{"id":6888,"date":"2026-09-13T20:46:54","date_gmt":"2026-09-13T15:16:54","guid":{"rendered":"https:\/\/mymockmate.com\/notes\/?p=6888"},"modified":"2026-09-13T20:46:59","modified_gmt":"2026-09-13T15:16:59","slug":"class-9-maths-exercise-set-6-3-area-of-sectors-and-segments","status":"publish","type":"post","link":"https:\/\/mymockmate.com\/notes\/class-9-maths-exercise-set-6-3-area-of-sectors-and-segments\/","title":{"rendered":"Class 9 | Maths | Exercise Set 6.3 \u2013 Area of Sectors and Segments"},"content":{"rendered":"<div class=\"mymoc-top mymoc-entity-placement\" id=\"mymoc-3388908276\"><div id=\"mymoc-2086562338\"><a href=\"https:\/\/amzn.to\/440rYSB\" aria-label=\"ssd\"><img src=\"https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd.png\" alt=\"\" srcset=\"https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd.png 1303w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd-300x50.png 300w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd-1024x171.png 1024w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd-768x128.png 768w\" sizes=\"(max-width: 1303px) 100vw, 1303px\" width=\"1303\" height=\"218\"><\/a><\/div><\/div>\n<h2 class=\"wp-block-heading\">Quick Facts<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Area of a circle:<\/strong><br><strong>A = &pi;r&sup2;<\/strong><\/li>\n\n\n\n<li><strong>Area of a sector:<\/strong><br><strong>A = (&theta;\/360&deg;) &times; &pi;r&sup2;<\/strong><\/li>\n\n\n\n<li><strong>Area of a quadrant:<\/strong><br><strong>A = &frac14;&pi;r&sup2;<\/strong><\/li>\n\n\n\n<li><strong>Area of a semicircle:<\/strong><br><strong>A = &frac12;&pi;r&sup2;<\/strong><\/li>\n\n\n\n<li><strong>Area of a segment:<\/strong><br><strong>Area of segment = Area of sector &minus; Area of triangle<\/strong><\/li>\n\n\n\n<li><strong>Major sector angle:<\/strong><br><strong>360&deg; &minus; minor sector angle<\/strong><\/li>\n\n\n\n<li>A <strong>sector<\/strong> is the region enclosed by two radii and an arc.<\/li>\n\n\n\n<li>A <strong>segment<\/strong> is the region bounded by an arc and the chord joining the endpoints of that arc.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Story<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\">Imagine the minute hand of a clock moving around the clock face. It sweeps out a portion of a circle. If the hand moves for a few minutes, it does not cover the whole circular region&mdash;it covers a <strong>sector<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now imagine drawing a chord across a circle. The smaller region between the chord and the corresponding arc is called a <strong>segment<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">These ideas help us calculate the areas covered by clock hands, car wipers, circular gardens, curved windows and many other objects.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In Exercise 6.3, we move from the familiar area of a circle to sectors and segments and learn how angles determine the portion of the circular area involved. The textbook introduces the sector-area formula using the rotational symmetry of a circle.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Key Terms<\/h1>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Term<\/th><th>Meaning<\/th><\/tr><\/thead><tbody><tr><td><strong>Circle<\/strong><\/td><td>A set of points at a fixed distance from a centre.<\/td><\/tr><tr><td><strong>Radius (r)<\/strong><\/td><td>Distance from the centre to any point on the circle.<\/td><\/tr><tr><td><strong>Sector<\/strong><\/td><td>A part of a circular region enclosed by two radii and an arc.<\/td><\/tr><tr><td><strong>Quadrant<\/strong><\/td><td>A sector with a central angle of 90&deg;.<\/td><\/tr><tr><td><strong>Minor Sector<\/strong><\/td><td>The smaller sector formed by an angle less than 180&deg;.<\/td><\/tr><tr><td><strong>Major Sector<\/strong><\/td><td>The larger sector formed by the remaining part of the circle.<\/td><\/tr><tr><td><strong>Chord<\/strong><\/td><td>A line segment joining two points on a circle.<\/td><\/tr><tr><td><strong>Segment<\/strong><\/td><td>Region bounded by a chord and its corresponding arc.<\/td><\/tr><tr><td><strong>Minor Segment<\/strong><\/td><td>The smaller region between a chord and its minor arc.<\/td><\/tr><tr><td><strong>Major Segment<\/strong><\/td><td>The larger region between a chord and its major arc.<\/td><\/tr><tr><td><strong>Central Angle<\/strong><\/td><td>The angle made by two radii at the centre of a circle.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Questions<\/h1>\n\n\n\n<h2 class=\"wp-block-heading\">Question 1<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60&deg;.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Radius, <strong>r = 7 cm<\/strong><\/li>\n\n\n\n<li>Angle, <strong>&theta; = 60&deg;<\/strong><\/li>\n\n\n\n<li>&pi; = <strong>22\/7<\/strong><\/li>\n<\/ul>\n\n\n\n<div class=\"internal-linking-related-contents\"><a href=\"https:\/\/mymockmate.com\/notes\/class-9-maths-exercise-1-1-solutions-coordinate-geometry-ncert-solutions\/\" class=\"template-2\"><span class=\"cta\">Related Topic to Read more<\/span><span class=\"postTitle\">NCERT Class 9 Maths Exercise 1.1 Solutions | Coordinate Geometry NCERT Solutions<\/span><\/a><\/div><p class=\"wp-block-paragraph\">Formula:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area of sector = (&theta;\/360&deg;) &times; &pi;r&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substituting:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (60\/360) &times; (22\/7) &times; 7&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1\/6 &times; 22\/7 &times; 49<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 77\/3<\/p>\n\n\n\n<div class=\"internal-linking-related-contents\"><a href=\"https:\/\/mymockmate.com\/notes\/class-9-maths-exercise-1-2-solutions-coordinate-geometry-ncert-solutions\/\" class=\"template-2\"><span class=\"cta\">Related Topic to Read more<\/span><span class=\"postTitle\">NCERT Class 9 Maths Exercise 1.2 Solutions | Coordinate Geometry NCERT Solutions<\/span><\/a><\/div><p class=\"wp-block-paragraph\">= <strong>25&#8532; cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area of the sector = 25&#8532; cm&sup2;<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h2 class=\"wp-block-heading\">Question 2<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Find the area of a quadrant of a circle whose circumference is 44 cm.<\/strong><\/p><div class=\"mymoc-middle mymoc-entity-placement\" id=\"mymoc-1997537248\"><div id=\"mymoc-2946078712\"><a href=\"https:\/\/amzn.to\/4xnw9oY\" aria-label=\"head-phones\"><img src=\"https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/head-phones.png\" alt=\"\" srcset=\"https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/head-phones.png 1301w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/head-phones-300x80.png 300w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/head-phones-1024x274.png 1024w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/head-phones-768x205.png 768w\" sizes=\"(max-width: 1301px) 100vw, 1301px\" width=\"1301\" height=\"348\"><\/a><\/div><\/div>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Circumference = 44 cm<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Formula:<\/p>\n\n\n\n<div class=\"internal-linking-related-contents\"><a href=\"https:\/\/mymockmate.com\/notes\/class-9-maths-end-of-chapter-exercise-solutions-coordinate-geometry-ncert-solutions\/\" class=\"template-2\"><span class=\"cta\">Related Topic to Read more<\/span><span class=\"postTitle\">NCERT Class 9 Maths End of Chapter Exercise Solutions | Coordinate Geometry<\/span><\/a><\/div><p class=\"wp-block-paragraph\"><strong>Circumference = 2&pi;r<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">44 = 2 &times; 22\/7 &times; r<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">44 = 44r\/7<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>r = 7 cm<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A quadrant is one-fourth of a circle.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of quadrant = &frac14;&pi;r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &frac14; &times; 22\/7 &times; 7&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &frac14; &times; 154<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>38.5 cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area of the quadrant = 38.5 cm&sup2;<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h2 class=\"wp-block-heading\">Question 3<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The minute hand completes:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>360&deg; in 60 minutes<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore, in 10 minutes it sweeps:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&theta; = (10\/60) &times; 360&deg;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>60&deg;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The radius of the swept sector is the length of the minute hand:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>r = 7 cm<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area swept:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (60\/360) &times; 22\/7 &times; 7&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1\/6 &times; 154<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>77\/3 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>25&#8532; cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area swept = 25&#8532; cm&sup2;<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h2 class=\"wp-block-heading\">Question 4<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A chord of a circle of radius 10 cm subtends 90&deg; at the centre. Find the area of the corresponding:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(i) minor sector<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(ii) major sector<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Use <strong>&pi; &asymp; 3.14<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(i) Minor Sector<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>r = 10 cm<\/li>\n\n\n\n<li>&theta; = 90&deg;<\/li>\n\n\n\n<li>&pi; = 3.14<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Area of minor sector:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (90\/360) &times; 3.14 &times; 10&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &frac14; &times; 314<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>78.5 cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(ii) Major Sector<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The major sector angle is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">360&deg; &minus; 90&deg; = <strong>270&deg;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of major sector:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (270\/360) &times; 3.14 &times; 10&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &frac34; &times; 314<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>235.5 cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(i) Minor sector = 78.5 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(ii) Major sector = 235.5 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Check:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">78.5 + 235.5 = 314 cm&sup2;, which is the area of the complete circle.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Question 5<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A chord of a circle of radius 15 cm subtends an angle of 60&deg; at the centre. Find the areas of the corresponding minor and major segments of the circle.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Use <strong>&pi; &asymp; 3.14<\/strong> and <strong>&radic;3 &asymp; 1.73<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Area of the Minor Sector<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">r = 15 cm<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&theta; = 60&deg;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of minor sector:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (60\/360) &times; 3.14 &times; 15&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1\/6 &times; 3.14 &times; 225<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>117.75 cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Area of the Triangle<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The two radii and the chord form an equilateral triangle because the two radii are 15 cm and the included angle is 60&deg;.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of an equilateral triangle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area = (&radic;3\/4)a&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1.73\/4 &times; 15&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1.73\/4 &times; 225<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>97.3125 cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Minor Segment<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Minor segment = Minor sector &minus; Triangle<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 117.75 &minus; 97.3125<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>20.4375 cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Major Segment<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Area of complete circle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &pi;r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 3.14 &times; 225<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>706.5 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Major segment:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= Area of circle &minus; Minor segment<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 706.5 &minus; 20.4375<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>686.0625 cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Minor segment = 20.4375 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Major segment = 686.0625 cm&sup2;<\/strong><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\"><strong>Important:<\/strong> Do not subtract the triangle from the whole circle when finding the minor segment. The minor segment is specifically the <strong>minor sector minus the triangle<\/strong>.<\/p>\n<\/blockquote>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Question 6<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120&deg;. Find the total area cleaned at each sweep of the blades.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">For each wiper:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>r = 28 cm<\/li>\n\n\n\n<li>&theta; = 120&deg;<\/li>\n\n\n\n<li>&pi; = 22\/7<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Area swept by one wiper:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (120\/360) &times; 22\/7 &times; 28&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1\/3 &times; 22\/7 &times; 784<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>616\/3 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since there are two non-overlapping wipers:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Total area:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 2 &times; 616\/3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1232\/3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>410&#8532; cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Total area cleaned = 410&#8532; cm&sup2;<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Question 7<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A chord of a circle of radius r subtends an angle of 60&deg; at the centre. Show that the area of the corresponding minor segment is<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>&pi;r&sup2;\/6 &minus; (&radic;3\/4)r&sup2;.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Proof<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The minor segment is obtained by removing the triangle from the corresponding minor sector.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Area of the 60&deg; sector<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Area of sector:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (60\/360) &times; &pi;r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>&pi;r&sup2;\/6<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Area of the triangle<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The two radii are both r and the included angle is 60&deg;. Therefore, the triangle is equilateral with side r.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of the equilateral triangle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>&radic;3\/4 &times; r&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Therefore<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Area of minor segment:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= Area of sector &minus; Area of triangle<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &pi;r&sup2;\/6 &minus; &radic;3r&sup2;\/4<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Hence,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area of minor segment = &pi;r&sup2;\/6 &minus; (&radic;3\/4)r&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Proved.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Question 8<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is 3&radic;3\/(4&pi;) &asymp; 0.413.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">For an equilateral triangle inscribed in a circle, each side subtends 60&deg; at the centre.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The side of the equilateral triangle is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>a = &radic;3r<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of triangle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &radic;3\/4 &times; a&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &radic;3\/4 &times; (&radic;3r)&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &radic;3\/4 &times; 3r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>3&radic;3r&sup2;\/4<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of circle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>&pi;r&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ratio:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (3&radic;3r&sup2;\/4) &divide; &pi;r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>3&radic;3\/(4&pi;)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using &pi; &asymp; 22\/7 and &radic;3 &asymp; 1.73:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">3 &times; 1.73 \/ (4 &times; 22\/7)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&asymp; <strong>0.413<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Ratio = 3&radic;3\/(4&pi;) &asymp; 0.413<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Question 9<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is 2\/&pi; &asymp; 0.637.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">For a square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Diagonal = 2r<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">For a square:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Diagonal = side &times; &radic;2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">side &times; &radic;2 = 2r<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">side = &radic;2r<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of square:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (&radic;2r)&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>2r&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of circle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>&pi;r&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ratio:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 2r&sup2; \/ &pi;r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>2\/&pi;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using &pi; &asymp; 22\/7:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">2\/(22\/7)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 14\/22<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 7\/11<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&asymp; <strong>0.636<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, to three decimal places:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>&asymp; 0.637<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Ratio = 2\/&pi; &asymp; 0.637<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Question 10<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is 3&radic;3\/(2&pi;) &asymp; 0.827. Can you see why the answer is exactly twice the answer to Question 8?<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Join the centre of the circle to all six vertices of the regular hexagon.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This divides the hexagon into <strong>six equilateral triangles<\/strong>, each having side r.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of one equilateral triangle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= &radic;3\/4 &times; r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore, area of six triangles:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 6 &times; &radic;3\/4 &times; r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>3&radic;3r&sup2;\/2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of hexagon:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>3&radic;3r&sup2;\/2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area of circle:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>&pi;r&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ratio:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= (3&radic;3r&sup2;\/2) &divide; &pi;r&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>3&radic;3\/(2&pi;)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using &pi; &asymp; 22\/7 and &radic;3 &asymp; 1.73:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>approximately 0.827<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why is it twice Question 8?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Question 8 gives:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3&radic;3\/(4&pi;)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Question 10 gives:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3&radic;3\/(2&pi;)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>1\/2 = 2 &times; 1\/4<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">we have:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3&radic;3\/(2&pi;) = 2 &times; [3&radic;3\/(4&pi;)]<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Hence, the answer to Question 10 is <strong>exactly twice<\/strong> the answer to Question 8.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Answer<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Ratio = 3&radic;3\/(2&pi;) &asymp; 0.827<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Memory Tricks<\/h1>\n\n\n\n<h3 class=\"wp-block-heading\">1. Sector Formula<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Remember:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>&ldquo;Angle over 360, multiplied by &pi;r&sup2;.&rdquo;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area of sector = &theta;\/360 &times; &pi;r&sup2;<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h3 class=\"wp-block-heading\">2. Segment Formula<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Think:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>SEGMENT = SECTOR &minus; TRIANGLE<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Area of segment = Area of sector &minus; Area of triangle<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h3 class=\"wp-block-heading\">3. Common Circle Fractions<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Region<\/th><th>Angle<\/th><th>Fraction of Circle<\/th><\/tr><\/thead><tbody><tr><td>Full circle<\/td><td>360&deg;<\/td><td>1<\/td><\/tr><tr><td>Semicircle<\/td><td>180&deg;<\/td><td>1\/2<\/td><\/tr><tr><td>Quadrant<\/td><td>90&deg;<\/td><td>1\/4<\/td><\/tr><tr><td>60&deg; sector<\/td><td>60&deg;<\/td><td>1\/6<\/td><\/tr><tr><td>120&deg; sector<\/td><td>120&deg;<\/td><td>1\/3<\/td><\/tr><tr><td>270&deg; sector<\/td><td>270&deg;<\/td><td>3\/4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h3 class=\"wp-block-heading\">4. Inscribed Shapes Trick<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Equilateral triangle:<\/strong> area ratio = <strong>3&radic;3\/(4&pi;)<\/strong><\/li>\n\n\n\n<li><strong>Square:<\/strong> area ratio = <strong>2\/&pi;<\/strong><\/li>\n\n\n\n<li><strong>Regular hexagon:<\/strong> area ratio = <strong>3&radic;3\/(2&pi;)<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">The hexagon can be divided into <strong>6 equilateral triangles<\/strong>, which makes its area easy to calculate.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Case Study<\/h1>\n\n\n\n<h2 class=\"wp-block-heading\">Case Study: Designing a Circular Wiper System<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">A car has two non-overlapping wipers. Each blade has a length of <strong>28 cm<\/strong> and sweeps through <strong>120&deg;<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Questions<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>1. What type of circular region does one wiper sweep?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A. Circle<br>B. Sector<br>C. Segment<br>D. Semicircle<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer:<\/strong> <strong>Sector<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. What fraction of the complete circle is swept by one wiper?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">120&deg;\/360&deg; = <strong>1\/3<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3. What is the area swept by one wiper?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Area = 1\/3 &times; 22\/7 &times; 28&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>616\/3 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>4. What is the total area cleaned by both wipers?<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 2 &times; 616\/3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>1232\/3 cm&sup2;<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= <strong>410&#8532; cm&sup2;<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Learning Outcome<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This case shows how the <strong>sector-area formula<\/strong> can be applied to a practical situation. The same mathematical idea can be used for rotating arms, sprinkler systems, radar sweeps and other circular motions.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The wiper problem appears as Question 6 in the exercise set.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h1 class=\"wp-block-heading\">Quiz<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\">A sector has radius 7 cm and central angle 60&deg;. Using &pi; = 22\/7, what is its area?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 of 5<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">154 cm&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">B<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">25&#8532; cm&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">C<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">38.5 cm&sup2;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">D<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">77 cm&sup2;Next<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Give feedback<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\">\n\n\n\n<h3 class=\"wp-block-heading\">Master Exercise 6.3 with MyMockMate<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Learn the concept &rarr; Understand the formula &rarr; Follow the solved examples &rarr; Practise &rarr; Take the quiz &rarr; Build exam confidence.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&#128073; <strong>Visit <a href=\"http:\/\/www.mymockmate.com\">www.mymockmate.com<\/a><\/strong> for structured chapter-wise learning resources, practice material and exam-oriented preparation.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Keep learning. Keep practising. Keep improving with MyMockMate.<\/strong><\/p>\n<div class=\"mymoc-bottom mymoc-entity-placement\" id=\"mymoc-3464850988\"><div id=\"mymoc-2086562338\"><a href=\"https:\/\/amzn.to\/440rYSB\" aria-label=\"ssd\"><img src=\"https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd.png\" alt=\"\" srcset=\"https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd.png 1303w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd-300x50.png 300w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd-1024x171.png 1024w, https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/06\/ssd-768x128.png 768w\" sizes=\"(max-width: 1303px) 100vw, 1303px\" width=\"1303\" height=\"218\"><\/a><\/div><\/div>","protected":false},"excerpt":{"rendered":"<p>Quick Facts Story Imagine the minute hand of a clock moving around the clock face. It sweeps out a portion of a circle. If the&#8230;<\/p>\n","protected":false},"author":1,"featured_media":6889,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"postBodyCss":"","postBodyMargin":[],"postBodyPadding":[],"postBodyBackground":{"backgroundType":"classic","gradient":""},"footnotes":""},"categories":[4,11],"tags":[],"class_list":["post-6888","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-9","category-maths-class-9"],"featured_image_src":"https:\/\/mymockmate.com\/notes\/wp-content\/uploads\/2026\/09\/Class-9-Maths-Exercise-Set-6.3-\u2013-Area-of-Sectors-and-Segments.png","author_info":{"display_name":"Team Mymockmate","author_link":"https:\/\/mymockmate.com\/notes\/author\/bsm_adm\/"},"_links":{"self":[{"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/posts\/6888","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/comments?post=6888"}],"version-history":[{"count":1,"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/posts\/6888\/revisions"}],"predecessor-version":[{"id":6890,"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/posts\/6888\/revisions\/6890"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/media\/6889"}],"wp:attachment":[{"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/media?parent=6888"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/categories?post=6888"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mymockmate.com\/notes\/wp-json\/wp\/v2\/tags?post=6888"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}