{"id":856,"date":"2020-09-15T11:11:57","date_gmt":"2020-09-15T05:41:57","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=856"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-5-ex-5-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1<\/h2>\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:
\nRadius of 1st Circle, r1<\/sub> = 19 cm.
\nCircumference of 1st circle, C1<\/sub> = 2\u03c0r1<\/sub>
\n\\(=2 \\times \\frac{22}{7} \\times 19\\)
\nRadius of 2nd circle, r2<\/sub> = 9 cm.
\nCircumference of 2nd circle, C2<\/sub> = 2\u03c0r2<\/sub>
\n\\(=2 \\times \\frac{22}{7} \\times 9\\)
\nSum of the circumferences of the two circles
\n\\(=2 \\times \\frac{22}{7} \\times 19+2 \\times \\frac{22}{7} \\times 9\\)
\n\\(C_{3}=2 \\times \\frac{22}{7}(19+9)\\)
\n\\(=2 \\times \\frac{22}{7} \\times 28\\)
\n= 2 \u00d7 22 \u00d7 4
\n\u2234 C3<\/sub> = 176 cm.
\nCircumference of 3rd circle, C3<\/sub> = 176 cm. then, r3 <\/sub>= ?
\n2\u03c0r3<\/sub> = C3<\/sub>
\n\\(2 \\times \\frac{22}{7} \\times r_{3}=176\\)
\n\\(r_{3}=176 \\times \\frac{7}{22} \\times \\frac{1}{2}\\)
\n\u2234 r3<\/sub> = 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:
\nRadius of 1st Circle, r1<\/sub> = 8 cm.
\nArea of 1st Circle, A1<\/sub> = \u03c0 r1<\/sub>2<\/sup>
\n\\(=\\frac{22}{7} \\times(8)^{2}\\)
\n\\(A_{1}=\\frac{22}{7} \\times 64\\)
\nRadius of 2nd Circle, r2<\/sub> = 6 cm.
\nArea of 2nd Circle, A2<\/sub> = \u03c0 r2<\/sub>2<\/sup>
\n\\(=\\frac{22}{7} \\times(6)^{2}\\)
\n\\(A_{2}=\\frac{22}{7} \\times 36\\)
\nTotal area of two circles =
\n\"KSEEB<\/p>\n

Question 3.
\nThe figure given below 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.
\n\"KSEEB
\nSolution:
\ni) The diameter of the region representing Gold Score :
\n21
\nd = 21 cm. \u2234 Radius, \\(r=\\frac{21}{2} \\mathrm{cm}\\)
\n\u2234 Area of the region representing Gold score
\n= \u03c0r2<\/sup>
\n\"KSEEB
\nA = 346.5 sq. cm.<\/p>\n

ii) Radius of the region representing Gold score and Red score:
\nr = 10.5 + 10.5 = 21 cm.
\n\u2234 Area of the region representing Red score: Area of the region representing Gold and Red Score \u2014 Area of the region representing Gold score.
\n= \u03c0r2<\/sup> – \u03c0r2<\/sup>
\n\"KSEEB
\n\u2234 A = 1039.5 sq.cm.<\/p>\n

iii) Total radii of the region representing Gold, Red and Blue region :
\nr =10.5 + 10.5 + 10.5 = 31.5 cm
\n\\(=\\frac{63}{2} \\mathrm{cm}\\)
\n\u2234 Area of the region representing Blue region:
\n[ Area of the region representing Gold, Red and Blue regions ] \u2014 Area of the region representing Gold and Red region
\n\u2234 A = \u03c0r2<\/sup> – \u03c0r2<\/sup>
\n\"KSEEB<\/p>\n

iv) Similarly, to find Area of the region representing Black region :
\n\u2234 A = \u03c0r2<\/sup> – \u03c0r2<\/sup>
\n\"KSEEB
\n\u2234 A = 2425.5 sq. cm.<\/p>\n

v) Area of the region representing white:
\n\u2234 A = \u03c0r2<\/sup> – \u03c0r2<\/sup>
\n\"KSEEB
\n\u2234 A = 3118.5 sq. cm<\/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:
\nDiameter of the Wheel of the car = 80 cm.
\nRadius of the wheel = 40 cm.
\nCircumference of the Wheel = 2\u03c0r
\n\\(=2 \\times \\frac{22}{7} \\times 40\\)
\n\\(=\\frac{44 \\times 40}{7}\\)
\n= 251.4 cm
\nSpeed of the car = 66 km\/hr.
\n\\(=\\frac{66 \\times 100000}{60} \\mathrm{cm} . \/ \\mathrm{min}\\)
\n= 110000 cm\/min.
\nDistance travelled by car in 10 minutes
\n= 110000 \u00d7 10
\n= 1100000 cm.
\nLet the number of revolutions of the wheel be \u2019n’, then
\nn \u00d7 251.4 = 1100000
\n\u2234 \\(\\mathrm{n}=\\frac{1100000}{251.4}\\)
\nn = 4375
\n\u2234 Car can complete 4,375 revolutions in 10 minutes.<\/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
\nA) 2 units
\nB) n units
\nC) 4 units
\nD) 7 units
\nSolution:
\nA) 2 units
\nCircumference of a circle = Area of the circle.
\n2\u03c0r = \u03c0r2<\/sup>
\n2 \u00d7 \u03c0 \u00d7 r = \u03c0 \u00d7 r \u00d7 r
\n\u2234 2 = r
\n\u2234 r = 2 units.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1. Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1 …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/856"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=856"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/856\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=856"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=856"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=856"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}