{"id":6088,"date":"2023-09-19T16:53:40","date_gmt":"2023-09-19T16:53:40","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6088"},"modified":"2023-09-19T16:53:40","modified_gmt":"2023-09-19T16:53:40","slug":"ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-1\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 12 (Areas Related to Circles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 12 Areas Related to Circles <\/strong>Exercise 12.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 12 Areas Related to Circles<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 12.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 12.3<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 12.1<\/span><\/strong><\/h2>\n<p><span style=\"color: #000000;\"><strong>Unless stated otherwise, take \u03c0 = 22\/7.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. The radii of the 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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 \u00a0<\/strong>Let the radius of the third circle be R.<\/span><br \/>\n<span style=\"color: #000000;\">Circumference of the circle with radius R = 2\u03c0R <\/span><br \/>\n<span style=\"color: #000000;\">Circumference of the circle with radius 19 cm = 2\u03c0 \u00d7 19 = 38\u03c0 cm <\/span><br \/>\n<span style=\"color: #000000;\">Circumference of the circle with radius 9 cm = 2\u03c0 \u00d7 9 = 18\u03c0 cm <\/span><br \/>\n<span style=\"color: #000000;\">Sum of the circumference of two circles = 38\u03c0 + 18\u03c0 = 56\u03c0 cm <\/span><br \/>\n<span style=\"color: #000000;\">Circumference of the third circle = 2\u03c0R = 56\u03c0 <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 2\u03c0R = 56\u03c0 cm <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 R = 28 cm <\/span><br \/>\n<span style=\"color: #000000;\">The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. The radii of the 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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Let the radius of the third circle be R.<br \/>\nArea of the circle with radius R = \u03c0R<sup>2<br \/>\n<\/sup>Area of the circle with radius 8 cm = \u03c0 \u00d7 8<sup>2\u00a0<\/sup>= 64\u03c0 cm<sup>2<br \/>\n<\/sup>Area\u00a0of the circle with radius 6 cm = \u03c0 \u00d7 6<sup>2\u00a0<\/sup>= 36\u03c0 cm<sup>2<br \/>\n<\/sup>Sum of the area of two circles = 64\u03c0 cm<sup>2\u00a0<\/sup>+<sup>\u00a0<\/sup>36\u03c0 cm<sup>2\u00a0<\/sup>= 100\u03c0<sup>\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>Area of the third circle = \u03c0R<sup>2<\/sup>\u00a0= 100\u03c0<sup>\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>\u21d2 \u03c0R<sup>2<\/sup>\u00a0= 100\u03c0<sup>\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>\u21d2 R<sup>2<\/sup>\u00a0= 100<sup>\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>\u21d2 R\u00a0= 10<sup>\u00a0<\/sup>cm<br \/>\nThus, the radius of the new circle is 10 cm. <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Fig. 12.3 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 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.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6611\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.1-Ans3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"160\" height=\"153\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0 \u00a0<\/strong>The radius of 1<sup>st<\/sup>\u00a0circle, r<sub>1<\/sub> = 21\/2 cm (as diameter D is given as 21 cm) <\/span><br \/>\n<span style=\"color: #000000;\">So, area of gold region = \u03c0 r<sub>1<\/sub><sup>2\u00a0<\/sup>= \u03c0(10.5)<sup>2\u00a0<\/sup>= 346.5 cm<sup>2<br \/>\n<\/sup>Now, it is given that each of the other bands is 10.5 cm wide, <\/span><br \/>\n<span style=\"color: #000000;\">So, the radius of 2<sup>nd<\/sup>\u00a0circle, r<sub>2<\/sub> = 10.5cm+10.5cm = 21 cm <\/span><br \/>\n<span style=\"color: #000000;\">Thus, <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 area of red region = Area of 2<sup>nd<\/sup>\u00a0circle \u2212 Area of gold region = (\u03c0r<sub>2<\/sub><sup>2<\/sup>\u2212346.5) cm<sup>2<br \/>\n<\/sup>= (\u03c0(21)<sup>2<\/sup>\u00a0\u2212 346.5) cm<sup>2<br \/>\n<\/sup>= 1386 \u2212 346.5 <\/span><br \/>\n<span style=\"color: #000000;\">= 1039.5 cm<sup>2<br \/>\n<\/sup>Similarly, <\/span><br \/>\n<span style=\"color: #000000;\">The radius of 3<sup>rd<\/sup>\u00a0circle, r<sub>3<\/sub> = 21 cm+10.5 cm = 31.5 cm <\/span><br \/>\n<span style=\"color: #000000;\">The radius of 4<sup>th<\/sup>\u00a0circle, r<sub>4<\/sub> = 31.5 cm+10.5 cm = 42 cm <\/span><br \/>\n<span style=\"color: #000000;\">The Radius of 5<sup>th<\/sup>\u00a0circle, r<sub>5<\/sub> = 42 cm+10.5 cm = 52.5 cm <\/span><br \/>\n<span style=\"color: #000000;\">For the area of n<sup>th\u00a0<\/sup>region, <\/span><br \/>\n<span style=\"color: #000000;\">A = Area of circle n \u2013 Area of the circle (n-1) <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 area of the blue region (n=3) = Area of the third circle \u2013 Area of the second circle <\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0(31.5)<sup>2<\/sup>\u00a0\u2013 1386 cm<sup>2<br \/>\n<\/sup>= 3118.5 \u2013 1386 cm<sup>2<br \/>\n<\/sup>= 1732.5 cm<sup>2<br \/>\n<\/sup>\u2234 area of the black region (n=4) = Area of the fourth circle \u2013 Area of the third circle <\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0(42)<sup>2<\/sup>\u00a0\u2013 1386 cm<sup>2<br \/>\n<\/sup>= 5544 \u2013 3118.5 cm<sup>2<br \/>\n<\/sup>= 2425.5 cm<sup>2<br \/>\n<\/sup>\u2234 area of the white region (n=5) = Area of the fifth circle \u2013 Area of the fourth circle <\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0(52.5)<sup>2<\/sup>\u00a0\u2013 5544 cm<sup>2<br \/>\n<\/sup>= 8662.5 \u2013 5544 cm<sup>2<br \/>\n<\/sup>= 3118.5 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. 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 travelling at a speed of 66 km per hour?<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The radius of car\u2019s wheel = 80\/2 = 40 cm (as D = 80 cm)<\/span><br \/>\n<span style=\"color: #000000;\">So, the circumference of wheels = 2\u03c0r = 80 \u03c0 cm<\/span><br \/>\n<span style=\"color: #000000;\">Now, in one revolution, the distance covered = circumference of the wheel = 80 \u03c0 cm<\/span><br \/>\n<span style=\"color: #000000;\">It is given that the distance covered by the car in 1 hr = 66km <\/span><br \/>\n<span style=\"color: #000000;\">Converting km into cm, we get, <\/span><br \/>\n<span style=\"color: #000000;\">Distance covered by the car in 1hr = (66\u00d710<sup>5<\/sup>) cm <\/span><br \/>\n<span style=\"color: #000000;\">In 10 minutes, the distance covered will be = (66\u00d710<sup>5<\/sup>\u00d710)\/60 = 1100000 cm\/s <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 distance covered by car = 11\u00d710<sup>5<\/sup> cm <\/span><br \/>\n<span style=\"color: #000000;\">Now, the no. of revolutions of the wheels = (Distance covered by the car\/Circumference of the wheels) <\/span><br \/>\n<span style=\"color: #000000;\">= (11 \u00d7 10<sup>5<\/sup>)\/80 \u03c0 <\/span><br \/>\n<span style=\"color: #000000;\">= 4375.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Tick the correct solution 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<br \/>\n<\/strong>(A) 2 units<br \/>\n(B) \u03c0 units<br \/>\n(C) 4 units<br \/>\n(D) 7 units<br \/>\n<strong>Solution &#8211;\u00a0 <\/strong>Since the perimeter of the circle = area of the circle,<br \/>\n2\u03c0r = \u03c0r<sup>2<br \/>\n<\/sup>r = 2<br \/>\nSo, option (A) is correct, i.e., the radius of the circle is 2 units.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 12 (Areas Related to Circles)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 12 Areas Related to Circles Exercise 12.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 12 Areas Related to Circles NCERT Class [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1105,1106,1044,1049,1048],"class_list":["post-6088","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-12-areas-related-to-circles-solutions","tag-ncert-class-10-mathematics-exercise-12-1-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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