{"id":28742,"date":"2021-07-09T18:58:36","date_gmt":"2021-07-09T13:28:36","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28742"},"modified":"2022-03-02T10:30:13","modified_gmt":"2022-03-02T05:00:13","slug":"ncert-solutions-for-class-10-maths-chapter-12-ex-12-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-12-ex-12-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nThe radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
\nSolution:
\n\"NCERT
\nGiven: radius of 1st<\/sup> circle (R1<\/sub>) = 19 cm
\n\u2234 Circumference of 1st<\/sup> circle = 2\u03c0R1<\/sub> = 2\u03c0(19) cm
\nRadius of 2nd<\/sup> circle (R2<\/sub>) = 9 cm
\n\u2234 Circumference of 2nd<\/sup> circle = 2\u03c0R2<\/sub> = 2\u03c0(9) cm
\nLet radius of 3rd<\/sup> circle be R3<\/sub>
\nCircumference of 3rd<\/sup> circle = 2\u03c0R3<\/sub>
\nAccording to question,
\n2\u03c0R1<\/sub>\u00a0+ 2\u03c0R2<\/sub> = 2\u03c0R3<\/sub>
\n\u21d2 2\u03c0(R1<\/sub> + R2<\/sub>) = 2\u03c0R3<\/sub>
\n\u21d2 R1<\/sub> + R2<\/sub> = R3<\/sub>
\n\u21d2 19 + 9 = R3<\/sub>
\n\u21d2 R3<\/sub> = 28 cm
\nSo, the radius of the circle whose circumference is equal to the two circles of radius 19 cm is 28 cm.<\/p>\n

Question 2.
\nThe radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
\nSolution:
\n\"NCERT
\nGiven: radius of 1st<\/sup> circle (R1<\/sub>) = 8 cm
\nArea of 1st<\/sup> circle = \u03c0R1<\/sub>2<\/sup> = \u03c0(8)2<\/sup>cm2<\/sup>
\nRadius of 2nd<\/sup> circle (R2<\/sub>) = 6 cm
\nArea of 2nd<\/sup> circle = \u03c0R2<\/sub>2<\/sup> = \u03c0(6)2<\/sup> cm2<\/sup>
\nLet radius of 3rd<\/sup> circle be R3<\/sub>
\nArea of 3rd<\/sup> circle = \u03c0R3<\/sub>2<\/sup>
\nAccording to question,
\n\u03c0R,2<\/sup> + \u03c0R2<\/sub>2<\/sup> – \u03c0R3<\/sub>2<\/sup>
\n\u21d2 R1<\/sub>2<\/sup> + R2<\/sub>2<\/sup> = R3<\/sub>2\u00a0<\/sup>
\n\u21d2 (8)2<\/sup> + i6)2<\/sup> – R3<\/sub>2<\/sup>
\n\u21d2 64 + 36 = R3<\/sub>2<\/sup> \u21d2 R3<\/sub>=\u00a0\u00a0\\( \\sqrt{100} \\) = 10 cm
\nTherefore, radius of the required circle is 10 cm.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nThe figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing 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.
\n\"NCERT
\nSolution:
\nDiameter of region representing Gold = 21 cm
\nRadius of Gold = \\(\\frac { 21 }{ 2 }\\) cm = 10.5cm
\nArea of region Gold = 2\u03c0r\u00b2
\n= \\(\\frac { 22 }{ 2 }\\) x (10.5)\u00b2
\n= 346.5 cm\u00b2
\nThe Red band is 10.5 wide
\nRadius of (Red + Gold) = 10.5 + 10.5 cm
\n= 21 cm
\nArea of (Red + Gold) = \u03c0r\u00b2 = \\(\\frac { 22 }{ 2 }\\)
\n= 1386 cm\u00b2
\nArea of Red band =Area of (Red + Gold) – Area of Gold
\n= 1386 – 346.5 cm
\n= 1039.5 cm\u00b2
\nAgain,
\nRadius of (Gold + Red + Blue)
\n= 21 + 10.5 cm = 31.5 cm
\nArea of (Gold + Red + Blue)
\n= \u03c0r\u00b2
\n= \\(\\frac { 22 }{ 2 }\\) x 31.5 x 31.5
\n= 3118.5 cm\u00b2
\nAre of Blue Band = Area of (Gold + Red + Blue) – Area of (Gold + Red)
\n= 3118.5 – 1386
\n= 1732.5 cm\u00b2
\nAgain,
\nRadius of (Gold + Red + Blue + Black) = 31,5 + 10.5 = 42 cm
\nArea of (Gold + Red + Blue + Black) = \u03c0r\u00b2
\n= \\(\\frac { 22 }{ 2 }\\) – (42)\u00b2
\n= \\(\\frac { 22 }{ 2 }\\) x 42 x 42 cm\u00b2
\n= 5544 cm\u00b2
\nArea of Black Band = Area of (Gold + Red + Blue + Black) – Area of (Gold + Red + Blue)
\n= 5544 – 3118.5 cm\u00b2
\n= 2425.5 cm\u00b2
\nAgain,
\nRadius of (Gold + Red + Blue + Black + White)
\n= 42 + 10.5
\n= 52.5
\nArea of (Gold + Red + Blue + Black + White)
\n= \u03c0r\u00b2
\n= \\(\\frac { 22 }{ 2 }\\) x (52.5)\u00b2
\n= \\(\\frac { 22 }{ 2 }\\) x 52.5 x 52.5 cm\u00b2
\n= 8662.5 cm\u00b2
\nArea of White Band = Area of (Gold + Red + Blue + White) – Area of (Gold + Red + Blue + Black)
\n= 8662.5 – 5544 cm\u00b2
\n= 3118.5 cm\u00b2<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nThe 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 travelling at a speed of 66 km per hour?
\nSolution:
\nDistance travelled by the wheel in one revolution
\n= \u03c0 x diameter of the wheel
\n= \\(\\frac { 22 }{ 2 }\\) x 80
\nTotal distance travelled by the wheel in one minute
\n= \\(\\frac{66 \\times 1000 \\times 100}{60}\\) cm
\n\u2234 Number of revolutions made by the wheel in one minute
\n\\(\\frac{66 \\times 1000 \\times 10}{60 \\times \\frac{22}{7} \\times 80}\\) = \\(\\frac{66 \\times 1000 \\times 100 \\times 7}{60 \\times 22 \\times 80}\\) = 437.5
\n\u2234 Number of revolutions made by the wheel m 10 minutes
\n= (437.5 x 10) = 4375<\/p>\n

Question 5.
\nTick 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
\nSolution:
\nLet radius of the circle = r units
\nPerimeter of the circle = 2\u03c0r
\nArea of the circle = \u03c0r2<\/sup>
\nAccording to question,
\nPerimeter of the circle = Area of the circle
\n\u21d2\u00a02\u03c0r =\u00a0\u03c0r2<\/sup>
\n\u21d2 r = 2 units
\nHence, option (a) is correct.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.1 Question 1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-12-ex-12-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts. 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