{"id":2957,"date":"2022-10-23T15:03:05","date_gmt":"2022-10-23T15:03:05","guid":{"rendered":"https:\/\/careerssutra.com\/?p=2957"},"modified":"2022-11-17T01:56:15","modified_gmt":"2022-11-17T01:56:15","slug":"ncert-solutions-for-class-10-maths-chapter-12-areas-related-to-circles","status":"publish","type":"post","link":"https:\/\/careerssutra.com\/ncert-solutions-for-class-10-maths-chapter-12-areas-related-to-circles\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles"},"content":{"rendered":"\n<h5 class=\"wp-block-heading\"><strong>NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles<\/strong><\/h5>\n\n\n\n<p>Circle-Related Area The Class 10 chapter builds on the Class 9 chapter &#8216;Circles,&#8217; which you have already completed. There are many questions in NCERT Chapter 12 Maths Class 10 that use Class 9 concepts. There are three exercises with 35 questions in NCERT Class 10 Maths book chapter 12. The NCERT solutions for Class 10 Maths chapter 12 Areas Related to Circles include the correct answer as well as solutions to all of the questions.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"NCERT-Solutions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-Excercise-121----\"><strong>NCERT Solutions for Class 10 Maths Chapter 12<\/strong><strong>&nbsp;Areas Related to Circles Excercise: 12.1<\/strong><\/h2>\n\n\n\n<p><strong>Q1&nbsp;<\/strong>The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Circumference of 1st circle is given :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%202%5Cpi%20r_1%5C%5C%5C%5C%3D%5C%202%5Cpi%20%5Ctimes%2019%5C%20%3D%5C%2038%20%5Cpi%5C%20cm\" alt=\"\\\\=\\ 2\\pi r_1\\\\\\\\=\\ 2\\pi \\times 19\\ =\\ 38 \\pi\\ cm\"\/><\/figure>\n\n\n\n<p>And the circumference of the 2nd circle is:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%202%5Cpi%20%5Ctimes%209%5C%20%3D%5C%2018%20%5Cpi%5C%20cm\" alt=\"=\\ 2\\pi \\times 9\\ =\\ 18 \\pi\\ cm\"><\/p>\n\n\n\n<p>Thus the circumference of the new circle is\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2018%5Cpi%5C%20+%5C%2038%5Cpi%5C%20%3D%5C%2056%5Cpi%5C%20cm\" alt=\"=\\ 18\\pi\\ +\\ 38\\pi\\ =\\ 56\\pi\\ cm\"><\/p>\n\n\n\n<p>or\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?2%5Cpi%20r%20%3D%5C%2056%5Cpi\" alt=\"2\\pi r =\\ 56\\pi\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"r\\ =\\ 28\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?r%5C%20%3D%5C%2028%5C%20cm\"><\/p>\n\n\n\n<p>Hence the radius of the new circle is 28 cm.<\/p>\n\n\n\n<p><strong>Q2&nbsp;<\/strong>The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>We know that the area of the circle is:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2\" alt=\"=\\ \\pi r^2\"><\/p>\n\n\n\n<p>Thus area of a 1st circle:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2_1\" alt=\"=\\ \\pi r^2_1\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\pi \\times 8^2\\ =\\ 64 \\pi\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20%5Ctimes%208%5E2%5C%20%3D%5C%2064%20%5Cpi%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And the area of a 2nd circle is :&nbsp;<img decoding=\"async\" alt=\"=\\ \\pi r^2_2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2_2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\pi \\times 6^2\\ =\\ 36\\pi\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20%5Ctimes%206%5E2%5C%20%3D%5C%2036%5Cpi%5C%20cm%5E2\"><\/p>\n\n\n\n<p>According to the question area of the new circle is&nbsp;<img decoding=\"async\" alt=\"=\\ 64\\pi\\ +\\ 36\\pi\\ =\\ 100\\pi\\ cm^2\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2022\/4\/1\/1648791439223.png\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"\\pi r^2\\ =\\ 100 \\pi\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cpi%20r%5E2%5C%20%3D%5C%20100%20%5Cpi\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"r\\ =\\ 10\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?r%5C%20%3D%5C%2010%5C%20cm\"><\/p>\n\n\n\n<p>Hence the area of the new circle is 10 cm.<\/p>\n\n\n\n<p><strong>Q3&nbsp;<\/strong>Fig. depicts an archery target marked with its five scoring regions from the center outwards as Gold, Red, Blue, Black, and White. The diameter of the region representing the Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2022\/4\/1\/1648791533512.png\" alt=\"1648791533512\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The radius of each scoring region can be found by adding 10.5 in respective colors.<\/p>\n\n\n\n<p>The area of the golden region is:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2%5C%20%3D%5C%20%5Cpi%20%5Ctimes%2010.5%5E2%5C%20%3D%5C%20346.5%5C%20cm%5E2\" alt=\"=\\ \\pi r^2\\ =\\ \\pi \\times 10.5^2\\ =\\ 346.5\\ cm^2\"><\/p>\n\n\n\n<p>Similarly area of the red region is = Area of a red score &#8211; Area of the golden region<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cpi%20%5Ctimes%2021%5E2%5C%20-%5C%20%5Cpi%20%5Ctimes%2010.5%5E2%5C%5C%5C%5C%3D%5C%201039.5%5C%20cm%5E2\" alt=\"\\\\=\\ \\pi \\times 21^2\\ -\\ \\pi \\times 10.5^2\\\\\\\\=\\ 1039.5\\ cm^2\"\/><\/figure>\n\n\n\n<p>Then the area of the blue region is:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cpi%20%5Ctimes%20%2831.5%29%5E2%5C%20-%5C%20%5Cpi%20%5Ctimes%2021%5E2%5C%5C%5C%5C%3D%5C%201732.5%5C%20cm%5E2\" alt=\"\\\\=\\ \\pi \\times (31.5)^2\\ -\\ \\pi \\times 21^2\\\\\\\\=\\ 1732.5\\ cm^2\"\/><\/figure>\n\n\n\n<p>And the area of the black region is:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cpi%20%5Ctimes%2042%5E2%5C%20-%5C%20%5Cpi%20%5Ctimes%20%2831.5%29%5E2%5C%5C%5C%5C%3D%5C%202425.5%5C%20cm%5E2\" alt=\"\\\\=\\ \\pi \\times 42^2\\ -\\ \\pi \\times (31.5)^2\\\\\\\\=\\ 2425.5\\ cm^2\"\/><\/figure>\n\n\n\n<p>Area of the white region is:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cpi%20%5Ctimes%2052.5%5E2%5C%20-%5C%20%5Cpi%5Ctimes%2042%5E2%5C%5C%5C%5C%3D%5C%203118.5%5C%20cm%5E2\" alt=\"\\\\=\\ \\pi \\times 52.5^2\\ -\\ \\pi\\times 42^2\\\\\\\\=\\ 3118.5\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q4&nbsp;<\/strong>The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Let the number of revolution of the wheel be n.<\/p>\n\n\n\n<p>The circumference of the wheel is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%202%5Cpi%20r%5C%5C%5C%5C%3D%5C%202%5Cpi%20%5Ctimes%2040%5C%20%3D%5C%2080%5Cpi%5C%20cm\" alt=\"\\\\=\\ 2\\pi r\\\\\\\\=\\ 2\\pi \\times 40\\ =\\ 80\\pi\\ cm\"\/><\/figure>\n\n\n\n<p>Now, the speed of the car is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%2066%5C%20Km\/hr%5C%20%5C%5C%5C%5C%3D%5C%2066%5Ctimes%20%5Cfrac%7B100000%7D%7B60%7D%5C%20cm\/min%5C%20%5C%5C%5C%5C%3D%5C%20110000%5C%20cm\/min.\" alt=\"\\\\=\\ 66\\ Km\/hr\\ \\\\\\\\=\\ 66\\times \\frac{100000}{60}\\ cm\/min\\ \\\\\\\\=\\ 110000\\ cm\/min.\"\/><\/figure>\n\n\n\n<p>Thus distance traveled in 10 minutes is:&nbsp;<img decoding=\"async\" alt=\"=\\ 1100000\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%201100000%5C%20cm\"><\/p>\n\n\n\n<p>According to the question, we get;<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?n%5Ctimes%2080%5Cpi%5C%20%3D%5C%201100000\" alt=\"n\\times 80\\pi\\ =\\ 1100000\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"n\\ =\\ 4375\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?n%5C%20%3D%5C%204375\"><\/p>\n\n\n\n<p>Hence the number of revolutions made by the wheel is 4372.<\/p>\n\n\n\n<p><strong>Q5&nbsp;<\/strong>Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is<\/p>\n\n\n\n<p>(A) 2 units<br>(B)&nbsp;<img decoding=\"async\" alt=\"\\pi\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cpi\">&nbsp;units<br>(C) 4 units<br>(D) 7 units<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Let the radius of the circle be r.<\/p>\n\n\n\n<p>Then according to question, we can write:-<\/p>\n\n\n\n<p>Perimeter = Area<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C2%5Cpi%20r%5C%20%3D%5C%20%5Cpi%20r%5E2%5C%5C%5C%5Cr%5C%20%3D%5C%202%5C%20units\" alt=\"\\\\2\\pi r\\ =\\ \\pi r^2\\\\\\\\r\\ =\\ 2\\ units\"\/><\/figure>\n\n\n\n<p>Hence option (A) is correct.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"----NCERT-Solutions-for-Class-10-Maths-Chapter-12-Areas-Related-to-Circles-Excercise-122----\"><strong>NCERT Solutions for Class 10 Maths Chapter 12&nbsp;<\/strong><strong>Areas Related to Circles Excercise: 12.2<\/strong><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Q1&nbsp;<\/strong>Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60\u00b0.<\/h3>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>We know that the area of a sector having radius r and angle\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5CTheta\" alt=\"\\Theta\">\u00a0is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\times \\pi r^2\"\/><\/figure>\n\n\n\n<p>Thus the area of the given sector is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cfrac%7B60%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%206%5E2%5C%5C%5C%5C%3D%5C%20%5Cfrac%7B132%7D%7B7%7D%5C%20cm%5E2\" alt=\"\\\\=\\ \\frac{60^{\\circ}}{360^{\\circ}}\\times \\pi \\times 6^2\\\\\\\\=\\ \\frac{132}{7}\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q2&nbsp;<\/strong>Find the area of a quadrant of a circle whose circumference is 22 cm.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>We are given the circumference of the circle.<\/p>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C2%5Cpi%20r%5C%20%3D%5C%2022%5C%5C%5C%5Cr%5C%20%3D%5C%20%5Cfrac%7B11%7D%7B%5Cpi%7D%5C%20cm\" alt=\"\\\\2\\pi r\\ =\\ 22\\\\\\\\r\\ =\\ \\frac{11}{\\pi}\\ cm\"\/><\/figure>\n\n\n\n<p>Also, we know that the area of a sector is given by :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\times \\pi r^2\"\/><\/figure>\n\n\n\n<p>It is given that we need to find the area of a quadrant thus&nbsp;<img decoding=\"async\" alt=\"\\Theta\\ =\\ 90^{\\circ}\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5CTheta%5C%20%3D%5C%2090%5E%7B%5Ccirc%7D\"><\/p>\n\n\n\n<p>Hence the area becomes:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cfrac%7B90%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Cleft%20%28%20%5Cfrac%7B11%7D%7B%5Cpi%7D%20%5Cright%20%29%5E2%5C%5C%5C%5C%3D%5C%20%5Cfrac%7B77%7D%7B8%7D%5C%20cm%5E2\" alt=\"\\\\=\\ \\frac{90}{360^{\\circ}}\\times \\pi \\left ( \\frac{11}{\\pi} \\right )^2\\\\\\\\=\\ \\frac{77}{8}\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q3&nbsp;<\/strong>The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The minute hand rotates 360&nbsp;<sup>o&nbsp;<\/sup>in one hour.<\/p>\n\n\n\n<p>We need to find rotation in 5 min. :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B360%5E%7B%5Ccirc%7D%7D%7B60%7D%5Ctimes%205%5C%20%3D%5C%2030%5E%7B%5Ccirc%7D\" alt=\"=\\ \\frac{360^{\\circ}}{60}\\times 5\\ =\\ 30^{\\circ}\"\/><\/figure>\n\n\n\n<p>The area of the sector is given by :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5CArea%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20r%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B30%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%2014%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B154%7D%7B3%7D%5C%20cm%5E2\" alt=\"\\\\Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\times \\pi r^2\\\\\\\\Area=\\ \\frac{30}{360^{\\circ}}\\times \\pi \\times 14^2\\\\\\\\Area=\\ \\frac{154}{3}\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area swept by minute hand in 5 minutes is.<\/p>\n\n\n\n<p><strong>Q4&nbsp;<\/strong>A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding :&nbsp;(i) minor segment<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The angle in the minor sector is 90\u00a0<sup>o<\/sup>.<\/p>\n\n\n\n<p>Thus the area of the sector is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5CArea%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20r%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%2010%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B1100%7D%7B14%7D%5C%20cm%5E2%5C%20%3D%5C%2078.5%5C%20cm%5E2\" alt=\"\\\\Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\times \\pi r^2\\\\\\\\Area=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\times \\pi \\times 10^2\\\\\\\\Area=\\ \\frac{1100}{14}\\ cm^2\\ =\\ 78.5\\ cm^2\"\/><\/figure>\n\n\n\n<p>Now the area of a triangle is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B1%7D%7B2%7Dbh%5C%20%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2010%5Ctimes%2010%5C%20%3D%5C%2050%5C%20cm%5E2\" alt=\"Area\\ =\\ \\frac{1}{2}bh\\ =\\ \\frac{1}{2}\\times 10\\times 10\\ =\\ 50\\ cm^2\"\/><\/figure>\n\n\n\n<p>Thus the area of minor segment = Area of the sector &#8211; Area of a triangle<\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 78.5\\ -\\ 50\\ =\\ 28.5\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2078.5%5C%20-%5C%2050%5C%20%3D%5C%2028.5%5C%20cm%5E2\"><\/p>\n\n\n\n<p><strong>Q4&nbsp;<\/strong>A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding :&nbsp;(ii) major sector. (Use \u03c0 = 3.14)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The area of the major sector can be found directly by using the formula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\times \\pi r^2\"\/><\/figure>\n\n\n\n<p>In the case of this, the angle is 360\u00a0<sup>o\u00a0<\/sup>&#8211; 90\u00a0<sup>o\u00a0<\/sup>= 270\u00a0<sup>o<\/sup>. <\/p>\n\n\n\n<p>Thus the area is : &#8211;<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cfrac%7B270%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%2010%5Ctimes%2010%5C%5C%5C%5C%3D%5C%20%5Cfrac%7B3300%7D%7B14%7D%5C%20%3D%5C%20235.7%5C%20cm%5E2\" alt=\"\\\\=\\ \\frac{270^{\\circ}}{360^{\\circ}}\\times \\pi \\times 10\\times 10\\\\\\\\=\\ \\frac{3300}{14}\\ =\\ 235.7\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q5&nbsp;<\/strong>In a circle of radius 21 cm, an arc subtends an angle of 60\u00b0 at the center. Find:&nbsp;(i) the length of the arc<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The length of the arc is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Length%5C%20of%5C%20arc%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%20%5Ctimes%202%5Cpi%20r\" alt=\"Length\\ of\\ arc\\ =\\ \\frac{\\Theta}{360^{\\circ}} \\times 2\\pi r\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cfrac%7B60%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%20%5Ctimes%202%5Ctimes%20%5Cpi%20%5Ctimes%2021%5C%5C%5C%5C%3D%5C%2022%5C%20cm\" alt=\"\\\\=\\ \\frac{60^{\\circ}}{360^{\\circ}} \\times 2\\times \\pi \\times 21\\\\\\\\=\\ 22\\ cm\"\/><\/figure>\n\n\n\n<p>Hence the length of the arc is 22 cm.<\/p>\n\n\n\n<p><strong>Q5\u00a0<\/strong>In a circle of radius 21 cm, an arc subtends an angle of 60\u00b0 at the center. Find:\u00a0(ii) area of the sector formed by the arc. <\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>We know that the area of the sector is given by:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cfrac%7B60%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%2021%5E2%5C%5C%5C%5C%3D%5C%20231%5C%20cm%5E2\" alt=\"\\\\=\\ \\frac{60}{360^{\\circ}}\\ \\times \\pi \\times 21^2\\\\\\\\=\\ 231\\ cm^2\"\/><\/figure>\n\n\n\n<p>Thus the area of the sector is 231 cm\u00a0<sup>2<\/sup>.<\/p>\n\n\n\n<p><strong>Q5&nbsp;<\/strong>In a circle of radius 21 cm, an arc subtends an angle of 60\u00b0 at the center. Find:&nbsp;(iii) area of the segment formed by the corresponding chord<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>For the area of the segment, we need to subtract the area of the triangle attached with the area of arc.<\/p>\n\n\n\n<p>Thus consider the triangle :-<\/p>\n\n\n\n<p>It is given that the angle of arc is 60&nbsp;<sup>o&nbsp;<\/sup>, or we can say that all angles are 60&nbsp;<sup>o&nbsp;<\/sup>(since two sides are equal). Hence it is an equilateral triangle.<\/p>\n\n\n\n<p>Area of triangle is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%20a%5E2%5C%5C%5C%5C%5C%5C%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%2021%5E2%5C%5C%5C%5C%3D%5C%20%5Cfrac%7B441%5Csqrt%7B3%7D%7D%7B4%7D%5C%20cm%5E2\" alt=\"\\\\=\\ \\frac{\\sqrt{3}}{4}\\times a^2\\\\\\\\\\\\=\\ \\frac{\\sqrt{3}}{4}\\times 21^2\\\\\\\\=\\ \\frac{441\\sqrt{3}}{4}\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of segment is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cleft%20%28231%5C%20-%5C%20%5Cfrac%7B441%5Csqrt%7B3%7D%7D%7B4%7D%20%5Cright%20%29%5C%20cm%5E2\" alt=\"=\\ \\left (231\\ -\\ \\frac{441\\sqrt{3}}{4} \\right )\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q6&nbsp;<\/strong>A chord of a circle of radius 15 cm subtends an angle of 60\u00b0 at the center. Find the areas of the corresponding minor and major segments of the circle.<br>(Use&nbsp;&nbsp;and&nbsp;&nbsp;)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The area of the sector is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20%5Cfrac%7B%5CTheta%20%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20r%5E2%5C%5C%5C%5C%3D%5C%20%5Cfrac%7B60%5E%7B%5Ccirc%7D%20%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%2015%5E2%5C%5C%5C%5C%3D%5C%20117.85%5C%20cm%5E2\" alt=\"\\\\=\\ \\frac{\\Theta }{360^{\\circ}}\\times \\pi r^2\\\\\\\\=\\ \\frac{60^{\\circ} }{360^{\\circ}}\\times \\pi \\times 15^2\\\\\\\\=\\ 117.85\\ cm^2\"\/><\/figure>\n\n\n\n<p>Now consider the triangle, the angle of the sector is 60\u00a0<sup>0<\/sup>.<\/p>\n\n\n\n<p>This implies it is an equilateral triangle. (As two sides are equal so will have the same angle. This possible only when all angles are equal i.e., 60&nbsp;<sup>o&nbsp;<\/sup>.)<\/p>\n\n\n\n<p>Thus, the area of the triangle is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%2015%5E2\" alt=\"=\\ \\frac{\\sqrt{3}}{4}\\times 15^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 56.25 \\sqrt{3}\\ =\\ 97.31\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2056.25%20%5Csqrt%7B3%7D%5C%20%3D%5C%2097.31%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Hence area of the minor segment:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C117.85%5C%20-%5C%2097.31%5C%20%3D%5C%2020.53%5C%20cm%5E2\" alt=\"=\\117.85\\ -\\ 97.31\\ =\\ 20.53\\ cm^2\"><\/p>\n\n\n\n<p>And the area of the major segment is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2%5C%20-%5C%2020.53\" alt=\"=\\ \\pi r^2\\ -\\ 20.53\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\pi\\times 15^2\\ -\\ 20.53\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%5Ctimes%2015%5E2%5C%20-%5C%2020.53\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 707.14\\ -\\ 20.53\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20707.14%5C%20-%5C%2020.53\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 686.6\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20686.6%5C%20cm%5E2\"><\/p>\n\n\n\n<p><strong>Q7&nbsp;<\/strong>A chord of a circle of radius 12 cm subtends an angle of 120\u00b0 at the center. Find the area of the corresponding segment of the circle.<br>(Use&nbsp;&nbsp;and&nbsp;&nbsp;)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>For the area of the segment, we need the area of sector and area of the associated triangle.<\/p>\n\n\n\n<p>So, the area of the sector is:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B120%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%2012%5E2\" alt=\"=\\ \\frac{120^{\\circ}}{360^{\\circ}}\\times \\pi \\times 12^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 150.72\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20150.72%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Now, consider the triangle :-<\/p>\n\n\n\n<p>Draw a perpendicular from the center of the circle on the base of the triangle (let it be h).<\/p>\n\n\n\n<p>Using geometry we can write,<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7Bh%7D%7Br%7D%5C%20%3D%5C%20%5Ccos%2060%5E%7B%5Ccirc%7D\" alt=\"\\frac{h}{r}\\ =\\ \\cos 60^{\\circ}\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"h\\ =\\ 6\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?h%5C%20%3D%5C%206%5C%20cm\"><\/p>\n\n\n\n<p>Similarly,&nbsp;<img decoding=\"async\" alt=\"\\frac{\\frac{b}{2}}{r}\\ =\\ \\sin 60^{\\circ}\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7B%5Cfrac%7Bb%7D%7B2%7D%7D%7Br%7D%5C%20%3D%5C%20%5Csin%2060%5E%7B%5Ccirc%7D\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"b\\ =\\ 12\\sqrt{3}\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?b%5C%20%3D%5C%2012%5Csqrt%7B3%7D%5C%20cm\"><\/p>\n\n\n\n<p>Thus the area of the triangle is:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2012%5Csqrt%7B3%7D%5Ctimes%206\" alt=\"=\\ \\frac{1}{2}\\times 12\\sqrt{3}\\times 6\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 62.28\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2062.28%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Hence the area of segment is:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20150.72%5C%20-%5C%2062.28%5C%20%3D%5C%2088.44%5C%20cm%5E2\" alt=\"=\\ 150.72\\ -\\ 62.28\\ =\\ 88.44\\ cm^2\">.<\/p>\n\n\n\n<p><strong>Q8&nbsp;<\/strong>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.). Find&nbsp;(i) the area of that part of the field in which the horse can graze.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091304687.png\" alt=\"1636091304687\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The part grazed by the horse is given by = Area of sector<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%205%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\ \\times \\pi \\times 5^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2019.62%5C%20m%5E2\" alt=\"=\\ 19.62\\ m^2\"\/><\/figure>\n\n\n\n<p><strong>Q8&nbsp;<\/strong>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.). Find&nbsp;(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use&nbsp;&nbsp;)<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091352998.png\" alt=\"1636091352998\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>When the length of the rope is 10 m, the area grazed will be:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%2010%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\ \\times \\pi \\times 10^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2025%20%5Cpi%5C%20m%5E2\" alt=\"=\\ 25 \\pi\\ m^2\"\/><\/figure>\n\n\n\n<p>Hence the change in the grazing area is given by :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2025%20%5Cpi%5C%20-%5C%20%5Cfrac%7B25%20%5Cpi%7D%7B4%7D%5C%20%3D%5C%2058.85%5C%20m%5E2\" alt=\"=\\ 25 \\pi\\ -\\ \\frac{25 \\pi}{4}\\ =\\ 58.85\\ m^2\"\/><\/figure>\n\n\n\n<p><strong>Q9&nbsp;<\/strong>A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. Find:&nbsp;(i) the total length of the silver wire required.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091383511.png\" alt=\"1636091383511\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The total wire required will be for 5 diameters and the circumference of the brooch.<\/p>\n\n\n\n<p>The circumference of the brooch :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%202%5Cpi%20r%5C%5C%5C%5C%3D%5C%202%5Ctimes%20%5Cpi%20%5Ctimes%20%5Cfrac%7B35%7D%7B2%7D%5C%5C%5C%5C%3D%5C%20110%5C%20mm\" alt=\"\\\\=\\ 2\\pi r\\\\\\\\=\\ 2\\times \\pi \\times \\frac{35}{2}\\\\\\\\=\\ 110\\ mm\"\/><\/figure>\n\n\n\n<p>Hence the total wire required will be:-&nbsp;<sub>&nbsp;.<\/sub><\/p>\n\n\n\n<p><strong>Q9&nbsp;<\/strong>A brooch is made with silver wire in the form of a circle with a diameter of 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 :&nbsp;(ii) the area of each sector of the brooch.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091389686.png\" alt=\"1636091389686\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The total number of lines present in the brooch is 10 (line starting from the centre).<\/p>\n\n\n\n<p>Thus the angle of each sector is 36&nbsp;<sup>o&nbsp;<\/sup>.<\/p>\n\n\n\n<p>The area of the sector is given by:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5CArea%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B36%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%20%5Cleft%20%28%20%5Cfrac%7B35%7D%7B2%7D%20%5Cright%20%29%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B385%7D%7B4%7D%5C%20mm%5E2\" alt=\"\\\\Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\\\\\\\\Area=\\ \\frac{36^{\\circ}}{360^{\\circ}}\\ \\times \\pi \\times \\left ( \\frac{35}{2} \\right )^2\\\\\\\\Area=\\ \\frac{385}{4}\\ mm^2\"\/><\/figure>\n\n\n\n<p><strong>Q10&nbsp;<\/strong>An umbrella has 8 ribs that are equally spaced (see Fig. ). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091407785.png\" alt=\"1636091407785\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>It is given that the umbrella has 8 ribs so the angle of each sector is 45\u00a0<sup>o<\/sup>.<\/p>\n\n\n\n<p>Thus the area of the sector is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5CArea%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B45%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%2045%5E2%5C%5C%5C%5CArea%3D%5C%20%5Cfrac%7B22275%7D%7B28%7D%5C%20cm%5E2\" alt=\"\\\\Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\\\\\\\\Area=\\ \\frac{45^{\\circ}}{360^{\\circ}}\\ \\times \\pi \\times 45^2\\\\\\\\Area=\\ \\frac{22275}{28}\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area between two consecutive ribs is. <\/p>\n\n\n\n<p><strong>Q11&nbsp;<\/strong>A car has two wipers that 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.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The area cleaned by one wiper is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{115^{\\circ}}{360^{\\circ}}\\ \\times \\pi \\times 25^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B115%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%2025%5E2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{158125}{252}\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B158125%7D%7B252%7D%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Hence the required area (area cleaned by both blades) is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%202%5Ctimes%20%5Cfrac%7B158125%7D%7B252%7D%5C%20%3D%5C%20%5Cfrac%7B158125%7D%7B126%7D%5C%20cm%5E2\" alt=\"=\\ 2\\times \\frac{158125}{252}\\ =\\ \\frac{158125}{126}\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q12&nbsp;<\/strong>To warn ships for underwater rocks, a lighthouse spreads a red-colored 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. (Use \u03c0 = 3.14)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The area of the sector is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<p>In this case, the angle is 80\u00a0<sup>o<\/sup>. <\/p>\n\n\n\n<p>Thus the area is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B80%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%2016.5%5E2\" alt=\"=\\ \\frac{80^{\\circ}}{360^{\\circ}}\\ \\times \\pi \\times 16.5^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 189.97\\ Km^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20189.97%5C%20Km%5E2\"><\/p>\n\n\n\n<p><strong>Q13\u00a0<\/strong>A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm\u00a0<sup>2<\/sup>. (Use\u00a0\u00a0)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The angle of each of the six sectors is 60&nbsp;<sup>o&nbsp;<\/sup>at the center.&nbsp;<img decoding=\"async\" alt=\"\\left ( \\because \\frac{360^{\\circ}}{6}\\ =\\ 60^{\\circ} \\right )\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cleft%20%28%20%5Cbecause%20%5Cfrac%7B360%5E%7B%5Ccirc%7D%7D%7B6%7D%5C%20%3D%5C%2060%5E%7B%5Ccirc%7D%20%5Cright%20%29\"><\/p>\n\n\n\n<p>Area of the sector is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{60^{\\circ}}{360^{\\circ}}\\ \\times \\pi \\times 78^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B60%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20%5Ctimes%2078%5E2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 410.66\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20410.66%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And the area of the equilateral triangle associated with segment :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%20a%5E2\" alt=\"=\\ \\frac{\\sqrt{3}}{4}\\times a^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{\\sqrt{3}}{4}\\times 28^2\\ =\\ 333.2\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%2028%5E2%5C%20%3D%5C%20333.2%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Hence the area of segment is:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20410.66%5C%20-%5C%20333.2%5C%20%3D%5C%2077.46%5C%20cm%5E2\" alt=\"=\\ 410.66\\ -\\ 333.2\\ =\\ 77.46\\ cm^2\"><\/p>\n\n\n\n<p>Thus the total area of design is:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C6%5Ctimes%2077.46%5C%20%3D%5C%20464.76%5C%20cm%5E2\" alt=\"=\\6\\times 77.46\\ =\\ 464.76\\ cm^2\"><\/p>\n\n\n\n<p>So, the total cost for the design is :-\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%200.35%5Ctimes%20464.76%5C%20%3D%5C%20Rs.%20162.66\" alt=\"=\\ 0.35\\times 464.76\\ =\\ Rs. 162.66\"><\/p>\n\n\n\n<p><strong>Q14\u00a0<\/strong>Tick the correct answer in the following:<\/p>\n\n\n\n<p>Area of a sector of angle p (in degrees) of a circle with radius R is<\/p>\n\n\n\n<p>(A)&nbsp;<img decoding=\"async\" alt=\"\\frac{p}{180}\\times 2\\pi R\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7Bp%7D%7B180%7D%5Ctimes%202%5Cpi%20R\"><\/p>\n\n\n\n<p>(B)&nbsp;<img decoding=\"async\" alt=\"\\frac{p}{180}\\times \\pi R^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7Bp%7D%7B180%7D%5Ctimes%20%5Cpi%20R%5E2\"><\/p>\n\n\n\n<p>(C)&nbsp;<img decoding=\"async\" alt=\"\\frac{p}{360}\\times 2\\pi R\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7Bp%7D%7B360%7D%5Ctimes%202%5Cpi%20R\"><\/p>\n\n\n\n<p>(D)&nbsp;<img decoding=\"async\" alt=\"\\frac{p}{720}\\times 2\\pi R^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7Bp%7D%7B720%7D%5Ctimes%202%5Cpi%20R%5E2\"><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>We know that the area of the sector is given by :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?Area%5C%20%3D%5C%20%5Cfrac%7B%5CTheta%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"Area\\ =\\ \\frac{\\Theta}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7Bp%7D%7B360%5E%7B%5Ccirc%7D%7D%5C%20%5Ctimes%20%5Cpi%20r%5E2\" alt=\"=\\ \\frac{p}{360^{\\circ}}\\ \\times \\pi r^2\"\/><\/figure>\n\n\n\n<p>Hence option (d) is correct.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"-----NCERT-Solutions-for-Class-10-Maths-Chapter-12---Areas-Related-to-Circles-Excercise-123-----\"><strong>NCERT Solutions for Class 10 Maths Chapter 12&nbsp;<\/strong><strong>Areas Related to Circles Excercise: 12.3<\/strong><\/h2>\n\n\n\n<p><strong>Q1&nbsp;<\/strong>Find the area of the shaded region in Fig, if PQ = 24 cm, PR = 7 cm, and O is the center of the circle.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2022\/4\/1\/1648791656559.png\" alt=\"1648791656559\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>We know that&nbsp;<img decoding=\"async\" alt=\"\\angle\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cangle\">&nbsp;RPQ is 90&nbsp;<sup>o&nbsp;<\/sup>as ROQ is the diameter.<\/p>\n\n\n\n<p>RQ can be found using the Pythagoras theorem.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?RP%5E2%5C%20+%5C%20PQ%5E2%5C%20%3D%5C%20QR%5E2\" alt=\"RP^2\\ +\\ PQ^2\\ =\\ QR^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"7^2\\ +\\ 24^2\\ =\\ QR^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?7%5E2%5C%20+%5C%2024%5E2%5C%20%3D%5C%20QR%5E2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"QR\\ =\\ \\sqrt{625}\\ =\\ 25\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?QR%5C%20%3D%5C%20%5Csqrt%7B625%7D%5C%20%3D%5C%2025%5C%20cm\"><\/p>\n\n\n\n<p>Now, the area of the shaded region is given by = Area of semicircle &#8211; Area of&nbsp;<img decoding=\"async\" alt=\"\\Delta\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5CDelta\">&nbsp;PQR<\/p>\n\n\n\n<p>Area of a semicircle is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2\" alt=\"=\\ \\frac{1}{2}\\pi r^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{1}{2}\\pi \\times \\left ( \\frac{25}{2} \\right )^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%5Ctimes%20%5Cleft%20%28%20%5Cfrac%7B25%7D%7B2%7D%20%5Cright%20%29%5E2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 245.53\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20245.53%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And, the area of triangle PQR is:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2024%5Ctimes%207%5C%20%3D%5C%2084%5C%20cm%5E2\" alt=\"=\\ \\frac{1}{2}\\times 24\\times 7\\ =\\ 84\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the shaded region is:\u00a0<\/p>\n\n\n\n<p><strong>Q2\u00a0<\/strong>Find the area of the shaded region in Fig, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The area of a shaded region can be easily found by using the formula of the area of the sector.<\/p>\n\n\n\n<p>Area of the shaded region is given by : Area of sector OAFC &#8211; Area of sector OBED<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B40%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%2814%29%5E2%5C%20-%5C%20%5Cfrac%7B40%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%287%29%5E2\" alt=\"=\\ \\frac{40^{\\circ}}{360^{\\circ}}\\times \\pi (14)^2\\ -\\ \\frac{40^{\\circ}}{360^{\\circ}}\\times \\pi (7)^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B616%7D%7B9%7D%5C%20-%5C%20%5Cfrac%7B154%7D%7B9%7D%5C%20%3D%5C%20%5Cfrac%7B462%7D%7B9%7D\" alt=\"=\\ \\frac{616}{9}\\ -\\ \\frac{154}{9}\\ =\\ \\frac{462}{9}\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2051.33%5C%20cm%5E2\" alt=\"=\\ 51.33\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q3&nbsp;<\/strong>Find the area of the shaded region in Fig, if ABCD is a square of side 14 cm and APD and BPC are semicircles.&nbsp;<img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2022\/4\/1\/1648791700601.png\" alt=\"1648791700601\"><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Area of the shaded region is given by = Area of the square &#8211; Area of two semicircles.<\/p>\n\n\n\n<p>Area of square is :&nbsp;<img decoding=\"async\" alt=\"=\\ 14^2\\ =\\ 196\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2014%5E2%5C%20%3D%5C%20196%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And the area of the semicircle is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2\" alt=\"=\\ \\frac{1}{2}\\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%5Ctimes%207%5E2\" alt=\"=\\ \\frac{1}{2}\\pi \\times 7^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2077%5C%20cm%5E2\" alt=\"=\\ 77\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the shaded region is given by:\u00a0<\/p>\n\n\n\n<p><strong>Q4&nbsp;<\/strong>Find the area of the shaded region in Fig, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.&nbsp;<img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2022\/4\/1\/1648791737051.png\" alt=\"1648791737051\"><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Area of the shaded region is given by = Area of triangle + Area of the circle &#8211; Area of the sector<\/p>\n\n\n\n<p>Area of the sector is : &#8211;<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B60%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%5Ctimes%206%5E2\" alt=\"=\\ \\frac{60^{\\circ}}{360^{\\circ}}\\times \\pi\\times 6^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{132}{7}\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B132%7D%7B7%7D%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And, the area of the triangle is:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7Da%5E2%5C%20%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%2012%5E2%5C%20%3D%5C%2036%5Csqrt%7B3%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{\\sqrt{3}}{4}a^2\\ =\\ \\frac{\\sqrt{3}}{4}\\times 12^2\\ =\\ 36\\sqrt{3}\\ cm^2\"\/><\/figure>\n\n\n\n<p>And, the area of the circle is:\u00a0<img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2\" alt=\"=\\ \\pi r^2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\pi \\times 6^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20%5Ctimes%206%5E2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{792}{7}\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B792%7D%7B7%7D%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Hence the area of the shaded region is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2036%5Csqrt%7B3%7D%5C%20+%5C%20%5Cfrac%7B792%7D%7B7%7D%5C%20-%5C%20%5Cfrac%7B132%7D%7B7%7D\" alt=\"=\\ 36\\sqrt{3}\\ +\\ \\frac{792}{7}\\ -\\ \\frac{132}{7}\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\left ( 36\\sqrt{3}\\ +\\ \\frac{660}{7} \\right )\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cleft%20%28%2036%5Csqrt%7B3%7D%5C%20+%5C%20%5Cfrac%7B660%7D%7B7%7D%20%5Cright%20%29%5C%20cm%5E2\"><\/p>\n\n\n\n<p><strong>Q5&nbsp;<\/strong>From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. Find the area of the remaining portion of the square.&nbsp;<img decoding=\"async\" src=\"https:\/\/d2pduerm2meudp.cloudfront.net\/media\/uploads\/2019\/06\/26\/25793.PNG\" alt=\"25793\"><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Consider the quadrant in the given figure:- We have an angle of the sector as 90&nbsp;<sup>o&nbsp;<\/sup>and radius 1 cm.<\/p>\n\n\n\n<p>Thus the area of the quadrant is :-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%5Ctimes%201%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\times \\pi\\times 1^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{22}{28}\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B22%7D%7B28%7D%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And the area of the square is :&nbsp;<img decoding=\"async\" alt=\"=\\ side^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20side%5E2\"><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%204%5E2\" alt=\"=\\ 4^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2016%5C%20cm%5E2\" alt=\"=\\ 16\\ cm^2\"\/><\/figure>\n\n\n\n<p>And, the area of the circle is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2%5C%20%3D%5C%20%5Cpi%20%5Ctimes%201%5E2%5C%20%3D%5C%20%5Cpi%5C%20cm%5E2\" alt=\"=\\ \\pi r^2\\ =\\ \\pi \\times 1^2\\ =\\ \\pi\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the shaded region is: = Area of the square &#8211; Area of the circle &#8211; 4 (Area of quadrant)<\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 16\\ -\\ \\frac{22}{7}\\ -\\ 4\\left ( \\frac{22}{28} \\right )\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2016%5C%20-%5C%20%5Cfrac%7B22%7D%7B7%7D%5C%20-%5C%204%5Cleft%20%28%20%5Cfrac%7B22%7D%7B28%7D%20%5Cright%20%29\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{68}{7}\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B68%7D%7B7%7D%5C%20cm%5E2\"><\/p>\n\n\n\n<p><strong>Q6&nbsp;<\/strong>In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. Find the area of the design.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Assume the center of the circle to be point C and AD as the median of the equilateral triangle.<\/p>\n\n\n\n<p>Then we can write:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?AO%5C%20%3D%5C%20%5Cfrac%7B2%7D%7B3%7DAD\" alt=\"AO\\ =\\ \\frac{2}{3}AD\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"32\\ =\\ \\frac{2}{3}AD\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?32%5C%20%3D%5C%20%5Cfrac%7B2%7D%7B3%7DAD\"><\/p>\n\n\n\n<p>Thus&nbsp;<img decoding=\"async\" alt=\"AD\\ =\\ 48\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?AD%5C%20%3D%5C%2048%5C%20cm\"><\/p>\n\n\n\n<p>Consider&nbsp;<img decoding=\"async\" alt=\"\\Delta\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5CDelta\">&nbsp;ABD,<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?AB%5E2%5C%20%3D%5C%20AD%5E2%5C%20+%5C%20BD%5E2\" alt=\"AB^2\\ =\\ AD^2\\ +\\ BD^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"AB^2\\ =\\ 48^2\\ +\\ \\left ( \\frac{AB}{2} \\right )^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?AB%5E2%5C%20%3D%5C%2048%5E2%5C%20+%5C%20%5Cleft%20%28%20%5Cfrac%7BAB%7D%7B2%7D%20%5Cright%20%29%5E2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"AB\\ =\\ 32\\sqrt{3}\\ cm\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?AB%5C%20%3D%5C%2032%5Csqrt%7B3%7D%5C%20cm\"><\/p>\n\n\n\n<p>Thus the area of an equilateral triangle is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%20%5Cleft%20%28%2032%5Csqrt%7B3%7D%20%5Cright%20%29%5E2\" alt=\"=\\ \\frac{\\sqrt{3}}{4}\\times \\left ( 32\\sqrt{3} \\right )^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ 768\\sqrt{3}\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20768%5Csqrt%7B3%7D%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And the area of the circle is :&nbsp;<img decoding=\"async\" alt=\"=\\ \\pi r^2\\ =\\ \\pi\\times 32^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2%5C%20%3D%5C%20%5Cpi%5Ctimes%2032%5E2\"><\/p>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{22528}{7} cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B22528%7D%7B7%7D%20cm%5E2\"><\/p>\n\n\n\n<p>Hence the area of the design is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cleft%20%28%20%5Cfrac%7B22528%7D%7B7%7D%5C%20-%5C%20768%20%5Csqrt%7B3%7D%20%5Cright%20%29%5C%20cm%5E2\" alt=\"=\\ \\left ( \\frac{22528}{7}\\ -\\ 768 \\sqrt{3} \\right )\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q7.&nbsp;<\/strong>In Fig, ABCD is a square of side 14 cm. With centers A, B, C, and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>It is clear from the figure that the area of all sectors is equal (due to symmetry).<\/p>\n\n\n\n<p>Also, the angle of the sector is 90&nbsp;<sup>o&nbsp;<\/sup>and the radius is 7 cm.<\/p>\n\n\n\n<p>Thus the area of the sector is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%287%29%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\times \\pi (7)^2\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{77}{2}\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B77%7D%7B2%7D%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And, the area of the square is :&nbsp;<img decoding=\"async\" alt=\"=\\ a^2\\ =\\ 14^2\\ =\\ 196\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20a%5E2%5C%20%3D%5C%2014%5E2%5C%20%3D%5C%20196%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Hence the area of the shaded region is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20196%5C%20-%5C%204%5Ctimes%20%5Cfrac%7B77%7D%7B2%7D\" alt=\"=\\ 196\\ -\\ 4\\times \\frac{77}{2}\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2042%5C%20cm%5E2\" alt=\"=\\ 42\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q8&nbsp;<\/strong>Fig. depicts a racing track whose left and right ends are semicircular.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091567493.png\" alt=\"1636091567493\"\/><\/figure>\n\n\n\n<p><a href=\"http:\/\/learn.careers360.com\/ncert\/question-fig-1226-depicts-a-racing-track-whose-left-and-right-ends-are-semicircular-the-distance-between-the-two-inner-parallel-line-segments-is-60-m-and-they-are-each-106-m-long-if-the-track-is-10-m-wide-find-i-the-distance-around-the-track-along-its\/\"><\/a><\/p>\n\n\n\n<p>The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :<\/p>\n\n\n\n<p>(i) the distance around the track along its inner edge<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The distance around the track is Length of two straight lines + Length of two arcs.<\/p>\n\n\n\n<p>Length of the arc is &#8211;<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202%5Cpi%20r\" alt=\"=\\ \\frac{1}{2}\\times 2\\pi r\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202%5Cpi%20%5Ctimes%2030%5C%20%3D%5C%2030%5Cpi%5C%20m\" alt=\"=\\ \\frac{1}{2}\\times 2\\pi \\times 30\\ =\\ 30\\pi\\ m\"\/><\/figure>\n\n\n\n<p>Thus the length of the inner track is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20106%5C%20+%5C%2030%5Cpi%5C%20+%5C%20106%5C%20+%5C%2030%5Cpi\" alt=\"=\\ 106\\ +\\ 30\\pi\\ +\\ 106\\ +\\ 30\\pi\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20212%5C%20+%5C%2060%5Cpi\" alt=\"=\\ 212\\ +\\ 60\\pi\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B2804%7D%7B7%7D%5C%20m\" alt=\"=\\ \\frac{2804}{7}\\ m\"\/><\/figure>\n\n\n\n<p><strong>Q8&nbsp;<\/strong>Fig. depicts a racing track whose left and right ends are semicircular.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091587830.png\" alt=\"1636091587830\"\/><\/figure>\n\n\n\n<p><a href=\"http:\/\/learn.careers360.com\/ncert\/question-fig-1226-depicts-a-racing-track-whose-left-and-right-ends-are-semicircular-the-distance-between-the-two-inner-parallel-line-segments-is-60-m-and-they-are-each-106-m-long-if-the-track-is-10-m-wide-find-ii-the-area-of-the-track\/\"><\/a><\/p>\n\n\n\n<p>The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :<\/p>\n\n\n\n<p>(ii) the area of the track.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The area of track = Area of outer structure &#8211; Area of inner structure.<\/p>\n\n\n\n<p>Area of outer structure is: = Area of square + Area of 2 semicircles<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20106%5Ctimes%2080%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2840%29%5E2%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2840%29%5E2%5C%20m%5E2\" alt=\"=\\ 106\\times 80\\ +\\ \\frac{1}{2}\\pi (40)^2\\ +\\ \\frac{1}{2}\\pi (40)^2\\ m^2\"\/><\/figure>\n\n\n\n<p>And, area of inner structure: = Area of inner square + Area of 2 inner semicircles<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20106%5Ctimes%2060%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2830%29%5E2%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2830%29%5E2%5C%20m%5E2\" alt=\"=\\ 106\\times 60\\ +\\ \\frac{1}{2}\\pi (30)^2\\ +\\ \\frac{1}{2}\\pi (30)^2\\ m^2\"\/><\/figure>\n\n\n\n<p>Thus the area of the track is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5C%5C%3D%5C%20106%5Ctimes%2080%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2840%29%5E2%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2840%29%5E2%5C%20-%5C%20%5Cleft%20%28%20106%5Ctimes%2060%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2830%29%5E2%5C%20+%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%2830%29%5E2%20%5Cright%20%29%5C%5C%5C%5C%3D%5C%204320%5C%20m%5E2\" alt=\"\\\\=\\ 106\\times 80\\ +\\ \\frac{1}{2}\\pi (40)^2\\ +\\ \\frac{1}{2}\\pi (40)^2\\ -\\ \\left ( 106\\times 60\\ +\\ \\frac{1}{2}\\pi (30)^2\\ +\\ \\frac{1}{2}\\pi (30)^2 \\right )\\\\\\\\=\\ 4320\\ m^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the track is 4320 m&nbsp;<sup>2&nbsp;<\/sup>.<\/p>\n\n\n\n<p><strong>Q9&nbsp;<\/strong>In Fig, AB, and CD are two diameters of a circle (with center O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091578284.png\" alt=\"1636091578284\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Firstly, the area of the smaller circle is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2\" alt=\"=\\ \\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20%5Ctimes%20%5Cleft%20%28%20%5Cfrac%7B7%7D%7B2%7D%20%5Cright%20%29%5E2\" alt=\"=\\ \\pi \\times \\left ( \\frac{7}{2} \\right )^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B77%7D%7B2%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{77}{2}\\ cm^2\"\/><\/figure>\n\n\n\n<p>Now, the area of&nbsp;<img decoding=\"async\" alt=\"\\Delta ABC\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5CDelta%20ABC\">&nbsp;:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20AB%5Ctimes%20OC\" alt=\"=\\ \\frac{1}{2}\\times AB\\times OC\"\/><\/figure>\n\n\n\n<p>or&nbsp;<img decoding=\"async\" alt=\"=\\ \\frac{1}{2}\\times 14\\times 7\\ =\\ 49\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2014%5Ctimes%207%5C%20%3D%5C%2049%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And, the area of the bigger semicircle is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2\" alt=\"=\\ \\frac{1}{2}\\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20%5Ctimes%207%5E2\" alt=\"=\\ \\frac{1}{2}\\pi \\times 7^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2077%5C%20cm%5E2\" alt=\"=\\ 77\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the shaded region is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B77%7D%7B2%7D%5C%20+%5C%2077%5C%20-%5C%2049\" alt=\"=\\ \\frac{77}{2}\\ +\\ 77\\ -\\ 49\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2066.5%5C%20cm%5E2\" alt=\"=\\ 66.5\\ cm^2\"\/><\/figure>\n\n\n\n<p>Therefore the area of the shaded region is 66.5 cm&nbsp;<sup>2&nbsp;<\/sup>.<\/p>\n\n\n\n<p><strong>Q10&nbsp;<\/strong>The area of an equilateral triangle ABC is 17320.5 cm&nbsp;<sup>2&nbsp;<\/sup>. With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle (see Fig.). Find the area of the shaded region. (Use&nbsp;&nbsp;and&nbsp;&nbsp;)<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091615101.png\" alt=\"1636091615101\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Area of an equilateral triangle is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%20a%5E2\" alt=\"=\\ \\frac{\\sqrt{3}}{4}\\times a^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5Ctimes%20a%5E2%5C%20%3D%5C%2017320.5\" alt=\"\\frac{\\sqrt{3}}{4}\\times a^2\\ =\\ 17320.5\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?a%5C%20%3D%5C%20200%5C%20cm\" alt=\"a\\ =\\ 200\\ cm\"\/><\/figure>\n\n\n\n<p>Now, consider the sector:- Angle of the sector is 60&nbsp;<sup>o&nbsp;<\/sup>and the radius is 100 cm.<\/p>\n\n\n\n<p>Thus the area of the sector:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B60%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%20100%5E2\" alt=\"=\\ \\frac{60^{\\circ}}{360^{\\circ}}\\times \\pi \\times 100^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B15700%7D%7B3%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{15700}{3}\\ cm^2\"\/><\/figure>\n\n\n\n<p>Thus the area of the shaded region is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2017320.5%5C%20-%5C%203%5Ctimes%20%5Cfrac%7B15700%7D%7B3%7D\" alt=\"=\\ 17320.5\\ -\\ 3\\times \\frac{15700}{3}\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%201620.5%5C%20cm%5E2\" alt=\"=\\ 1620.5\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q11&nbsp;<\/strong>On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig.). Find the area of the remaining portion of the handkerchief.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091628689.png\" alt=\"1636091628689\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Since one side of the square has 3 circles, thus the side of the square is 42 cm.<\/p>\n\n\n\n<p>Area of the square :&nbsp;<img decoding=\"async\" alt=\"=\\ a^2\\ =\\ 42^2\\ =\\ 1764\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20a%5E2%5C%20%3D%5C%2042%5E2%5C%20%3D%5C%201764%5C%20cm%5E2\"><\/p>\n\n\n\n<p>And, area of a circle :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cpi%20r%5E2%5C%20%3D%5C%20%5Cfrac%7B22%7D%7B7%7D%5Ctimes%207%5E2\" alt=\"=\\ \\pi r^2\\ =\\ \\frac{22}{7}\\times 7^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20154%5C%20cm%5E2\" alt=\"=\\ 154\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the remaining portion is :&nbsp;<img decoding=\"async\" alt=\"=\\ 1764\\ -\\ 9\\times 154\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%201764%5C%20-%5C%209%5Ctimes%20154\"><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20378%5C%20cm%5E2\" alt=\"=\\ 378\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q12&nbsp;<\/strong>In Fig, OACB is a quadrant of a circle with center O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091641153.png\" alt=\"1636091641153\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>The quadrant OACB is a sector with angle 90&nbsp;<sup>o&nbsp;<\/sup>and radius 3.5 cm.<\/p>\n\n\n\n<p>Thus the area of the quadrant is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%203.5%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\times \\pi \\times 3.5^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B77%7D%7B8%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{77}{8}\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the quadrant is&nbsp;<sub><\/sub>&nbsp;.<\/p>\n\n\n\n<p><strong>Q12&nbsp;<\/strong>In Fig, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the&nbsp;(ii) shaded region.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091648074.png\" alt=\"1636091648074\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>For area of shaded region we need to find area of the triangle.<\/p>\n\n\n\n<p>Area of triangle is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%203.5%5Ctimes%202\" alt=\"=\\ \\frac{1}{2}\\times 3.5\\times 2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%203.5%5C%20cm%5E2\" alt=\"=\\ 3.5\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the shaded region is = Area of the quadrant &#8211; Area of triangle<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B77%7D%7B8%7D%5C%20-%5C%203.5\" alt=\"=\\ \\frac{77}{8}\\ -\\ 3.5\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B49%7D%7B8%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{49}{8}\\ cm^2\"\/><\/figure>\n\n\n\n<p><strong>Q13&nbsp;<\/strong>In Fig, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use&nbsp;&nbsp;)<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091662355.png\" alt=\"1636091662355\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>In the given figure we need to find the radius of the circle:-<\/p>\n\n\n\n<p>Consider&nbsp;<img decoding=\"async\" alt=\"\\Delta\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5CDelta\">&nbsp;OAB,<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?OB%5E2%5C%20%3D%5C%20OA%5E2%5C%20+%5C%20AB%5E2\" alt=\"OB^2\\ =\\ OA^2\\ +\\ AB^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2020%5E2%5C%20+%5C%2020%5E2\" alt=\"=\\ 20^2\\ +\\ 20^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?OB%5C%20%3D%5C%2020%5Csqrt%7B2%7D%5C%20cm\" alt=\"OB\\ =\\ 20\\sqrt{2}\\ cm\"\/><\/figure>\n\n\n\n<p>Thus area of quadrant:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%203.14%5Ctimes%20%2820%5Csqrt%7B2%7D%29%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\times 3.14\\times (20\\sqrt{2})^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20628%5C%20cm%5E2\" alt=\"=\\ 628\\ cm^2\"\/><\/figure>\n\n\n\n<p>Also, the area of the square is :&nbsp;<img decoding=\"async\" alt=\"=\\ 20^2\\ =\\ 400\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2020%5E2%5C%20%3D%5C%20400%5C%20cm%5E2\"><\/p>\n\n\n\n<p>Area of the shaded region is :&nbsp;<img decoding=\"async\" alt=\"=\\ 628\\ -\\ 400\\ =\\ 228\\ cm^2\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20628%5C%20-%5C%20400%5C%20%3D%5C%20228%5C%20cm%5E2\"><\/p>\n\n\n\n<p><strong>Q14&nbsp;<\/strong>AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If&nbsp;&nbsp;, find the area of the shaded region.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091673181.png\" alt=\"1636091673181\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Area of the shaded region is = Area of larger sector &#8211; Area of smaller sector<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B30%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%2021%5E2%5C%20-%5C%20%5Cfrac%7B30%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%207%5E2\" alt=\"=\\ \\frac{30^{\\circ}}{360^{\\circ}}\\times \\pi \\times 21^2\\ -\\ \\frac{30^{\\circ}}{360^{\\circ}}\\times \\pi \\times 7^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B30%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%20%2821%5E2%5C%20-%5C%207%5E2%29\" alt=\"=\\ \\frac{30^{\\circ}}{360^{\\circ}}\\times \\pi \\times (21^2\\ -\\ 7^2)\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B308%7D%7B3%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{308}{3}\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the shaded region is&nbsp;<img decoding=\"async\" alt=\"\\frac{308}{3}\\ cm^2.\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5Cfrac%7B308%7D%7B3%7D%5C%20cm%5E2.\"><\/p>\n\n\n\n<p><strong>Q15&nbsp;<\/strong>In Fig, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091683741.png\" alt=\"1636091683741\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>Consider&nbsp;<img decoding=\"async\" alt=\"\\Delta\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%5CDelta\">&nbsp;ABC,<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?BC%5E2%5C%20%3D%5C%20AC%5E2%5C%20+%5C%20AB%5E2\" alt=\"BC^2\\ =\\ AC^2\\ +\\ AB^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2014%5E2%5C%20+%5C%2014%5E2\" alt=\"=\\ 14^2\\ +\\ 14^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?BC%5C%20%3D%5C%2014%5Csqrt%7B2%7D%5C%20cm\" alt=\"BC\\ =\\ 14\\sqrt{2}\\ cm\"\/><\/figure>\n\n\n\n<p>Area of triangle is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2014%5Ctimes%2014\" alt=\"=\\ \\frac{1}{2}\\times 14\\times 14\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%2098%5C%20cm%5E2\" alt=\"=\\ 98\\ cm^2\"\/><\/figure>\n\n\n\n<p>Now, area of sector is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%2014%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\times \\pi \\times 14^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20154%5C%20cm%5E2\" alt=\"=\\ 154\\ cm^2\"\/><\/figure>\n\n\n\n<p>And area of semicircle is : &#8211;<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%20%5Cpi%20r%5E2\" alt=\"=\\ \\frac{1}{2} \\pi r^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Ctimes%20%287%5Csqrt%7B2%7D%29%5E2\" alt=\"=\\ \\frac{1}{2} \\pi \\times (7\\sqrt{2})^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20154%5C%20cm%5E2\" alt=\"=\\ 154\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the shaded region is :&nbsp;<sub><\/sub><\/p>\n\n\n\n<p><strong>Q16&nbsp;<\/strong>Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cache.careers360.mobi\/media\/articles\/uploads\/froala_editor\/images\/2021\/11\/5\/1636091693677.png\" alt=\"1636091693677\"\/><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>It is clear from the figure that the required area (designed area) is the area of the intersection of two sectors.<\/p>\n\n\n\n<p>Area of the sector is:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B90%5E%7B%5Ccirc%7D%7D%7B360%5E%7B%5Ccirc%7D%7D%5Ctimes%20%5Cpi%20%5Ctimes%208%5E2\" alt=\"=\\ \\frac{90^{\\circ}}{360^{\\circ}}\\times \\pi \\times 8^2\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B352%7D%7B7%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{352}{7}\\ cm^2\"\/><\/figure>\n\n\n\n<p>And, area of the triangle:-<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%208%5Ctimes%208%5C%20%3D%5C%2032%5C%20cm%5E2\" alt=\"=\\ \\frac{1}{2}\\times 8\\times 8\\ =\\ 32\\ cm^2\"\/><\/figure>\n\n\n\n<p>Hence the area of the designed region is :<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%202%5Cleft%20%28%20%5Cfrac%7B352%7D%7B7%7D%5C%20-%5C%2032%20%5Cright%20%29\" alt=\"=\\ 2\\left ( \\frac{352}{7}\\ -\\ 32 \\right )\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/entrancecorner.codecogs.com\/gif.latex?%3D%5C%20%5Cfrac%7B256%7D%7B7%7D%5C%20cm%5E2\" alt=\"=\\ \\frac{256}{7}\\ cm^2\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Circle-Related Area The Class 10 chapter builds on the Class 9 chapter &#8216;Circles,&#8217; which you have already completed. There are many questions in NCERT Chapter 12 Maths Class 10 that use Class 9 concepts. There are three exercises with 35 questions in NCERT &#8230; <a title=\"NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles\" class=\"read-more\" href=\"https:\/\/careerssutra.com\/ncert-solutions-for-class-10-maths-chapter-12-areas-related-to-circles\/\" aria-label=\"More on NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","footnotes":""},"categories":[1],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false},"uagb_author_info":{"display_name":"admin","author_link":"https:\/\/careerssutra.com\/author\/admin\/"},"uagb_comment_info":0,"uagb_excerpt":"NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Circle-Related Area The Class 10 chapter builds on the Class 9 chapter &#8216;Circles,&#8217; which you have already completed. There are many questions in NCERT Chapter 12 Maths Class 10 that use Class 9 concepts. There are three exercises with 35 questions in NCERT&hellip;","_links":{"self":[{"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/posts\/2957"}],"collection":[{"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/comments?post=2957"}],"version-history":[{"count":3,"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/posts\/2957\/revisions"}],"predecessor-version":[{"id":2972,"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/posts\/2957\/revisions\/2972"}],"wp:attachment":[{"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/media?parent=2957"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/categories?post=2957"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/careerssutra.com\/wp-json\/wp\/v2\/tags?post=2957"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}