radius 21 cm, an arc subtends an angle of 60\u00b0 at the centre. Find:<br \/>\n<\/strong><strong>(i)<\/strong> the length of the arc<br \/>\n<strong>(ii)<\/strong> area of the sector formed by the arc<br \/>\n<strong>(iii)<\/strong> area of the segment formed by the corresponding chord<br \/>\n<strong>Solution &#8211;\u00a0 <\/strong>Let&#8217;s draw a figure to visualize the problem.<strong><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6614\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"259\" height=\"246\" \/><br \/>\n<\/strong>Here, r = 21 cm, \u03b8 = 60\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Visually it\u2019s clear from the figure that, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the segment APB = Area of sector AOPB &#8211; Area of \u0394AOB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)\u00a0<\/strong>Length of the Arc, APB = \u03b8\/360\u00b0 \u00d7 2\u03c0r<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 60\u00b0\/360\u00b0 \u00d7 2 \u00d7 22\/7 \u00d7 21 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 22 cm <\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Area of the sector, AOBP = \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 60\u00b0\/360\u00b0 \u00d7 22\/7 \u00d7 21 \u00d7 21 cm<sup>2<br \/>\n<\/sup>= 231 cm<sup>2<\/sup><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> Area of the segment = Area of the sector AOBP &#8211; Area of the triangle AOB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">To find the area of the segment, we need to find the area of \u0394AOB<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6615\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans5I.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"235\" height=\"210\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394AOB, draw OM \u22a5 AB. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Consider \u0394OAM and \u0394OMB, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OA = OB (radii of the circle) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OM = OM (common side) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220OMA = \u2220OMB = 90\u00b0 (Since OM \u22a5 AB) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, \u0394OMB \u2245 \u0394OMA (By\u00a0RHS Congruency)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">So, AM = MB (Corresponding parts of the\u00a0congruent triangles are always equal) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220AOM = \u2220BOM = 1\/2 \u00d7 60\u00b0 = 30\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394AOM, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">cos 30\u00b0 = OM\/OA and sin 30\u00b0 = AM\/OA <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u221a3\/2 = OM\/r and 1\/2 = AM\/r <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OM = (\u221a3\/2) r and AM = (1\/2) r <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = 2AM <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = 2 \u00d7 (1\/2) r <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = r <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, area of \u0394AOB = 1\/2 \u00d7 AB \u00d7 OM <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 \u00d7 r \u00d7 (\u221a3\/2) r <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 \u00d7 21 cm \u00d7 (\u221a3\/2) \u00d7 21 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 441\u221a3\/4 cm<sup>2<br \/>\n<\/sup>Area of the segment formed by the chord\u00a0= Area of the sector AOBP &#8211; Area of the triangle AOB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (231 &#8211; 441\u221a3\/4) cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. A chord of a circle of radius 15 cm subtends an angle of 60\u00b0 at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use \u03c0 = 3.14 and \u221a3 = 1.73)<br \/>\n<\/strong><\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6629\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"254\" height=\"251\" \/><br \/>\n<\/span><strong>Given &#8211;\u00a0 <\/strong>Radius of the circle = 15 cm<br \/>\n\u03b8 = 60\u00b0<br \/>\nSo,<br \/>\nThe area of sector OAPB = (60\u00b0\/360\u00b0) \u00d7 \u03c0r<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 225\/6 \u03c0 cm<sup>2<br \/>\n<\/sup>Now, \u0394AOB is equilateral because two sides are circle radii and hence equal, and one angle is 60\u00b0.<br \/>\nAlternatively, the area of \u0394AOB = (\u221a3\/4) \u00d7 a<sup>2<br \/>\n<\/sup>= (\u221a3\/4) \u00d7 15<sup>2<br \/>\n<\/sup>= 97.31 cm<sup>2<br \/>\n<\/sup>Therefore, the area of \u0394AOB = 97.31 cm<sup>2<br \/>\n<\/sup>Now, Area of OAPB \u2013 Area of \u0394AOB = Area of minor segment APB<br \/>\nAlternatively, the area of minor segment APB = ((225\/6)\u03c0 \u2013 97.31) cm<sup>2\u00a0<\/sup><br \/>\n= 225\/6 \u00d7 3.14 &#8211; 97.31 cm<sup>2<br \/>\n<\/sup>= 117.75 &#8211; 97.31 cm<sup>2<br \/>\n<\/sup>=\u00a0 20.43 cm<sup>2<br \/>\n<\/sup>Area of circle \u2013 Area of segment APB = Area of major segment<br \/>\nAlternatively, the area of major segment = (\u03c0 \u00d7 15<sup>2<\/sup>) \u2013 20.4<br \/>\n= 3.14 \u00d7 225 &#8211; 20.4 cm<sup>2<\/sup><br \/>\n= 706.5 &#8211; 20.4 cm<sup>2<br \/>\n<\/sup>= 686.06 cm<sup>2<\/sup><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. A chord of a circle of radius 12 cm subtends an angle of 120\u00b0 at the centre. Find the area of the corresponding segment of the circle. (Use \u03c0 = 3.14 and \u221a3 = 1.73)<br \/>\n<\/strong><\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6630\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans-7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"332\" height=\"254\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans-7.png 332w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans-7-300x230.png 300w\" sizes=\"auto, (max-width: 332px) 100vw, 332px\" \/><br \/>\nGiven &#8211;\u00a0<\/strong><\/span>Radius, r = 12 cm<br \/>\nDraw a perpendicular OD on chord AB now, and it will bisect the chord AB.<br \/>\nSo, AD = DB<br \/>\nNow, the area of the minor sector = (\u03b8\/360\u00b0) \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= (120\/360) \u00d7 3.14 \u00d7 12<sup>2<br \/>\n<\/sup>= 1\/3 \u00d7 3.14 \u00d7 144<br \/>\n= 150.72 cm<sup>2<br \/>\n<\/sup>Now, considering the \u0394AOB,<br \/>\n\u2220 OAB = 180\u00b0 &#8211; (90\u00b0 + 60\u00b0)<br \/>\n= 30\u00b0<br \/>\nNow, cos 30\u00b0 = AD\/OA<br \/>\n\u221a3\/2 = AD\/12<br \/>\nAD = 6\u221a3 cm<br \/>\nWe know that OD bisects AB .<br \/>\nSo, AB = 2 \u00d7 AD = 12\u221a3 cm<br \/>\nAs a result, sin 30\u00b0 = OD\/OA<br \/>\nOr,<br \/>\n\u00bd = OD\/12<br \/>\nTherefore, OD = 6 cm<br \/>\nAlternatively, the area of \u0394AOB = \u00bd \u00d7 base \u00d7 height<br \/>\nBase = AB = 12\u221a3 and<br \/>\nHere, height = OD = 6<br \/>\nAlternatively, the area of \u0394AOB = \u00bd \u00d7 12\u221a3 \u00d7 6<br \/>\n= 36\u221a3 cm<sup>2<\/sup><br \/>\n= 36 \u00d7 <span style=\"color: #000000; font-family: Georgia, Palatino;\">1.73 <\/span>cm<sup>2<\/sup><br \/>\n= 62.28 cm<sup>2<br \/>\n<\/sup>Therefore, area of the Minor sector \u2013 area of \u0394AOB = Area of the corresponding Minor segment<br \/>\n= 150.72 cm<sup>2<\/sup>\u2013 62.28 cm<sup>2<br \/>\n<\/sup>= 88.44 cm<sup>2<\/sup><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 12.11). Find<br \/>\n<\/strong><\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)<\/strong> the area of that part of the field in which the horse can graze.<br \/>\n<\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use \u03c0 = 3.14)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6631\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"175\" height=\"176\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-8.png 265w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-8-150x150.png 150w\" sizes=\"auto, (max-width: 175px) 100vw, 175px\" \/><br \/>\n<\/span><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">As the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with \u03b8 = 90\u00b0) of the field with a radius 5 m. <\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> From the figure it\u2019s clear that the horse can graze an area of a sector of a circle with radius (r) 5m and an angle with a degree measure 90\u00b0 [Since its a square field] <\/span><br \/>\n<span style=\"color: #000000;\">Length of the rope = 5 m <\/span><br \/>\n<span style=\"color: #000000;\">Area of the field the horse can graze =\u00a0Area of the sector with\u00a0\u03b8 = 90\u00b0 and r =\u00a05 <\/span><br \/>\n<span style=\"color: #000000;\">= \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 90\u00b0\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 1\/4\u00a0\u00d7 \u03c0 \u00d7 (5 m)<sup>2<br \/>\n<\/sup>= 25\/4 \u00d7 3.14 m<sup>2<br \/>\n<\/sup>= 19.625 m<sup>2<\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)\u00a0<\/strong>The area that can be grazed by the horse when the length of the rope is 10m, is the area of the sector of a circle with a radius (r) 10 m and an angle with a degree measure 90\u00b0. <\/span><br \/>\n<span style=\"color: #000000;\">Area of the field the horse can graze = Area of the\u00a0sector with\u00a0\u03b8 = 90\u00b0 and r = 10 <\/span><br \/>\n<span style=\"color: #000000;\">= 90\u00b0\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 1\/4 \u00d7 3.14 \u00d7 100 m<sup>2<br \/>\n<\/sup>= 78.5 m<sup>2<br \/>\n<\/sup>Increase in the grazing area <\/span><br \/>\n<span style=\"color: #000000;\">= 78.5 m<sup>2<\/sup>\u00a0&#8211; 19.625 m<sup>2<br \/>\n<\/sup>= 58.875 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>9. A brooch is made with silver wire in the form of a circle with a diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors, as shown in Fig. 12.12. Find:<br \/>\n<\/strong><strong>(i)<\/strong> the total length of the silver wire required.<strong><br \/>\n<\/strong><strong>(ii) <\/strong>the area of each sector of the brooch.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6633\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"147\" height=\"130\" \/><br \/>\n<strong>Solution &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)\u00a0<\/strong>Diameter of the brooch (d) = 35 mm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Total length of silver wire required = circumference of brooch + 5 \u00d7 diameter <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0d + 5d <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (\u03c0 + 5) \u00d7 35 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (22\/7 + 5) \u00d7 35 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (22 + 35)\/7 \u00d7 35 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 57 \u00d7 5 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 285 mm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)\u00a0<\/strong>Radius of the brooch (r) = 35\/2 mm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">The wire divides the brooch into 10 equal sectors. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">So, angle of the sector (\u03b8) = 360\u00b0\/10 = 36\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234\u00a0Area of each sector of the brooch\u00a0= 36\u00b0\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 1\/10 \u00d7 22\/7 \u00d7 35\/2 mm \u00d7 35\/2 mm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 96.25 mm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6634\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"245\" height=\"180\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-10.png 314w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-10-300x221.png 300w\" sizes=\"auto, (max-width: 245px) 100vw, 245px\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>There are 8 equally spaced ribs in an umbrella, so the umbrella is assumed to be a flat circle.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 The angle between 2 consecutive ribs (\u03b8) = 360\u00b0\/8 = 45\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">The radius of the flat circle (r) is given as 45 cm. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">The area between 2 consecutive ribs of the umbrella =\u00a0Area of a sector\u00a0with an angle of 45\u00b0 = \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 45\u00b0\/360\u00b0 \u00d7 22\/7 \u00d7 45 cm \u00d7 45 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/8 \u00d7 22\/7 \u00d7 45 cm \u00d7 45 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 22275\/28 cm<sup>2<br \/>\n<\/sup>= 795.535 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115\u00b0. Find the total area cleaned at each sweep of the blades.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Let&#8217;s draw a figure according to the given question.<br \/>\n<strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6635\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans-11.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"279\" height=\"174\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans-11.png 361w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans-11-300x187.png 300w\" sizes=\"auto, (max-width: 279px) 100vw, 279px\" \/><br \/>\n<\/strong><strong>Given,<br \/>\n<\/strong>Radius (r) = 25 cm<br \/>\nSector angle (\u03b8) = 115\u00b0<br \/>\nSince there are 2 blades,<br \/>\nThe total area of the sector made by wiper = 2 \u00d7 (\u03b8\/360\u00b0) \u00d7 \u03c0 r<sup>2<br \/>\n<\/sup>= 2 \u00d7 (115\/360) \u00d7 (22\/7) \u00d7 25<sup>2<br \/>\n<\/sup>= 2 \u00d7 158125\/252 cm<sup>2<br \/>\n<\/sup>= 158125\/126 = 1254.96 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>12. To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80\u00b0 to a distance of 16.5 km. Find the area of the sea over which the ships are warned. <\/strong><strong>(Use \u03c0 = 3.14)<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>The lighthouse spreads red coloured light over a sector of a\u00a0circle with a radius 16.5 km and an angle of\u00a0degree measure 80\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the sea over which the ships are warned = Area of the sector of the\u00a0circle with radius 16.5 km with an angle of degree measure 80\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 80\u00b0\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 2\/9\u03c0r<sup>2<br \/>\n<\/sup>= 2\/9 \u00d7 3.14 \u00d7 16.5 km \u00d7 16.5 km <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 189.97 km<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>13. A round table cover has six equal designs, as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of \u20b9 0.35 per cm<sup>2<\/sup>. (Use \u221a3 = 1.7)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6636\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Q-13.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"155\" height=\"176\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6637\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.2-Ans-13.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"184\" height=\"186\" \/><br \/>\n<\/strong>Total number of equal designs = 6 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AOB= 360\u00b0\/6 = 60\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">The radius of the cover = 28 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Cost of making design = \u20b9 0.35 per cm<sup>2<br \/>\n<\/sup>Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60\u00b0, \u0394AOB is an equilateral triangle. So, its area will be (\u221a3\/4) \u00d7 a<sup>2<\/sup> sq. units <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Here, a = OA <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Area of equilateral \u0394AOB = (\u221a3\/4) \u00d7 28<sup>2<br \/>\n<\/sup>= 1.7 \/ 4 \u00d7 28 \u00d7 28 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 333.2 cm<sup>2<br \/>\n<\/sup>Area of sector ACB = (60\u00b0\/360\u00b0) \u00d7 \u03c0r<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 1\/6 \u00d7 3.14 \u00d7 28 \u00d7 28<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 410.66 cm<sup>2<br \/>\n<\/sup>So, the area of a single design = the area of sector ACB \u2013 the area of \u0394AOB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 410.66 cm<sup>2<\/sup>\u00a0\u2013 333.2 cm<sup>2<br \/>\n<\/sup>= 77.46 cm<sup>2<br \/>\n<\/sup>\u2234 area of 6 designs = 6 \u00d7 77.46 cm<sup>2<br \/>\n<\/sup>= 464.76 cm<sup>2<br \/>\n<\/sup>So, total cost of making design = 464.76 cm<sup>2\u00a0<\/sup>\u00d7 Rs.0.35 per cm<sup>2<br \/>\n<\/sup>= Rs. 162.66<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>14. Tick the correct solution in the following:<br \/>\n<\/strong><strong>The area of a sector of angle p (in degrees) of a circle with radius R is<br \/>\n<\/strong><strong>(A)<\/strong> p\/180 \u00d7 2\u03c0R<br \/>\n<strong>(B)<\/strong> p\/180 \u00d7 \u03c0 R<sup>2<br \/>\n<\/sup><strong>(C)<\/strong> p\/360 \u00d7 2\u03c0R<br \/>\n<strong>(D)<\/strong> p\/720 \u00d7 2\u03c0R<sup>2<br \/>\n<\/sup><strong>Solution &#8211;\u00a0 <\/strong>The area of a sector = (\u03b8\/360\u00b0) \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>Given, \u03b8 = p<br \/>\nSo, the area of sector = p\/360 \u00d7 \u03c0R<sup>2<br \/>\n<\/sup>Multiplying and dividing by 2 simultaneously,<br \/>\n= (p\/360) \u00d7 2\/2 \u00d7 \u03c0R<sup>2<br \/>\n<\/sup>= (p\/360) \u00d7 \u03c0R<sup>2<\/sup><br \/>\n= (p\/720) \u00d7 2\u03c0R<sup>2<br \/>\n<\/sup>So, option (D) is correct.<\/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; 12 (Areas Related to Circles)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 12 Areas Related to Circles Exercise 12.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 12 Areas Related to Circles NCERT Class [&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":[1045,1105,1107,1044,1049,1048],"class_list":["post-6089","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-12-areas-related-to-circles-solutions","tag-ncert-class-10-mathematics-exercise-12-2-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) - 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