{"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":"
Question 1.
\nSome figures are given below.
\n
\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<\/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
\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
\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
\nSolution:
\n(i) In quadrilateral three angles are 40\u00b0, 140\u00b0 and 100\u00b0
\n
\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
\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
\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
\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
\n(ii) In the given figure, find x + y + z + w.
\n
\nSolution:
\n(i) In \u2206ABC, Sides AB, BC, CA are produce in order in one side.
\n
\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
\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 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":"\n