{"id":44221,"date":"2022-05-31T03:30:33","date_gmt":"2022-05-30T22:00:33","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44221"},"modified":"2023-01-25T11:30:38","modified_gmt":"2023-01-25T06:00:38","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1<\/h2>\n

Question 1.
\nSome figures are given below.
\n\"ML
\nClassify each of them on the basis of the following:
\n(a) Simple curve
\n(b) Simple closed curve
\n(c) Polygon
\n(d) Convex polygon
\n(e) Concave polygon
\nSolution:
\n(a) (i), (ii), (iii), (v) and (vi) are simple curves.
\n(b) (iii), (v), (vi) are simple closed curves.
\n(c) (iii) and (vi) are polygons.
\n(d) (iii) is a convex polygon.
\n(e) (v) is a concave polygon.<\/p>\n

Question 2.
\nHow many diagonals does each of the following have?
\n(a) A convex quadrilateral
\n(b) A regular hexagon
\nSolution:
\n(a) A convex quadrilateral: It has two diagonals.
\n(b) A regular hexagon: It has 9 diagonals as shown.
\n\"ML<\/p>\n

Question 3.
\nFind the sum of measures of all interior angles of a polygon with number of sides:
\n(i) 8
\n(ii) 12
\nSolution:
\n(i) Sum of measures of all interior angles of
\n8-sided polygon = (2n – 4) \u00d7 90\u00b0
\n= (2 \u00d7 8 – 4) \u00d7 90\u00b0
\n= 12 \u00d7 90\u00b0 = 1080\u00b0
\n(ii) Sum of measures of all interior angles of
\n12-sided polygon = (2n – 4) \u00d7 90\u00b0
\n= (2 \u00d7 12 – 4) \u00d7 90\u00b0
\n= 18 \u00d7 90\u00b0= 1800\u00b0<\/p>\n

Question 4.
\nFind the number of sides of a regular polygon whose each exterior angles has a measure of
\n(i) 24\u00b0
\n(ii) 60\u00b0
\n(iii) 72\u00b0
\nSolution:
\n(i) Let number of sides of the polygon = n
\nEach exterior angle = 24\u00b0
\n\u2234 n = \\(\\frac{360^{\\circ}}{24^{\\circ}}\\) = 15 sides
\n\u2234 Polygon is of 15 sides.
\n(ii) Each interior angle of the polygon = 60\u00b0
\nLet number of sides of the polygon = n
\n\u2234 n = \\(\\frac{360^{\\circ}}{60^{\\circ}}\\) = 6
\n\u2234 Number of sides = 6
\n(iii) Each interior angle of the polygon = 72\u00b0
\nLet number of sides of the polygon = n
\n\u2234 n = \\(\\frac{360^{\\circ}}{72^{\\circ}}\\) = 5
\n\u2234 Number of sides = 5<\/p>\n

Question 5.
\nFind the number of sides of a regular polygon if each of its interior angles is
\n(i) 90\u00b0
\n(ii) 108\u00b0
\n(iii) 165\u00b0
\nSolution:
\n(i) Each interior angle = 90\u00b0
\nLet number of sides of the regular polgyon = n
\n\u2234 90\u00b0 = \\(\\frac{2 n-4}{n}\\) \u00d7 90\u00b0
\n\u21d2 \\(\\frac{2 n-4}{n}=\\frac{90^{\\circ}}{90^{\\circ}}\\) = 1
\n\u21d2 2n – 4 = n
\n\u21d2 2n – n = 4
\n\u21d2 n = 4
\n\u21d2 n = 4
\n\u2234 It is a square.
\n(ii) Each interior angle = 108\u00b0
\nLet number of sides of the regular polygon = n
\n\u2234 108\u00b0 = \\(\\frac{2 n-4}{n}\\) \u00d7 90\u00b0
\n\u21d2 \\(\\frac{2 n-4}{n}=\\frac{108^{\\circ}}{90^{\\circ}}=\\frac{6}{5}\\)
\n\u21d2 10n – 20 = 6n \u21d2 10n – 6n = 20
\n\u21d2 4n = 20
\n\u21d2 n = \\(\\frac{20}{4}\\) = 5
\n\u2234 It is a pentagon.
\n(iii) Each interior angle = 165\u00b0
\nLet number of sides of the regular polygon = n
\n\u2234 165\u00b0 = \\(\\frac{2 n-4}{n}\\) \u00d7 90\u00b0
\n\u21d2 \\(\\frac{2 n-4}{n}=\\frac{165^{\\circ}}{90^{\\circ}}=\\frac{11}{6}\\)
\n\u21d2 12n – 24 = 11n
\n\u21d2 12n – 11n = 24
\n\u21d2 n = 24
\n\u2234 It is 24-sided polygon.<\/p>\n

Question 6.
\nFind the number of sides in a polygon if the sum of its interior angles is:
\n(i) 1260\u00b0
\n(ii) 1980\u00b0
\n(iii) 3420\u00b0
\nSolution:
\nWe know that, sum of interior angles of polygon
\nis given by (2n – 4) at right angles.
\n(i) 1260\u00b0
\n\u2234 1260 = (2n – 4) \u00d7 90
\n\u21d2 \\(\\frac{1260}{90}\\) = 2n – 4
\n\u21d2 14 = 2n – 4
\n\u21d2 n = 9
\n(ii) 1980\u00b0
\n\u2234 1980 = (2n – 4) \u00d7 90
\n\u21d2 \\(\\frac{1980}{90}\\) = 2n – 4
\n\u21d2 22 = 2n – 4
\n\u21d2 n = 13.
\n(iii) 3420\u00b0
\n\u2234 3420 = (272 – 4) \u00d7 90 3420
\n\u21d2 \\(\\frac{3420}{90}\\) = 2n – 4
\n\u21d2 38 = 2n – 4
\n\u21d2 n = 21<\/p>\n

Question 7.
\nIf the angles of a pentagon are in the ratio 7 : 8 : 11 : 13 : 15, find the angles.
\nSolution:
\nRatio in the angles of a polygon = 7 : 8 : 11 : 13 : 15
\nSum of angles of a pentagon = (2n – 4) \u00d7 90\u00b0
\n= (2 \u00d7 5 – 4) \u00d7 90\u00b0
\n= 6 \u00d7 90\u00b0 = 540\u00b0
\nLet the angles of the pentagon be
\n7x, 8x, 11x, 13x, 15x
\n\u2234 7x + 8x + 11x + 13x + 15x = 540\u00b0
\n\u21d2 54x = 540\u00b0 \u21d2 x = \\(\\frac{540^{\\circ}}{54}\\) = 10\u00b0
\n\u2234 Angles are 7 \u00d7 10\u00b0 = 70\u00b0, 8 \u00d7 10\u00b0 = 80\u00b0,
\n11 \u00d7 10\u00b0 = 110\u00b0, 13 \u00d7 10\u00b0 = 130\u00b0 and 15 \u00d7 10\u00b0= 150\u00b0
\n\u2234 Angles are 70\u00b0, 80\u00b0, 110\u00b0, 130\u00b0 and 150\u00b0<\/p>\n

Question 8.
\nThe angles of a pentagon are x\u00b0, (x – 10)\u00b0, (x + 20)\u00b0, (2x – 44)\u00b0 and (2x – 70\u00b0) Calculate x.
\nSolution:
\nAngles of a pentaon are x\u00b0, (x – 10)\u00b0, (x + 20)\u00b0,
\n(2x – 44)\u00b0 and (2x – 70\u00b0)
\nBut sum of angles of a pentagon
\n= (2n – 4) \u00d7 90\u00b0
\n= (2 \u00d7 5 – 4) \u00d7 90\u00b0
\n= 6 \u00d7 90\u00b0 = 540\u00b0
\n\u2234 x + x – 10\u00b0 + x + 20\u00b0 + 2x – 44\u00b0 + 2x – 70\u00b0 = 540\u00b0
\n\u21d2 7x – 104\u00b0 = 540\u00b0
\n\u21d2 7x = 540\u00b0 + 104\u00b0 = 644\u00b0
\n\u21d2 x = \\(\\frac{644^{\\circ}}{7}\\) = 92\u00b0
\n\u2234 x = 92\u00b0<\/p>\n

Question 9.
\nThe exterior angles of a pentagon are in ratio 1 : 2 : 3 : 4 : 5. Find all the interior angles of the pentagon.
\nSolution:
\nLet the exterior angles of the pentagon are x, 2x, 3x, 4x and 5x.
\nWe know that sum of exterior angles of polygon is 360\u00b0.
\n\u2234 x + 2x + 3x + 4x + 5x = 360\u00b0
\n\u21d2 15x = 360\u00b0
\n\u21d2 x = \\(\\frac{360^{\\circ}}{15}\\)
\n\u21d2 x = 24\u00b0
\n\u2234 Exterior angles are 24\u00b0, 48\u00b0, 72\u00b0, 96\u00b0, 120\u00b0
\nInterior angles are 180\u00b0 – 24\u00b0, 180\u00b0 – 48\u00b0, 180\u00b0 – 72\u00b0, 180\u00b0 – 96\u00b0,
\n180\u00b0 – 120\u00b0 i.e. 156\u00b0, 132\u00b0, 108\u00b0, 84\u00b0, 60\u00b0.<\/p>\n

Question 10.
\nIn a quadrilateral ABCD, AB || DC. If \u2220A : \u2220D = 2:3 and \u2220B : \u2220C = \u22207 : 8, find the measure of each angle.
\nSolution:
\nAs AB || CD
\n\"ML
\n\u2220A + \u2220D = 180\u00b0 and \u2220B + \u2220C = 180\u00b0
\n\u21d2 2x + 3x = 180\u00b0 and 7y + 8y = 180\u00b0
\n5x = 180\u00b0 and 15y = 180\u00b0
\nx = 36\u00b0 and y = 12\u00b0
\n\u2234 \u2220A = 2 \u00d7 36 = 72\u00b0
\nand \u2220D = 3 \u00d7 36 = 108\u00b0
\n\u2220B = 7y = 7 \u00d7 12 = 84\u00b0
\nand \u2220C = 8y = 8 \u00d7 12 = 96\u00b0<\/p>\n

Question 11.
\nFrom the adjoining figure, find
\n(i) x
\n(ii) \u2220DAB
\n(iii) \u2220ADB
\n\"ML
\nSolution:
\n(i) ABCD is a quadrilateral
\n\u2234 \u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0
\n\u21d2 (3x + 4) + (50 + x) + (5x + 8) + (3x + 10) = 360
\n\u21d2 3x + 4 + 50 + x + 5x + 8 + 3x + 10 = 360\u00b0
\n\u21d2 12x + 72 = 360\u00b0
\n\u21d2 12x = 288
\n\u21d2 x = 24
\n(ii) \u2220DAB = (3x + 4) = 3 \u00d7 24 + 4 = 76\u00b0
\n(iii) \u2220ADB = 180\u00b0- (76\u00b0 + 50\u00b0) = 54\u00b0 (\u2235 ABD is a \u2206)<\/p>\n

Question 12.
\nFind the angle measure x in the following figures:
\n\"ML
\nSolution:
\n(i) In quadrilateral three angles are 40\u00b0, 140\u00b0 and 100\u00b0
\n\"ML
\nBut sum of Four angles = 360\u00b0
\n\u21d2 40\u00b0 + 140\u00b0+ 100\u00b0 + x = 360\u00b0
\n\u21d2 280\u00b0 + x = 360\u00b0
\n\u21d2 x = 360\u00b0 – 280\u00b0 = 80\u00b0
\n(ii) In the given figure, ABCDE is a pentagon.
\nWhere side AB is produced to both sides0
\n\"ML
\n\u22201 + 60\u00b0 = 180\u00b0 (Linear pair)
\n\u22201 = 180\u00b0- 60\u00b0= 120\u00b0
\nSimilarly \u22202 + 80\u00b0 = 180\u00b0
\n\u2234 \u22202 = 180\u00b0- 80\u00b0= 100\u00b0
\nNow, sum of angles of a pentagon = (2n – 4) \u00d7 90\u00b0
\n= (2 \u00d7 5 – 4) \u00d7 90\u00b0 = 6 \u00d7 90\u00b0 = 540\u00b0
\n\u2234 \u2220A + \u2220B + \u2220C + \u2220D + \u2220E = 540\u00b0
\n\u21d2 120\u00b0 + 100\u00b0 + x + 40\u00b0 + x = 540\u00b0
\n\u21d2 260\u00b0 + 2x = 540\u00b0
\n\u21d2 2x = 540\u00b0 – 260\u00b0 = 280\u00b0
\n\u21d2 x = \\(\\frac{280^{\\circ}}{2}\\) = 140\u00b0
\n(iii) In the given figure, ABCD is a quadrilateral
\nwhose side AB is produced is both sides \u2220A = 90\u00b0
\n\"ML
\nBut \u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0
\n(Sum of angles of a quadrilateral)
\n\u21d2 90\u00b0+ 60\u00b0+ 110\u00b0 + x = 360\u00b0
\n\u21d2 260\u00b0 + x = 360\u00b0
\n\u21d2 x = 360\u00b0 – 260\u00b0 = 100\u00b0
\n\u2234 x = 100\u00b0
\n(iv) In the given figure, ABCD is a quadrilateral
\nwhose side AB is produced to E.
\n\"ML
\n\u2220A = 90\u00b0, \u2220C = 83\u00b0, \u2220D = 110\u00b0
\n\u2220B + x = 180\u00b0 (Lienar pair)
\n\u2220B = 180\u00b0 – x
\nBut \u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0
\n\u21d2 90\u00b0 + (180\u00b0 – x) + 83\u00b0 + 110\u00b0 = 360\u00b0
\n(Sum of angles of a quadrilateral)
\n\u21d2 283\u00b0+ 180\u00b0 – x = 360\u00b0
\n\u21d2 x = 283\u00b0 + 180\u00b0- 360\u00b0
\n\u21d2 x = 463\u00b0 – 360\u00b0= 103\u00b0<\/p>\n

Question 13.
\n(i) In the given figure, find x + y + z.
\n\"ML
\n(ii) In the given figure, find x + y + z + w.
\n\"ML
\nSolution:
\n(i) In \u2206ABC, Sides AB, BC, CA are produce in order in one side.
\n\"ML
\n\u2220B = 70\u00b0, \u2220C = 90\u00b0
\n\u2234 \u2220A = 180\u00b0 – (\u2220B + \u2220C)
\n= 180\u00b0- (70\u00b0+ 90\u00b0)
\n= 180\u00b0- 160\u00b0 = 20\u00b0
\nBut x + 90\u00b0 = 180\u00b0 (Linear pair)
\n\u2234 x = 180\u00b0 – 90\u00b0 = 90\u00b0
\nSimilarly, y = 180\u00b0 – 70\u00b0 = 110\u00b0
\ny = 180\u00b0 – 20\u00b0= 160\u00b0
\nx + y + z = 90\u00b0 + 110\u00b0+ 160\u00b0 = 360\u00b0
\n(ii) In the given figure,
\nABCD is a quadrilateral whose sides are produced in order.
\n\"ML
\n\u2220A = 130\u00b0, \u2220B = 80\u00b0, \u2220C = 70\u00b0
\n\u2234 \u2220D = 360\u00b0 – (\u2220A + \u2220B + \u2220C)
\n= 360\u00b0 – (130\u00b0 + 80\u00b0 + 70\u00b0)
\n= 360\u00b0 – 280\u00b0 = 80\u00b0
\nNow, x + 130\u00b0 = 180\u00b0 (Linear pair)
\n\u2234 x = 180\u00b0- 130\u00b0 = 50\u00b0
\nSimilarly, y = 180\u00b0 – 80\u00b0 = 100\u00b0
\nz = 180\u00b0 – 70\u00b0= 110\u00b0
\nw = 180\u00b0-80\u00b0= 100\u00b0
\n\u2234 x + y + z + w = 52\u00b0 + 100\u00b0 +110\u00b0+ 100\u00b0
\n= 360\u00b0<\/p>\n

Question 14.
\nA heptagon has three equal angles each of 120\u00b0 and four equal angles. Find the size of equal angles.
\nSolution:
\nSum of angles of a heptagon = (2 \u00d7 n – 4) \u00d7 90\u00b0
\n= (2 \u00d7 7 – 4) \u00d7 90\u00b0
\n= 10 \u00d7 90\u00b0 = 900\u00b0
\nSum of three angles are each equal i.e. 120\u00b0
\n= 120\u00b0 \u00d7 3 = 360\u00b0
\nSum of remaining 4 equal angles
\n= 900\u00b0 – 360\u00b0 = 540\u00b0
\n\u2234 Each angle = \\(\\frac{540^{\\circ}}{4}\\) = 135\u00b0<\/p>\n

Question 15.
\nThe ratio between an exterior angle and the interior angle of a regular polygon is 1 : 5. Find
\n(i) the measure of each exterior angle
\n(ii) the measure of each interior angle
\n(iii) the number of sides in the polygon.
\nSolution:
\nRatio between an exterior and an interior angle = 1 : 5
\nLet exterior angle = x
\nThen interior angle = 5x
\nBut sum of interior angle and exterior angle = 180\u00b0
\n\u2234 x + 5x = 180\u00b0
\n\u21d2 6x = 180\u00b0
\n\u21d2 \\(\\frac{180^{\\circ}}{6}\\) = 30\u00b0
\n(i) Measure of exterior angle x = 30\u00b0
\n(ii) and measure of interior angle = 5x = 5 \u00d7 30\u00b0 = 150\u00b0
\n(iii) Let number of sides = n, then
\n\\(\\frac{2 n-4}{n}\\) \u00d7 90\u00b0 = 150\u00b0
\n\u21d2 \\(\\frac{2 n-4}{n}=\\frac{150^{\\circ}}{90}=\\frac{5}{3}\\)
\n\u21d2 6n – 12 = 5n
\n\u21d2 6n – 5n = 12
\n\u21d2 n = 12
\n\u2234 Number of sides = 12<\/p>\n

Question 16.
\nEach interior angle of a regular polygon is double of its exterior angle. Find the number of sides in the polygon.
\nSolution:
\nIn a polygon,
\nLet exterior angle = x
\nThen interior angle = 2x
\nBut sum of interior angle and exterior angle = 180\u00b0
\n\u2234 2x + x = 180\u00b0
\n\u21d2 3x = 180\u00b0
\n\u21d2 x = \\(\\frac{180^{\\circ}}{3}\\) = 60\u00b0
\n\u2234 Interior angle = 2 \u00d7 60\u00b0 = 120\u00b0
\nLet number of sides of the polygon = x
\nThen \\(\\frac{2 n-4}{n}\\) \u00d7 90\u00b0 = 120\u00b0
\n\u21d2 \\(\\frac{2 n-4}{n}=\\frac{120^{\\circ}}{90}=\\frac{4}{3}\\)
\n\u21d2 6n – 12 = 4n
\n\u21d2 6n – 4n = 12
\n\u21d2 2n = 12
\n\u21d2 n = \\(\\frac{12}{2}\\) = 6
\n\u2234 Number of sides = 6<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 Question 1. Some figures are given below. Classify each of them on the basis of the following: (a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon Solution: (a) (i), (ii), (iii), (v) and (vi) are … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 - CBSE Library<\/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:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 Question 1. Some figures are given below. Classify each of them on the basis of the following: (a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon Solution: (a) (i), (ii), (iii), (v) and (vi) are ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/\" \/>\n<meta property=\"og:site_name\" content=\"CBSE Library\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/aplustopper\/\" \/>\n<meta property=\"article:published_time\" content=\"2022-05-30T22:00:33+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-01-25T06:00:38+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/cbselibrary.com\/#organization\",\"name\":\"Aplus Topper\",\"url\":\"https:\/\/cbselibrary.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/aplustopper\/\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"contentUrl\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"width\":1585,\"height\":375,\"caption\":\"Aplus Topper\"},\"image\":{\"@id\":\"https:\/\/cbselibrary.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/cbselibrary.com\/#website\",\"url\":\"https:\/\/cbselibrary.com\/\",\"name\":\"CBSE Library\",\"description\":\"Improve your Grades\",\"publisher\":{\"@id\":\"https:\/\/cbselibrary.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/cbselibrary.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png\",\"contentUrl\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png\",\"width\":333,\"height\":232,\"caption\":\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 Q1.1\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#webpage\",\"url\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/\",\"name\":\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 - CBSE Library\",\"isPartOf\":{\"@id\":\"https:\/\/cbselibrary.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#primaryimage\"},\"datePublished\":\"2022-05-30T22:00:33+00:00\",\"dateModified\":\"2023-01-25T06:00:38+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/cbselibrary.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1\"}]},{\"@type\":\"Article\",\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#webpage\"},\"author\":{\"@id\":\"https:\/\/cbselibrary.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794\"},\"headline\":\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1\",\"datePublished\":\"2022-05-30T22:00:33+00:00\",\"dateModified\":\"2023-01-25T06:00:38+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#webpage\"},\"wordCount\":1264,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/cbselibrary.com\/#organization\"},\"image\":{\"@id\":\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png\",\"articleSection\":[\"ICSE\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#respond\"]}]},{\"@type\":\"Person\",\"@id\":\"https:\/\/cbselibrary.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794\",\"name\":\"Prasanna\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"caption\":\"Prasanna\"},\"url\":\"https:\/\/cbselibrary.com\/author\/prasanna\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 - CBSE Library","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:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/","og_locale":"en_US","og_type":"article","og_title":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1","og_description":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 Question 1. Some figures are given below. Classify each of them on the basis of the following: (a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon Solution: (a) (i), (ii), (iii), (v) and (vi) are ... Read more","og_url":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/","og_site_name":"CBSE Library","article_publisher":"https:\/\/www.facebook.com\/aplustopper\/","article_published_time":"2022-05-30T22:00:33+00:00","article_modified_time":"2023-01-25T06:00:38+00:00","og_image":[{"url":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png"}],"twitter_card":"summary","twitter_misc":{"Written by":"Prasanna","Est. reading time":"6 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/cbselibrary.com\/#organization","name":"Aplus Topper","url":"https:\/\/cbselibrary.com\/","sameAs":["https:\/\/www.facebook.com\/aplustopper\/"],"logo":{"@type":"ImageObject","@id":"https:\/\/cbselibrary.com\/#logo","inLanguage":"en-US","url":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","contentUrl":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","width":1585,"height":375,"caption":"Aplus Topper"},"image":{"@id":"https:\/\/cbselibrary.com\/#logo"}},{"@type":"WebSite","@id":"https:\/\/cbselibrary.com\/#website","url":"https:\/\/cbselibrary.com\/","name":"CBSE Library","description":"Improve your Grades","publisher":{"@id":"https:\/\/cbselibrary.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/cbselibrary.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#primaryimage","inLanguage":"en-US","url":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png","contentUrl":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png","width":333,"height":232,"caption":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 Q1.1"},{"@type":"WebPage","@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#webpage","url":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/","name":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 - CBSE Library","isPartOf":{"@id":"https:\/\/cbselibrary.com\/#website"},"primaryImageOfPage":{"@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#primaryimage"},"datePublished":"2022-05-30T22:00:33+00:00","dateModified":"2023-01-25T06:00:38+00:00","breadcrumb":{"@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/cbselibrary.com\/"},{"@type":"ListItem","position":2,"name":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1"}]},{"@type":"Article","@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#article","isPartOf":{"@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#webpage"},"author":{"@id":"https:\/\/cbselibrary.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794"},"headline":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1","datePublished":"2022-05-30T22:00:33+00:00","dateModified":"2023-01-25T06:00:38+00:00","mainEntityOfPage":{"@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#webpage"},"wordCount":1264,"commentCount":0,"publisher":{"@id":"https:\/\/cbselibrary.com\/#organization"},"image":{"@id":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#primaryimage"},"thumbnailUrl":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-13-Understanding-Quadrilaterals-Ex-13.1-Q1.1.png","articleSection":["ICSE"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-13-ex-13-1\/#respond"]}]},{"@type":"Person","@id":"https:\/\/cbselibrary.com\/#\/schema\/person\/2533e4338ba14fc0e4001efcca2f8794","name":"Prasanna","image":{"@type":"ImageObject","@id":"https:\/\/cbselibrary.com\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","caption":"Prasanna"},"url":"https:\/\/cbselibrary.com\/author\/prasanna\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/posts\/44221"}],"collection":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/comments?post=44221"}],"version-history":[{"count":1,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/posts\/44221\/revisions"}],"predecessor-version":[{"id":152483,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/posts\/44221\/revisions\/152483"}],"wp:attachment":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/media?parent=44221"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/categories?post=44221"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/tags?post=44221"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}