{"id":6090,"date":"2023-09-20T10:10:19","date_gmt":"2023-09-20T10:10:19","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6090"},"modified":"2023-09-19T16:54:34","modified_gmt":"2023-09-19T16:54:34","slug":"ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Chapter &#8211; 12 (Areas Related to Circles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 12 Areas Related to Circles <\/strong>Exercise 12.3 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-12-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 12.1<\/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-12-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 12.2<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Exercise &#8211; 12.3<\/strong><\/span><\/h2>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6642\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-13.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"178\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>We know that, the angle in a semicircle is a right angle.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u2220RPQ = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Thus, \u0394RQP is a\u00a0right-angled triangle.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Given, PQ = 24 cm,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PR = 7 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Using\u00a0Pythagoras theorem,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">RQ<sup>2<\/sup>\u00a0= PR<sup>2\u00a0<\/sup>+ PQ<sup>2<br \/>\n<\/sup>RQ = \u221a7\u00b2 + 24\u00b2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u221a49 + 576<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u221a625<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Thus, PQ = 25 cm which is the diameter<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Radius (r) = 25\/2 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of shaded region = Area of semicircle RPQ &#8211;\u00a0Area of \u0394RQP<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 \u00d7 \u03c0r<sup>2\u00a0<\/sup>&#8211; 1\/2 \u00d7 PQ \u00d7 RP<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 [(22\/7 \u00d7 25\/2 \u00d7 25\/2) &#8211; (24 \u00d7 7)]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 [6875\/14 &#8211; 168]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 [(6875 &#8211; 2352)\/14]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 \u00d7 4523\/14<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 4523\/28 cm<sup>2<br \/>\n<\/sup>= 161.54 cm<sup>2\u00a0<\/sup>(approximately)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Find the area of the shaded region in Fig. 12.20, if the radii of the two concentric circles with centre O are 7 cm and 14 cm, respectively and AOC = 40\u00b0.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6644\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"244\" height=\"193\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-2.png 315w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-2-300x237.png 300w\" sizes=\"auto, (max-width: 244px) 100vw, 244px\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Given,<br \/>\nAngle made by sector = 40\u00b0,<br \/>\nRadius the inner circle = r = 7 cm, and<br \/>\nRadius of the outer circle = R = 14 cm<br \/>\nArea of shaded region ABDC = Area of sector ACO &#8211; Area of sector BDO <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03b8\/360\u00b0 \u00d7\u00a0\u03c0R<sup>2<\/sup>\u00a0&#8211; \u03b8\/360\u00b0 \u00d7\u00a0\u03c0r<sup>2<br \/>\n<\/sup>= \u03b8\/360\u00b0 \u03c0 (R<sup>2<\/sup>\u00a0&#8211; r<sup>2<\/sup>) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03b8\/360\u00b0 \u03c0 (R + r )(R &#8211; r) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 40\u00b0\/360\u00b0 \u00d7 22\/7 \u00d7 (14 + 7) (14 &#8211; 7) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/9 \u00d7 22\/7 \u00d7 21 \u00d7 7 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (22 \u00d7 21 \u00d7 7)\/(9 \u00d7 7)\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (22 \u00d7 7)\/3<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 154\/3 cm<sup>2<br \/>\n<\/sup>= 51.33 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6645\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"171\" height=\"155\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>From figure, it is clear that the diameter of both the semicircles = Side of the square = 14 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Radius of each semicircle (r) = 14\/2 = 7 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Semicircles APD and BPC are drawn using sides AD and BC respectively as their\u00a0diameter. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Diameter of each semicircle = 14 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Radius of each semicircle (r) = 14\/2 = 7 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of shaded region = Area of square ABCD &#8211; (Area of semicircle APD +\u00a0Area of semicircle BPC) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (side)<sup>2<\/sup>\u00a0&#8211; (1\/2\u03c0r<sup>2<\/sup>\u00a0+ 1\/2\u03c0r<sup>2<\/sup>) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (14)<sup>2<\/sup>\u00a0&#8211; \u03c0 \u00d7 (7)<sup>2<br \/>\n<\/sup>= 196 cm<sup>2<\/sup> &#8211; 22\/7 \u00d7 7 cm \u00d7 7 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 196 cm<sup>2<\/sup>\u00a0 &#8211; 154 cm<sup>2<\/sup>\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 42\u00a0cm<sup>2<\/sup>\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6646\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"177\" height=\"210\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>It is given that OAB is an equilateral triangle having each angle as 60\u00b0<br \/>\nThe area of the sector is common in both.<br \/>\nRadius of circle (r) = 6 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Side of equilateral \u0394OAB, (s) = 12 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">We know each interior angle of\u00a0equilateral \u0394\u00a0= 60\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Area of shaded region = Area of circle + Area of \u0394OAB &#8211; Area of sector OCDE <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0r<sup>2<\/sup>\u00a0+ \u221a3\/4 (side)<sup>2\u00a0<\/sup>&#8211; \u03b8\/360\u00b0 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= \u03c0(6 cm)<sup>2\u00a0<\/sup>+ \u221a3\/4 (12)<sup>2<\/sup>\u00a0&#8211; 60\u00b0\/360\u00b0 \u00d7 \u03c0 (6 cm)<sup>2<br \/>\n<\/sup>= 36\u03c0 cm\u00b2 + 36\u221a3 cm\u00b2 &#8211; 6\u03c0 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (30\u03c0 + 36\u221a3) cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (30 \u00d7 22\/7 + 36\u221a3) cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (36\u221a3 + 660\/7) cm\u00b2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. From 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 Fig. 12.23. Find the area of the remaining portion of the square.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6648\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"167\" height=\"154\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Side of the square = 4 cm<br \/>\nThe diameter of the circle which is cut out = 2 cm<br \/>\n\u2234 Radius of this circle (r) = 1 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Radius of all quadrants cut out (r) = 1 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Since all quadrants cut out are of the same radius thus, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of portions cut out of square =\u00a0Area of the circle + 4 \u00d7 (Area of each quadrant) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0r<sup>2<\/sup>\u00a0+ 4 (90\u00b0\/360\u00b0 \u00d7 \u03c0r<sup>2<\/sup>) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0r<sup>2<\/sup>\u00a0+ 4 \u00d7 \u03c0r<sup>2<\/sup>\/4 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0r<sup>2<\/sup>\u00a0+ \u03c0r<sup>2<br \/>\n<\/sup>= 2\u03c0r\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2\u03c0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2 \u00d7 22\/7 cm<sup>2\u00a0<\/sup>=\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 44\/7 cm<sup>2<br \/>\n<\/sup>Area of the remaining portion of the square = Area of square &#8211; Area of portion cut out <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (4 cm)<sup>2<\/sup>\u00a0&#8211; 44\/7 cm<sup>2<br \/>\n<\/sup>= 16 cm<sup>2<\/sup>\u00a0&#8211; 44\/7 cm<sup>2<br \/>\n<\/sup>= (112 &#8211; 44)\/7 cm<sup>2<br \/>\n<\/sup>= 68\/7 cm<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. In a circular table cover of radius 32 cm, a design is formed, leaving an equilateral triangle ABC in the middle, as shown in Fig. 12.24. Find the area of the design.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6649\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"172\" height=\"151\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Mark O as center of the circle. Join BO and CO.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6650\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Ans-6i.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"162\" height=\"161\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Ans-6i.png 271w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Ans-6i-150x150.png 150w\" sizes=\"auto, (max-width: 162px) 100vw, 162px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Since we know that equal chords of a circle subtend equal angles at the center and all sides of an\u00a0equilateral triangle\u00a0are equal,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 Each side of triangle ABC will subtend equal angles at the center. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u2220BOC = 360\u00b0\/3 = 120\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6651\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Ans-6ii.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"159\" height=\"151\" \/> <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Consider \u0394BOC. Drop a perpendicular from OM to BC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">We know perpendicular from the center of circle to a chord bisects it. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 BM = MC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OB = OC (radii) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OM = OM (common) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394OBM \u2245 \u0394OCM (by\u00a0SSS congruency) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220BOM = \u2220COM (by CPCT) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 2\u2220BOM = \u2220BOC = 120\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220BOM = 120\u00b0\/2 = 60\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">sin 60\u00b0 = BM\/BO = \u221a3\/2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 BM = \u221a3\/2 \u00d7 BO = \u221a3\/2 \u00d7 32 = 16\u221a3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 BC = 2BM = 32\u221a3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Using the formula of area of equilateral triangle\u00a0= \u221a3\/4 (side)<sup>2<br \/>\n<\/sup>We can find the area of \u0394ABC since a side BC of \u0394ABC is known. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Visually from the figure, it\u2019s clear that <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the design =\u00a0Area of circle &#8211; Area of \u0394ABC = \u03c0r\u00b2 &#8211; \u221a3\/4 (BC)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">This can be solved with ease as both the radius of the circle and BC are known.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Radius of circle (r) = 32 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">From figure, we observe area of design = Area of circle &#8211; Area of \u0394ABC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0r\u00b2 &#8211; \u221a3\/4 (BC)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 22\/7 \u00d7 (32)\u00b2 &#8211; \u221a3\/4 \u00d7 (32\u221a3)\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 22\/7 \u00d7 1024 &#8211; \u221a3\/4 \u00d7 1024 \u00d7 3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 22528\/7 &#8211; 768\u221a3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of design = (22528\/7 &#8211; 768\u221a3) cm\u00b2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. In Fig. 12.25, 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.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6652\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"196\" height=\"189\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Side of square = 14 cm<br \/>\nFour quadrants are included in the four sides of the square.<br \/>\n\u2234 radius of the circles = 14\/2 cm = 7 cm<br \/>\nArea of the square ABCD = 14<sup>2\u00a0<\/sup>= 196 cm<sup>2<br \/>\n<\/sup>Area of the quadrant = (\u03c0R<sup>2<\/sup>)\/4 cm<sup>2<\/sup>\u00a0= (22\/7) \u00d77<sup>2<\/sup>\/4 cm<sup>2<br \/>\n<\/sup>= 77\/2 cm<sup>2<br \/>\n<\/sup>Total area of the quadrant = 4 \u00d7 77\/2 cm<sup>2\u00a0<\/sup>= 154cm<sup>2<br \/>\n<\/sup>Area of the shaded region = Area of the square ABCD \u2013 Area of the quadrant<br \/>\n= 196 cm<sup>2\u00a0<\/sup>\u2013 154 cm<sup>2<br \/>\n<\/sup>= 42 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-6653 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"285\" height=\"90\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-8.png 611w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-8-300x95.png 300w\" sizes=\"auto, (max-width: 285px) 100vw, 285px\" \/> <\/strong><strong>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<br \/>\n<\/strong><strong>(i)<\/strong> the distance around the track along its inner edge<br \/>\n<strong>(ii)<\/strong> the area of the track.<br \/>\n<strong>Solution &#8211;<br \/>\n<\/strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6654\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Ans-8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"330\" height=\"193\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Ans-8.png 609w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Ans-8-300x175.png 300w\" sizes=\"auto, (max-width: 330px) 100vw, 330px\" \/><br \/>\n<strong>Given &#8211;<\/strong><br \/>\nWidth of track = 10 m<br \/>\nRadii of the inner semicircles HIJ and KLG (r\u2081) = 60\/2 = 30 m<br \/>\nRadii of the outer semicircles BCD and EFA = 30 m + 10 m = 40 m <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">JK = GH = 106 m <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">DJ = HB = 10 m <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) The distance around the track along its inner edge.<br \/>\n<\/strong>= GH + arc HIJ + JK + arc KLG <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 106 + 2\u03c0r\u2081\/2 + 106 + 2\u03c0r\u2081\/2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 106 + \u03c0 \u00d7 30 + 106 + \u03c0 \u00d7 30 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 212 + 1320\/7 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= (1484 + 1320)\/7 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2804\/7 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)\u00a0Radius of semicircle BCD = Radius of semicircle EFA<br \/>\n<\/strong>(r\u2082) = 30 m + 10 = 40 m <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the track = Area of rectangle ABHG + Area of rectange KJDE + (Area of semicircle BCD &#8211; Area of semicircle HIJ) + (Area of semicircle EFA &#8211; Area of semicircle KLG) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= {(106 \u00d7 10) + (106 \u00d7 10) + [1\/2\u03c0 (40)\u00b2 &#8211; 1\/2\u03c0 (30)\u00b2]<sup>\u00a0<\/sup>+ [1\/2\u03c0 (40)\u00b2 &#8211; 1\/2\u03c0 (30)\u00b2]} <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= {1060<sup>\u00a0<\/sup>+ 1060 + [1\/2\u03c0 (1600 &#8211; 900)] + [1\/2\u03c0 (1600 &#8211; 900)] } <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1060 + 1060 + \u03c0\/2 \u00d7 700 + \u03c0\/2 \u00d7 700 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2120 + 700\u03c0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2120 + 700 \u00d7 22\/7<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2120<sup>\u00a0<\/sup>+ 2200\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 4320 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>9. In Fig. 12.27, 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.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6655\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"150\" height=\"149\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-9.png 282w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-9-150x150.png 150w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>AB and CD are diameters of the\u00a0<a style=\"color: #000000;\" href=\"https:\/\/www.cuemath.com\/geometry\/circles\/\">circle<\/a>\u00a0with center O<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 OD = OC = OA = OB =\u00a0<a style=\"color: #000000;\" href=\"https:\/\/www.cuemath.com\/geometry\/radius\/\">Radius\u00a0<\/a>of the circle R = 7 cm\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 AB = 2R = 14 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Radius of shaded circular region, r = OD\/2 = 7\/2 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of the shaded smaller circular region = \u03c0r\u00b2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u03c0 (7\/2 cm)<sup>2<br \/>\n<\/sup>= 22\/7 \u00d7 7\/2 \u00d7 7\/2 cm<sup>2<br \/>\n<\/sup>= 77\/2 cm<sup>2<br \/>\n<\/sup>= 38.5 cm<sup>2<br \/>\n<\/sup>Area of the shaded segment of larger circular region = Area of semicircle ACB &#8211; Area of \u0394ABC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 \u03c0(OA)<sup>2<\/sup> &#8211; 1\/2 \u00d7 AB \u00d7 OC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 \u03c0R<sup>2<\/sup> &#8211; 1\/2 \u00d7 2R \u00d7 R <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 \u00d7 22\/7 \u00d7 7<sup>2<\/sup> &#8211; 1\/2 \u00d7 14\u00a0\u00d7 7 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 77 &#8211; 49\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 28 cm<sup>2<br \/>\n<\/sup>Area of the shaded region = Area of the shaded smaller circular region + Area of the shaded segment of larger circular region <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 38.5 cm<sup>2<\/sup>\u00a0+ 28 cm<sup>2<br \/>\n<\/sup>= 66.5 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>10. The area of an equilateral triangle ABC is 17320.5 cm<sup>2<\/sup>. With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region (Use \u03c0 = 3.14 and \u221a3 = 1.73205).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6656\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"166\" height=\"149\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-10.png 312w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-10-300x268.png 300w\" sizes=\"auto, (max-width: 166px) 100vw, 166px\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>ABC is an equilateral triangle.<br \/>\n\u2234 \u2220 A = \u2220 B = \u2220 C = 60\u00b0<br \/>\nThere are three sectors, each making 60\u00b0.<br \/>\nArea of \u0394ABC = 17320.5 cm<sup>2<br \/>\n<\/sup>\u21d2 \u221a3\/4 \u00d7 (side)<sup>2<\/sup>\u00a0= 17320.5<br \/>\n\u21d2 (side)<sup>2<\/sup> =17320.5 \u00d7 4\/1.73205<br \/>\n\u21d2 (side)<sup>2<\/sup> = 4 \u00d7 10<sup>4<br \/>\n<\/sup>\u21d2 side = 200 cm<br \/>\nRadius of the circles = 200\/2 cm = 100 cm<br \/>\nArea of the sector = (60\u00b0\/360\u00b0)\u00d7\u03c0 r<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 1\/6 \u00d7 3.14 \u00d7 (100)<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 15700\/3cm<sup>2<br \/>\n<\/sup>Area of 3 sectors = 3\u00d715700\/3 = 15700 cm<sup>2<br \/>\n<\/sup>Thus, the area of the shaded region = Area of an equilateral triangle ABC \u2013 Area of 3 sectors<br \/>\n= 17320.5-15700 cm<sup>2\u00a0<\/sup>= 1620.5 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>11. On a square handkerchief, nine circular designs, each of a radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6657\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-11.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"179\" height=\"150\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-11.png 327w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-11-300x250.png 300w\" sizes=\"auto, (max-width: 179px) 100vw, 179px\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Number of circular designs = 9<br \/>\nThe radius of the circular design = 7 cm<br \/>\nThere are three circles on one side of the square handkerchief.<br \/>\n\u2234 side of the square = 3 \u00d7 diameter of circle<br \/>\n= 3 \u00d7 14<br \/>\n= 42 cm<br \/>\nArea of the square = 42 \u00d7 42 cm<sup>2<\/sup><br \/>\n= 1764 cm<sup>2<br \/>\n<\/sup>Area of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= (22\/7) \u00d7 7 \u00d7 7<br \/>\n= 154 cm<sup>2<br \/>\n<\/sup>Total area of the design = 9 \u00d7 154<br \/>\n= 1386 cm<sup>2<br \/>\n<\/sup>Area of the remaining portion of the handkerchief = Area of the square \u2013 Total area of the design<br \/>\n= 1764 \u2013 1386<br \/>\n= 378 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and a radius 3.5 cm. If OD = 2 cm, find the area of the<br \/>\n<\/strong><strong>(i)<\/strong> quadrant OACB<br \/>\n<strong>(ii)<\/strong> shaded region<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6658\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-12.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"152\" height=\"131\" \/><br \/>\n<strong>Solution &#8211; <\/strong>Radius of the quadrant = 3.5 cm = 7\/2 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i)<\/strong>\u00a0Area of the quadrant OACB = (\u03c0R<sup>2<\/sup>)\/4 cm<sup>2<br \/>\n<\/sup>= (22\/7) \u00d7 (7\/2) \u00d7 (7\/2)\/4 cm<sup>2<br \/>\n<\/sup>= 77\/8 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> Area of the triangle BOD = (\u00bd) \u00d7 (7\/2) \u00d7 2 cm<sup>2<br \/>\n<\/sup>= 7\/2 cm<sup>2<br \/>\n<\/sup>Area of the shaded region = Area of the quadrant \u2013 Area of the triangle BOD<br \/>\n= (77\/8) &#8211; (7\/2) cm<sup>2<br \/>\n<\/sup>= 49\/8 cm<sup>2<br \/>\n<\/sup>= 6.125 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use \u03c0 = 3.14)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6660\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-13.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"153\" height=\"148\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-13.png 302w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-13-300x291.png 300w\" sizes=\"auto, (max-width: 153px) 100vw, 153px\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Side of square = OA = AB = 20 cm<br \/>\nThe radius of the quadrant = OB<br \/>\nOAB is the right-angled triangle<br \/>\nBy Pythagoras\u2019 theorem in \u0394OAB,<br \/>\nOB<sup>2\u00a0<\/sup>= AB<sup>2<\/sup>+OA<sup>2<br \/>\n<\/sup>\u21d2 OB<sup>2\u00a0<\/sup>= 20<sup>2\u00a0<\/sup>+20<sup>2<br \/>\n<\/sup>\u21d2 OB<sup>2\u00a0<\/sup>= 400 + 400<br \/>\n\u21d2 OB<sup>2\u00a0<\/sup>= 800<br \/>\n\u21d2 OB = 20\u221a2 cm<br \/>\nArea of the quadrant = (\u03c0R<sup>2<\/sup>)\/4 cm<sup>2<br \/>\n<\/sup>= (3.14\/4)\u00d7(20\u221a2)<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 628cm<sup>2<br \/>\n<\/sup>Area of the square = 20 \u00d7 20 = 400 cm<sup>2<br \/>\n<\/sup>Area of the shaded region = Area of the quadrant \u2013 Area of the square<br \/>\n= 628-400 cm<sup>2\u00a0<\/sup>= 228cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>14. AB and CD are, respectively, arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If \u2220AOB = 30\u00b0, find the area of the shaded region.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6661\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-14.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"136\" height=\"158\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-14.png 268w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-14-258x300.png 258w\" sizes=\"auto, (max-width: 136px) 100vw, 136px\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>The radius of the larger circle, R = 21 cm<br \/>\nThe radius of the smaller circle, r = 7 cm<br \/>\nAngle made by sectors of both concentric circles = 30\u00b0<br \/>\nArea of the larger sector = (30\u00b0\/360\u00b0) \u00d7 \u03c0R<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= (1\/12) \u00d7 (22\/7) \u00d7 21<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 231\/2cm<sup>2<br \/>\n<\/sup>Area of the smaller circle = (30\u00b0\/360\u00b0)\u00d7\u03c0r<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 1\/12 \u00d7 22\/7 \u00d7 7<sup>2\u00a0<\/sup>cm<sup>2<br \/>\n<\/sup>= 77\/6 cm<sup>2<br \/>\n<\/sup>Area of the shaded region = (231\/2) \u2013 (77\/6) cm<sup>2<br \/>\n<\/sup>= 616\/6 cm<sup>2<\/sup><br \/>\n= 308\/3cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm, and a semicircle is drawn with BC as a diameter. Find the area of the shaded region<\/strong>.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6662\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-15.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"167\" height=\"136\" \/><br \/>\n<strong>Solution &#8211; <\/strong>The radius of the quadrant ABC of the circle = 14 cm<br \/>\nAB = AC = 14 cm<br \/>\nBC is the diameter of the semicircle.<br \/>\nABC is the right-angled triangle.<br \/>\nBy Pythagoras\u2019 theorem in \u0394ABC,<br \/>\nBC<sup>2\u00a0<\/sup>= AB<sup>2\u00a0<\/sup>+AC<sup>2<br \/>\n<\/sup>\u21d2 BC<sup>2\u00a0<\/sup>= 14<sup>2\u00a0<\/sup>+14<sup>2<br \/>\n<\/sup>\u21d2 BC = 14\u221a2 cm<br \/>\nRadius of the semicircle = 14\u221a2\/2 cm = 7\u221a2 cm<br \/>\nArea of the \u0394ABC =( \u00bd) \u00d7 14 \u00d7 14 = 98 cm<sup>2<br \/>\n<\/sup>Area of the quadrant = (\u00bc) \u00d7 (22\/7) \u00d7 (14 \u00d7 14) = 154 cm<sup>2<br \/>\n<\/sup>Area of the semicircle = (\u00bd) \u00d7 (22\/7) \u00d7 7\u221a2 \u00d7 7\u221a2 = 154 cm<sup>2<br \/>\n<\/sup>Area of the shaded region =Area of the semicircle + Area of the \u0394ABC \u2013 Area of the quadrant<br \/>\n= 154 + 98 &#8211; 154 cm<sup>2<br \/>\n<\/sup>= 98cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6663\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-16.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"195\" height=\"161\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-16.png 316w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Quetion-16-300x248.png 300w\" sizes=\"auto, (max-width: 195px) 100vw, 195px\" \/><br \/>\n<\/strong>AB = BC = CD = AD = 8 cm<br \/>\nArea of \u0394ABC = Area of \u0394ADC = (\u00bd) \u00d7 8 \u00d7 8 = 32 cm<sup>2<br \/>\n<\/sup>Area of quadrant AECB = Area of quadrant AFCD = (\u00bc) \u00d7 22\/7 \u00d7 8<sup>2<br \/>\n<\/sup>= 352\/7 cm<sup>2<br \/>\n<\/sup>Area of shaded region = (Area of quadrant AECB \u2013 Area of \u0394ABC)<br \/>\n= (Area of quadrant AFCD \u2013 Area of \u0394ADC)<br \/>\n= (352\/7 &#8211; 32) + (352\/7 &#8211; 32) cm<sup>2<br \/>\n<\/sup>= 2 \u00d7 (352\/7 &#8211; 32) cm<sup>2<br \/>\n<\/sup>= 256\/7 cm<sup>2<\/sup><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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 12.3 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":[1045,1105,1108,1044,1049,1048],"class_list":["post-6090","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-12-3-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) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 10 Maths\u00a0Chapter - 12 (Areas Related to Circles)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 12 Areas Related to Circles Exercise 12.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 12.3\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-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 Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions Class 10 Maths\u00a0Chapter - 12 (Areas Related to Circles)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 12 Areas Related to Circles Exercise 12.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 12.3\" \/>\n<meta property=\"og:url\" content=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/\" \/>\n<meta property=\"og:site_name\" content=\"TheExamPillar NCERT\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-20T10:10:19+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-13.png\" \/>\n<meta name=\"author\" content=\"Admin\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@exampillar\" \/>\n<meta name=\"twitter:site\" content=\"@exampillar\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Admin\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"16 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/\"},\"author\":{\"name\":\"Admin\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"headline\":\"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3\",\"datePublished\":\"2023-09-20T10:10:19+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/\"},\"wordCount\":1805,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex12.3-Que-13.png\",\"keywords\":[\"Class 10 NCERT Mathematics Solutions\",\"NCERT Class 10 Mathematics\u00a0Chapter 12 Areas Related to Circles Solutions\",\"NCERT Class 10 Mathematics Exercise 12.3 Solutions\",\"NCERT Class 10 Mathematics Solutions Class 10 NCERT Solutions\",\"NCERT Solutions Class 10 Mathematics\",\"NCERT Solutions Class 10 Maths\"],\"articleSection\":[\"Class 10 Maths\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/\",\"name\":\"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 | TheExamPillar NCERT\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#primaryimage\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex12.3-Que-13.png\",\"datePublished\":\"2023-09-20T10:10:19+00:00\",\"description\":\"NCERT Solutions Class 10 Maths\u00a0Chapter - 12 (Areas Related to Circles)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 12 Areas Related to Circles Exercise 12.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 12.3\",\"breadcrumb\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#primaryimage\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex12.3-Que-13.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/09\\\/NCERT-Class-10-Maths-Ex12.3-Que-13.png\",\"width\":214,\"height\":178,\"caption\":\"NCERT Class 10 Maths Solution\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\\\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\",\"name\":\"TheExamPillar NCERT\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"alternateName\":\"NCERT Solution\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":[\"Person\",\"Organization\"],\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\",\"name\":\"Admin\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"width\":512,\"height\":512,\"caption\":\"Admin\"},\"logo\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\"},\"sameAs\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\"],\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/author\\\/ncert_eng_vikram\\\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 | TheExamPillar NCERT","description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 12 (Areas Related to Circles)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 12 Areas Related to Circles Exercise 12.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 12.3","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3","og_description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 12 (Areas Related to Circles)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 12 Areas Related to Circles Exercise 12.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 12.3","og_url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/","og_site_name":"TheExamPillar NCERT","article_published_time":"2023-09-20T10:10:19+00:00","og_image":[{"url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-13.png","type":"","width":"","height":""}],"author":"Admin","twitter_card":"summary_large_image","twitter_creator":"@exampillar","twitter_site":"@exampillar","twitter_misc":{"Written by":"Admin","Est. reading time":"16 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#article","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/"},"author":{"name":"Admin","@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"headline":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3","datePublished":"2023-09-20T10:10:19+00:00","mainEntityOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/"},"wordCount":1805,"commentCount":0,"publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#primaryimage"},"thumbnailUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-13.png","keywords":["Class 10 NCERT Mathematics Solutions","NCERT Class 10 Mathematics\u00a0Chapter 12 Areas Related to Circles Solutions","NCERT Class 10 Mathematics Exercise 12.3 Solutions","NCERT Class 10 Mathematics Solutions Class 10 NCERT Solutions","NCERT Solutions Class 10 Mathematics","NCERT Solutions Class 10 Maths"],"articleSection":["Class 10 Maths"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/","url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/","name":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 | TheExamPillar NCERT","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/#website"},"primaryImageOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#primaryimage"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#primaryimage"},"thumbnailUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-13.png","datePublished":"2023-09-20T10:10:19+00:00","description":"NCERT Solutions Class 10 Maths\u00a0Chapter - 12 (Areas Related to Circles)\u00a0The NCERT Solutions in English Language for Class 10 Mathematics Chapter - 12 Areas Related to Circles Exercise 12.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 12.3","breadcrumb":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#primaryimage","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-13.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex12.3-Que-13.png","width":214,"height":178,"caption":"NCERT Class 10 Maths Solution"},{"@type":"BreadcrumbList","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-12-areas-related-to-circles-ex-12-3\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/theexampillar.com\/ncert\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3"}]},{"@type":"WebSite","@id":"https:\/\/theexampillar.com\/ncert\/#website","url":"https:\/\/theexampillar.com\/ncert\/","name":"TheExamPillar NCERT","description":"","publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"alternateName":"NCERT Solution","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/theexampillar.com\/ncert\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":["Person","Organization"],"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1","name":"Admin","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","width":512,"height":512,"caption":"Admin"},"logo":{"@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png"},"sameAs":["https:\/\/theexampillar.com\/ncert"],"url":"https:\/\/theexampillar.com\/ncert\/author\/ncert_eng_vikram\/"}]}},"_links":{"self":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6090","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/comments?post=6090"}],"version-history":[{"count":6,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6090\/revisions"}],"predecessor-version":[{"id":6695,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/6090\/revisions\/6695"}],"wp:attachment":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/media?parent=6090"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/categories?post=6090"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/tags?post=6090"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}