{"id":926,"date":"2020-09-15T06:15:36","date_gmt":"2020-09-15T00:45:36","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=926"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-maths-solutions-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-5-ex-5-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3<\/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 5.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3<\/h2>\n

{Unless stated otherwise, use \\(\\pi=\\frac{22}{7}\\)}<\/p>\n

Question 1.
\nFind the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
\nSolution:
\n\"KSEEB
\nPQ = 24 cm, PR = 7 cm,
\n‘O’ is the centre of circle.
\nAngle in semicircle,
\n\u2220RPQ = 90\u00b0
\nIn \u22a5\u2206RPQ, \u2220P = 90\u00b0
\nRQ = RP2<\/sup> + PQ2<\/sup>
\n= (7)2<\/sup> + (24)2<\/sup>
\n= 49 + 576
\nRQ2<\/sup> = 625
\n\u2234 RQ = 25 cm.
\nDiameter, RQ = 25 cm.
\n\u2234 Radius, OR = OQ = \\(\\frac{25}{2}\\) cm.
\nArea of shaded part :
\n= Area of semicircle – Area of \u2206RPQ
\n\"KSEEB
\n= 161.3 sq.cm.<\/p>\n

Question 2.
\nFind the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and \u2220AOC = 40\u00b0.
\nSolution:
\n\"KSEEB
\nRadius of small circle, r = 7 cm
\nRadius of big circle, R = 14 cm
\n\u2220AOC = \u03b8 = 40\u00b0
\nArea of shaded part, ABCD =?
\nArea of ABCD
\n= Area of OAC – Area of OBD
\n\"KSEEB<\/p>\n

Question 3.
\nFind the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
\nSolution:
\n\"KSEEB
\nEach side of square ABCD = 14 cm.
\nAPD, APC are semicircles.
\nArea of the shaded region = ?
\ni) Total area of square
\n= (Side)2<\/sup>
\n= (14)2<\/sup>
\n= 196 sq.cm.<\/p>\n

ii) Each radius, r of semicircle
\nr = \\(\\frac{14}{2}\\) =7cm
\nArea of two semicircles :
\n\"KSEEB
\n\u2234 Area of shaded region:
\n= Area of square – Area of two semicircles
\n= 196 – 154
\n= 42 sq. cm<\/p>\n

Question 4.
\nFind the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
\nSolution:
\n\"KSEEB
\nOAB is an equilateral Triangle with side 12 Circle with radius of 6
\nEach angle of an equilateral triangle is 60\u00b0
\nAngle at the centre in segment, \u03b8 = 60\u00b0
\nArea of Major segment + Area of an equilateral triangle OAB =
\n\"KSEEB
\n= 94.28 + 36 \u00d7 1.73
\n= 94.28 + 62.28
\n= 156.56 sq. cm.<\/p>\n

Question 5.
\nFrom 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 the given figure. Find the area of the remaining portion of the square.
\nSolution:
\ni) Each side of square ABCD = 4 cm.
\n\"KSEEB
\n\u2234 Area of Square:
\n= (Side)2<\/sup>
\n= (4)2<\/sup>
\n= 16 sq.cm.<\/p>\n

(ii) Area of circle in the corner A is 1 cm.
\nRadius, r = 1 cm.
\n\u03b8 = 90\u00b0
\n\u2234 Area of quadrant of circle:
\n\"KSEEB
\n= 0.78 sq.cm.<\/p>\n

iii) Radius of circle having diameter 2 cm, r = 1 cm.
\n\u2234 Area of circle:
\n\"KSEEB
\n= 3.14 sq. cm.<\/p>\n

\u2234 Area of remaining part of Square :
\n= Area of Square – Area of 4 quadrants – Area of Circle.
\n= 16 – 4 \u00d7 0.78 – 3.14
\n= 16 – 3.12 – 3.14
\n= 16 – 6.26
\n= 9.74 sq.cm.<\/p>\n

Question 6.
\nIn a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design.
\n\"KSEEB
\nSolution:
\nRadius of the circle r = 32 cm.
\n\u2234 Area of the circle = \u03c0r2<\/sup>
\n\"KSEEB<\/p>\n

ii) \u2220BAC = 60\u00b0 \u2234 \u2220BOC = 120\u00b0
\n(\u2235 Angle at the centre is double than the angle in the circumference).
\nNow, in \u2206OBC, BC \u22a5 OD,
\n\u2220BOD = \u2220COD = 60\u00b0
\n\"KSEEB
\n\u2234 \u2206ABC is an equilateral triangle.
\nEach side of this triangle,
\na= 2 \u00d7 \\(16 \\sqrt{3}=32 \\sqrt{3} \\mathrm{cm}\\)
\nArea of equilateral triangle, \u2206ABC :
\n\"KSEEB
\n= 1328.64 sq.cm,<\/p>\n

iii) Area of shaded region :
\n= Area of circle – Area of an equilateral \u2206le<\/sup>
\n= 3218.28 – 1328.64
\n= 1889.6 sq.cm.<\/p>\n

Question 7.
\nIn the Figure given below, ABCD is a square of side 14 cm. With centres 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.
\nSolution:
\n\"KSEEB
\ni) Each side of square ABCD =14 cm.
\n\u2234 Area of square ABCD
\n= (Side)2<\/sup>
\n= (14)2<\/sup>
\n= 196 sq.cm.<\/p>\n

ii) Measure of radii of 4 circles, r = \\(\\frac{14}{2}=\\) = 7 cm.
\n\u2235 Distance between tangents which touches circles externally,
\nd = R + r = 7 + 7 = 14 cm.
\nArea of segment with centre, A = ?
\nr = 7 cm, \u03b8 = 90\u00b0
\n\"KSEEB
\n\u2234 Total area of 4 sectors
\n= 4 \u00d7 38.5 = 154 sq.cm.<\/p>\n

iii) The Area of track :
\n= (Area of GHIJ – Area of ABCD)
\n= 196 – 154
\n= 42 sq.cm.<\/p>\n

Question 8.
\nThe Figure given below depicts a racing track whose left and right ends are semicircular.
\n\"KSEEB
\nThe 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:
\ni) The distance around the track along its inner edge,
\nii) The area of the track.
\nSolution:
\n\"KSEEB
\nThe distance around the track along its inner edge:
\n= AB + Arc BEC + CD + Area of Arc DFA
\n\"KSEEB
\nThe Area of Track:
\n= (Area of GHIJ – Area of ABCD) + (Area of semicircle HKI – Area of semicircle BEC) + (Area of semicircle GIJ – Area of semicircle AFD)
\n\"KSEEB
\n= 2120 + 2200
\n= 4320 sq.m.
\n\u2234 Total area of Track = 4320 sq.m.<\/p>\n

Question 9.
\nIn the figure given below, AB and CD are two diameters of a circle (with centre 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.
\nSolution:
\n\"KSEEB
\ni) Radius of Big circle is the diameter of a small circle.
\n\u2234 Radius of small circle
\n= \\(\\frac{7}{2}\\) = 3.5 cm
\n\u2234 Area of smalle circle = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 (3.5)2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 12.25
\n= 3.14 \u00d7 12.25
\n= 38.47 sq.cm.<\/p>\n

ii) Area of segment OCB = ?
\nRadius, r = 7 cm, \u03b8 = 90\u00b0
\n\"KSEEB
\n\u2234 Area of sector OCB – Area of \u2206OBC
\n= 38.47 – 24.50
\n= 13.97 sq.cm.
\nSimilarly, Area of sector OAC – Area of \u2206OAC
\n= 38.47 – 24.50
\n= 13.97 sq.cm.
\n\u2234 Total area of shaded region :
\n= Area of small circle + Area of Minor segment OBC + Area of Minor segment OAC
\n= 38.47 + 13.97 + 13.97
\n= 66.41 sq.cm.<\/p>\n

Question 10.
\nThe area of an equilateral triangle ABC is 17320.5 cm2<\/sup>. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see the Figure). Find the area of the shaded region.
\n(Use \u03c0 = 314 and \\(\\sqrt{3}\\) = 1.73205).
\nSolution:
\n\"KSEEB
\ni) Area of an equilateral \u2206ABC :
\n= 17320.5 cm2<\/sup>
\n\\(\\frac{\\sqrt{3} a^{2}}{4}\\) = 17320.5
\n\"KSEEB<\/p>\n

ii) Area formed by centre A:
\nr = 100 cm, \u03b8 = 60\u00b0
\n\"KSEEB<\/p>\n

Similarly Area of sectors formed by centres B and C is 5233 sq.cm.
\n\u2234 Total Area of THREE sectors :
\n= 3 \u00d7 5233 = 15699 sq.cm.
\n\u2234 Area of the shaded region:
\n= Area of an equilateral Triangle – Area of three sectors.
\n= 17320.5 – 15699
\n= 1621.5 sq. cm.<\/p>\n

Question 11.
\nOn a square handkerchief, nine circular designs each of radius 7 cm are made (see the Figure). Find the area of the remaining portion of the handkerchief.
\nSolution:
\n\"KSEEB
\ni) Total area of 9 circular designs each of radius 7 cm
\n\"KSEEB<\/p>\n

ii) All the circles touches externally.
\n\u2234 Sum of the diameter of 3 circles in first Row =
\n14 + 14 + 14 = 42 cm.
\n\u2234 Length of each side of square ABCD,
\na = 42 cm.
\n\u2234 Area of square ABCD = a2<\/sup>
\n= (42)2<\/sup>
\n= 1764 sq.cm.
\nArea of remaining part of handkerchief:
\n= Area of a square – Area of 9 circles.
\n= 1764 – 1386
\n= 378 sq.cm.<\/p>\n

Question 12.
\nIn the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm If OD = 2 cm, find the area of the
\ni) quadrant OACB,
\nii) shaded region.
\nSolution:
\nRadius of Circle, OA = OB = 3.5 cm.
\nOD = 2 cm.
\ni) Area of a quadrant of a circle
\nii) Area of the shaded region
\n\"KSEEB
\ni) Area of quadrant (OACB) of a Circle
\nr = 3.5 cm, \u03b8 = 90\u00b0
\n\"KSEEB<\/p>\n

ii) In \u22a5\u2206BOD, \u2220BOD = 90\u00b0 OB = 3.5 cm; OD = 2 cm
\n\"KSEEB
\n\u2234 Area of shaded region,
\n= Area of quadrant OACB – Area of ABOD
\n= 9.61 – 3.5
\n= 6.11 sq.cm.<\/p>\n

Question 13.
\nIn Figure (i) given below, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use \u03c0 = 3.14)
\nSolution:
\n\"KSEEB
\ni) Each side of square OABC = 20 cm.
\n\u2234 OA = 20 cm, AB = 20 cm.
\n\u2234 Area of square PABC = (Side)2<\/sup>
\n= (20)2<\/sup>
\n= 400 sq.cm.<\/p>\n

ii) Diagonal OB is drawn in OABC square.
\nIn \u22a5\u2206OAB, OA = 20 cm.
\nAB = 20 cm.
\nOB =?
\nOB2<\/sup> = OA2<\/sup> + AB2<\/sup>
\n= (20)2<\/sup> + (20)2<\/sup>
\n= 400 + 400
\n= 800
\n\"KSEEB
\n\u2234 Area of shaded region = 628 – 400 = 228 cm2<\/sup>.<\/p>\n

Question 14.
\nAB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Figure given below). If \u2220AOB = 30\u00b0, find the area of the shaded region.
\nSolution:
\ni) Area of segment OAPB = ?
\nr = 21 cm, \u03b8 = 30\u00b0
\n\"KSEEB
\n= \\(\\frac{1}{12}\\) \u00d7 3.14 \u00d7 49
\n= 12.82 sq.cm.
\n\u2234 Area of the shaded region = Area of Sector OAPB – Area of segment OCQD
\n= 115.4 – 12.82
\n= 102.58 sq.cm.<\/p>\n

Question 15.
\nIn the given figure (i), 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.
\nSolution:
\n\"KSEEB
\ni) Area of Part II :
\nArea of Segment ABC – Area of \u2206ABC
\n\"KSEEB
\n= 56 sq.cm.<\/p>\n

ii) Area of shaded region Part III :
\n\"KSEEB<\/p>\n

Question 16.
\nCalculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each
\n\"KSEEB
\nSolution:
\n\"KSEEB
\ni) Area of Square, ABCD = a2<\/sup> = (8)2<\/sup> = 64 cm2<\/sup>.
\nii) Sum of Areas of Part II and Part III = Area of the segment with centre D and radius of 8 cm.
\n\"KSEEB<\/p>\n

iii) Area of Part I = (Sum of Part I, II and III) – (Sum of the area of Part II, III)
\n= Area of Square ABCD – (Sum of the area of part II and III)
\n\"KSEEB<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3, 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 5.3 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 5.3. Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3 …<\/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\/926"}],"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=926"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/926\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=926"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=926"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=926"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}