{"id":28782,"date":"2021-07-12T17:54:35","date_gmt":"2021-07-12T12:24:35","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28782"},"modified":"2022-03-02T10:30:11","modified_gmt":"2022-03-02T05:00:11","slug":"ncert-solutions-for-class-10-maths-chapter-12-ex-12-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-12-ex-12-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3"},"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 12.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFind the area of the shaded region in the given figure, if PQ = 24cm, PR = 7cm and O is the centre of the circle.
\nSolution:
\n\"NCERT<\/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 = 400.
\nSolution:
\n\u2220AOC = 40\u00b0 (given)
\nRadius of the sector AOC = 14 cm
\n\"NCERT<\/p>\n

\"NCERT<\/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.
\n\"NCERT
\nSolution:
\nABCD is a square
\nGiven: side of the square = 14 cm
\n\u2234 Area of the square = (side)\u00b2 = (14)\u00b2 = 196 cm\u00b2
\nRadius of the semicircle APD = \\(\\frac { 1 }{ 2 }\\)(side of square) = \\(\\frac { 1 }{ 2 }\\) x 14 = 7 cm
\nArea of the semicircle APD = \\(\\frac { 1 }{ 2 }\\) \u03c0r\u00b2 = \\(\\frac { 1 }{ 2 }\\) \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 7 \u00d7 7 = 11 \u00d7 7 = 77cm\u00b2
\nSimilarly, area of the semicircle BPC = 77 cm\u00b2
\nTotal area of both the semicircles = 77 + 77 = 154 cm\u00b2
\nArea of the shaded region = Area of square – area of both semicircles
\n= 196 – 154 = 42 cm\u00b2<\/p>\n

Question 4.
\nFind the area of the shaded region in the 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:
\nArea of the equilateral triangle OAB
\n\"NCERT<\/p>\n

\"NCERT<\/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 figure. Find the area of the remaining portion of the square.
\n\"NCERT
\nSolution:
\nIn the figure, area of the shaded region = area of the square ABCD with side 4 cm – area of the circle of radius 1 cm, centrally placed – area of the four quarter circles of radii 1 cm each, placed | at each corner of the square ABCD.
\n\"NCERT
\n\u2234 Area of the remaining portion of the square = \\(\\frac { 68 }{ 7 }\\) cm\u00b2<\/p>\n

Question 6.
\nIn a circular table cover of the radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).
\n\"NCERT
\nSolution:
\n\"NCERT<\/p>\n

Question 7.
\nIn the figure, 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.
\n\"NCERT
\nSolution:
\nEdge of the square ABCD = 14 cm
\nArea of square ABCD = (14)\u00b2 = 196 cm\u00b2
\nHere radius of each circle is 7 cm.
\n\"NCERT
\nArea of shaded region = Area of square – Area of 4 sectors
\n196 – 154 = 42cm\u00b2.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nThe given figure 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:
\n(i) the distance around the track along its inner edge.
\n(ii) the area of the track.
\n\"NCERT
\nSolution:
\n(i) ABCD and EFGH are two rectangles and corner has two semicircles. The distance around the track along its inner edge
\n\"NCERT
\n= 2 x length of rectangle + 2 x circumference of semicirlce
\n\"NCERT<\/p>\n

(ii) Area of the track = 2 [Area of rectangles] + 2 [Area of outer semicircle] – [Area of inner semicircles]
\n\"<\/p>\n

Question 9.
\nIn the figure, 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.
\n\"NCERT
\nSolution:
\nSince AB \u22a5 CD,
\ntherefore, \u2220COB = \u2220COA
\n\u2220DOA = \u2220DOB = 90\u00b0
\nIn the figure, area of the shaded region, area of the small circle of diameter (OD = OA = 7cm) + (area of the segment BMC with central angle BOC = 90c and radius (OB = OC = 7cm) + area of the segment ANC with central angle AOC = 90\u00b0 and radius (OA = OC) = 7cm
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nThe area of an equilateral triangle ABC is 17320.5 cm\u00b2. 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 figure). Find the area of the shaded region. (Use \u03c0 = 3.14 and \\(\\sqrt{3}\\) = 1.73205).
\nSolution:
\n\u2206ABC is an equilateral triangle. About the angular points A, B and C as centres three circles half the length of the side of the triangle are described.
\nArea of equilateral \u2206ABC(given) = 17320.5 cm\u00b2
\n\"NCERT
\nSince ABC is an equilateral triangle, therefore \u2220A = \u2220B = \u2220C = 60\u00b0
\n= 17320.5 cm\u00b2
\nThere are 3 equal sectors in the figure of central angles 60\u00b0 and radii 100 cm [\u2235 \\(\\frac { 200 }{ 2 }\\) = 10 cm]
\nThen, area of the shaded region = area of \u2206ABC – 3 (area of one sector of central angle 60\u00b0 and radius 100 cm)
\n= 17320.5 cm\u00b2 – [3 x \\(\\frac { 60\u00b0 }{ 360\u00b0 }\\) x \u03c0 x (100)\u00b2]
\n= 17320.5 cm\u00b2 – (\\(\\frac { 3 }{ 6 }\\) x 3.14 x 1000) cm\u00b2
\n= 17320.5 cm\u00b2 – (5000 x 3.14) cm\u00b2
\n= (17320.5 – 15700) cm\u00b2 = 1620.5 cm\u00b2
\n\u2234 area of the shaded region is 1620.5 cm\u00b2<\/p>\n

Question 11.
\nOn a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
\n\"NCERT
\nSolution:
\nABCD is a square
\nEdge of the square ABCD = 3 x diameter of circle
\n= 3 x 4 = 42 cm
\nArea of square ABCD = (42)\u00b2
\n= 1764 cm\u00b2
\nRadius of one circle = 7 cm
\nArea of one circle = \u03c0r\u00b2
\n= \\(\\frac { 22 }{ 7 }\\) x (7)\u00b2
\n= 154 cm\u00b2
\nArea of nine circles = 9 x Area of one circle
\n= 9 x 154 = 1386 cm\u00b2
\nArea of remaining portion = Area of square – Area of circle
\n= 1764 – 1386 = 378 cm\u00b2<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nIn the 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 (i) quadrant OACB, (ii) shaded region.
\nSolution:
\n(i) Let r be the radius of the circle and r = 35 cm \\(\\frac { 35 }{ 10 }\\) = \\(\\frac { 7 }{ 2 }\\) cm
\nRadius of the circle = Radius of quadrant of circle
\n= r = \\(\\frac { 7 }{ 2 }\\) cm
\nArea of the quadrant of the circle = \\(\\frac { 1 }{ 4 }\\)\u03c0r\u00b2
\n\"NCERT
\nSince it is quadrant of circle, therfore the central angle of it \u03b8 = \u2220AOB = 90\u00b0<\/p>\n

(ii) Area of the shaded portion = Area of the sector for area of the quadrant of circle) – area of \u2206AOD
\n\"NCERT<\/p>\n

Question 13.
\nIn the figure, 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:
\nOABC is a square and OA is 20 cm. Join OB. Now we have a triangle QAB.
\nBy Pythagoras theorem
\n(OB)\u00b2 = (OA)\u00b2 + (AB)\u00b2
\n(OB)\u00b2 = (20)\u00b2 + (20)\u00b2
\n= 400 + 400
\n\"NCERT
\nArea of square
\nOABC = (20)\u00b2 = 400 cm\u00b2.
\n\"NCERT
\nArea of shaded region = Area of sector OPBQ – Area of square OABC
\n= 628 – 400
\n= 228 cm\u00b2<\/p>\n

\"NCERT<\/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). If \u2220AOB = 30\u00b0, find the area of the shaded region.
\nSolution:
\n\"NCERT<\/p>\n

Question 15.
\nIn the figure, 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.
\n\"NCERT
\nSolution:
\nLet us find BC By Pythagoras theorem BC\u00b2 = AB\u00b2 + AC\u00b2
\nBC\u00b2 = 14\u00b2 + 14\u00b2
\nBC = 14\\(\\sqrt{2}\\) cm
\nRequired Area = Area of semicircle BRC – [Area of quadrant – area of \u2206ABC)
\nRequired Area = Area BCQB – (Area BACQB – Area of \u2206ABC)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nCalculate the area of the designed region in the figure common between the two quadrants of the circles of the radius 8 cm each.
\n\"NCERT
\nSolution:
\n\"NCERT
\nArea of square = (8)\u00b2 = 64 cm\u00b2
\nArea of shaded region = Area of both sectors – Area of square
\n= \\(\\frac { 704 }{ 7 }\\) – 64
\n= \\(\\frac { 704-448 }{ 7 }\\) = \\(\\frac { 256 }{ 7 }\\) cm\u00b2<\/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 12.3 Question 1. Find the area of the shaded region in the given figure, if PQ = 24cm, …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3<\/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 12.3 - 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-12-3\/\" \/>\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 12.3 - 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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