NCERT Solutions for Class 8 Maths<\/a> Chapter 3 Understanding Quadrilaterals Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3<\/h2>\n Question 1. \nGiven a parallelogram ABCD. Complete each statement along with the definition or property used. \n \n(i) AD = __________ \n(ii) \u2220DCB = __________ \n(iii) OC = __________ \n(iv) m\u2220DAB + m\u2220CDA = __________ \nSolution: \n(i) AD = BC (opposite sides are equal) \n(ii) \u2220DCB = \u2220DAB (opposite angles are equal) \n(iii) OC = OA (Diagonals bisect each other) \n(iv) m\u2220DAB + m\u2220CDA = 180\u00b0 (Adjacent angles are supplementary)<\/p>\n
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Question 2. \nConsider the following parallelograms. Find the values of the unknown x, y, z. \n <\/p>\n
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\nSolution: \n(i) \u2220y = 100\u00b0 (opposite angles of a parallelogram are equal) \n \n\u2220x + 100\u00b0 = 180\u00b0 (adjacent angles in a parallelogram) \n\u2220x = 180\u00b0 – 100\u00b0 = 80\u00b0 \n\u2220z = \u2220x = 80\u00b0 (opposite angles of parallelogram are equal) \n\u2234 \u2220x = 80\u00b0, \u2220y = 100\u00b0, \u2220z = 80\u00b0<\/p>\n
(ii) Opposite angles are equal. \n \n\u22201 = 50\u00b0 (opposite angles are equal) \n\u22201 + \u2220z = 180\u00b0 (Linear pair) \n50 + \u2220z = 180 \n\u2220z = 180\u00b0 – 50\u00b0= 130\u00b0 \nx + \u22201 + y + 50\u00b0 = 360\u00b0 \n\u21d2 x + 50\u00b0 + y + 50\u00b0 = 360\u00b0 \n\u21d2 x + y + 100\u00b0 = 360\u00b0 \n\u21d2 x + y = 360\u00b0 – 100\u00b0 \n\u21d2 x + y = 260\u00b0 \n\u2234 x = y (opposite angles of a parallelogram) \nx = \\(\\frac{260^{\\circ}}{2}\\) = 130\u00b0 \nThus x = 130\u00b0, y = 130\u00b0 and z = 130\u00b0.<\/p>\n
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(iii) x = 90\u00b0 (vertically opposite angles are equal) \n \nx + y + 30\u00b0 = 180\u00b0 \n(sum of the angles of a triangle = 180\u00b0) \n90\u00b0 + y + 30\u00b0 = 180\u00b0 \ny + 120\u00b0 = 180\u00b0 \n\u2220y = 180\u00b0 – 120\u00b0 = 60\u00b0 \nIn the parallelogram ABCD \nAD || BC and BD is a transversal. \n\u2234 y = z (alternate angles are equal) \nz = y = 60\u00b0 \nThus x = 90\u00b0, y = 60\u00b0 and z = 60\u00b0<\/p>\n
(iv) In the parallelogram ABCD \ny = 80\u00b0 (opposite angles are equal) \nAD || BC and CD is a transversal equal) \n \n\u2234 z = 80\u00b0 (corresponding angles are equal) \nx + 80\u00b0 = 180\u00b0 (sum of the adjacent angle is 180\u00b0) \nx = 180\u00b0 – 80\u00b0 = 100\u00b0 \nThus, x = 100\u00b0, y = 80\u00b0 and z = 80\u00b0<\/p>\n
(v) y = 112\u00b0 (in a parallelogram, opposite angles are equal) \n \nIn \u0394ACD, \nx + y + 40\u00b0 = 180\u00b0 (sum of the angles of a triangle is 180\u00b0) \n\u21d2 x + 112\u00b0 + 40\u00b0 = 180\u00b0 \n\u21d2 x + 152\u00b0= 180\u00b0 \n\u21d2 x = 180\u00b0 – 152\u00b0 = 28\u00b0 \n\u2220C + \u2220B = 180\u00b0 (sum of the adjacent angles of a parallelogram is 180\u00b0) \n\u21d2 40\u00b0 + z + 112\u00b0 = 180\u00b0 \n\u21d2 z + 152\u00b0 = 180\u00b0 \n\u21d2 z = 180\u00b0 – 152\u00b0 = 28\u00b0 \nThus, x = 28\u00b0, y = 112\u00b0 and z = 28\u00b0.<\/p>\n
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Question 3. \nCan a quadrilateral ABCD be a parallelogram if \n(i) \u2220D + \u2220B = 180\u00b0? \n(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm? \n(iii) \u2220A = 70\u00b0 and \u2220C = 65\u00b0? \nSolution: \n(i) In a quadrilateral ABCD \n\u2220D + \u2220B = 180\u00b0 can be, but need not be \n\u2234 The quadrilateral may be a parallelogram but not always<\/p>\n
(ii) In a quadrilateral ABCD \nAB = DC = 8 cm \nAD = 4 cm \nBC = 4.4 cm \n\u2234 Opposite sides AD and BC are not equal. \n\u2234 It cannot be a parallelogram.<\/p>\n
(iii) In a quadrilateral ABCD \n\u2220A = 70\u00b0 and \u2220C = 65\u00b0 \n\u2235 Opposite angles \u2220A \u2260 \u2220C \n\u2234 It cannot be a parallelogram.<\/p>\n
Question 4. \nDraw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure. \nSolution: \nIn the adjoining figure, ABCD is not a parallelogram such that opposite angles \u2220B and \u2220D are equal. It is a kite. \n <\/p>\n
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Question 5. \nThe measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram. \nSolution: \nLet ABCD be a parallelogram in which adjacent angles \u2220A and \u2220B are 3x and 2x respectively. \nSince adjacent angles are supplementary. \n\u2234 \u2220A + \u2220B = 180\u00b0 \n\u21d2 3x + 2x = 180\u00b0 \n\u21d2 5x = 180\u00b0 \n\u21d2 x = \\(\\frac{180^{\\circ}}{5}\\) = 36\u00b0 \n\u2220A = 3 \u00d7 36\u00b0 = 108\u00b0 \nand \u2220B = 2 \u00d7 36\u00b0 = 72\u00b0 \nSince opposite angles are equal. \n\u2234 \u2220D = \u2220B = 72\u00b0 and \u2220C = \u2220A = 108\u00b0 \n\u2234 \u2220A = 108\u00b0, \u2220B = 72\u00b0, \u2220C = 108\u00b0 and \u2220D = 72\u00b0<\/p>\n
Question 6. \nTwo adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram. \nSolution: \nLet ABCD be a parallelogram such that adjacent angles \u2220A = \u2220B \nSince \u2220A + \u2220B = 180\u00b0 \n\u2220A = \u2220B = \\(\\frac{180^{\\circ}}{2}\\) = 90\u00b0 \nSince opposite angles of a parallelogram are equal \n\u2234 \u2220A = \u2220C = 90\u00b0 and \u2220B = \u2220D = 90\u00b0 \nThus, \u2220A = 90\u00b0, \u2220B = 90\u00b0, \u2220C = 90\u00b0 and \u2220D = 90\u00b0<\/p>\n
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Question 7. \nThe adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them. \n \nSolution: \ny + z = 70\u00b0 \n(An exterior angles of a triangle is equal to the sum of the opposite interior angles) \n\u2235 In \u2206HOP \n\u2220HOP = 180\u00b0 – 70\u00b0 = 110\u00b0 (sum of the adjacent angle is 180\u00b0) \nNow, x = \u2220HOP = 110\u00b0 (opposite angles of a parallelogram are equal) \nEH || OP and PH is a transversal \n\u2234 y = 40\u00b0 (alternate angles are equal) \nFrom (1), \ny + z = 70\u00b0 \n\u21d2 40\u00b0 + z = 70\u00b0 \n\u21d2 z = 70\u00b0 – 40\u00b0 = 30\u00b0 \nThus x = 110\u00b0, y = 40\u00b0 and z = 30\u00b0<\/p>\n
Question 8. \nThe following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm) \n \nSolution: \n(i) GUNS is a parallelogram. \nGS = NU (opposite sides are equal) \n3x = 18 \n\u21d2 x = \\(\\frac{18}{3}\\) = 6 \nIn the parallelogram GUNS \nGU = SN (opposite sides are equal) \n\u21d2 3y – 1 = 26 \n\u21d2 3y = 26 + 1 \n\u21d2 3y = 27 \n\u21d2 y = \\(\\frac{27}{3}\\) = 9 \nThus, x = 6 cm and y = 9 cm.<\/p>\n
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(ii) RUNS is a parallelogram and the di-agonal RN and US bisect each other. \n\u2234 y + 7 = 20 \n\u21d2 y = 20 – 7 = 13 \nx + y = 16 \n\u21d2 x + 13 = 16 \n\u21d2 x = 16 – 13 = 3 \nThus, x = 3 cm and y = 13 cm<\/p>\n
Question 9. \nIn the figure given below both RISK and CLUE are parallelograms. Find the value of x. \n \nSolution: \nRISK is a parallelogram \n\u2220R + \u2220K = 180\u00b0 (adjacent angles of a parallelogram are supplementary) \n\u2234 \u2220R + 120\u00b0 = 180\u00b0 \n\u2220R = 180\u00b0 – 120\u00b0 = 60\u00b0 \nIn the parallelogram RISK \n\u2220R = \u2220S (opposite angles are equal) \n\u2220S = 60\u00b0 \nCLUE is also a parallelogram. \n\u2220E = \u2220L = 70\u00b0 (opposite angle of a parallelogram) \n\u2220E = 70\u00b0 \nNow, in triangle ESO \n\u2220E + \u2220S + x = 180\u00b0 \n\u21d2 70\u00b0 + 60\u00b0 + x = 180\u00b0 \n\u21d2 130\u00b0 + x = 180\u00b0 \n\u21d2 x = 180\u00b0 – 130\u00b0 \n\u21d2 x = 50\u00b0<\/p>\n
Question 10. \nExplain how this figure is a trapezium. Which of its two sides are parallel? \n \nSolution: \n\u2220KLM + \u2220NML = 80\u00b0 + 100\u00b0 = 180\u00b0 \n\u2234 KL || NM (The sum of consecutive interior angles is 180\u00b0) \n\u2234 The given figure KLMN is a trapezium.<\/p>\n
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Question 11. \nFind m\u2220C in the figure given below if AB || DC. \n \nSolution: \nAB || DC and BC is a transversal, \n\u2235 m\u2220B + m\u2220C = 180\u00b0 \nsum of interior angles is 180\u00b0 \nm\u2220C = 180\u00b0 – m\u2220B \n\u2234 m\u2220C = 180\u00b0 – 120\u00b0 = 60\u00b0<\/p>\n
Question 12. \nFind the measure of \u2220P and \u2220S if SP || QR in figure (if you find m\u2220R, is there more than one method to find m\u2220P) \n \nSolution: \nPQRS is a trapezium such that SP || RQ and PQ is a transversal. \n\u2234 m\u2220P + m\u2220Q = 180\u00b0 (Interior angles are supplementary) \nm\u2220P + 130\u00b0 = 180\u00b0 \nm\u2220P = 180\u00b0 – 130\u00b0 = 50\u00b0 \nAlso, m\u2220S + m\u2220R = 180\u00b0 \n\u21d2 m\u2220S + 90\u00b0 = 180\u00b0 \n\u21d2 m\u2220S = 180\u00b0 – 90\u00b0 \n\u21d2 m\u2220S = 90\u00b0 \nm\u2220P + m\u2220Q + m\u2220R + m\u2220S = 360\u00b0 (sum of the angles of a quadrilateral is 360\u00b0) \n\u21d2 m\u2220P + 130\u00b0 + 90\u00b0 + 90\u00b0 = 360\u00b0 \n\u21d2 m\u2220P + 130\u00b0 + 90\u00b0 + 90\u00b0 = 360\u00b0 \n\u21d2 m\u2220P + 130\u00b0 + 90\u00b0 + 90\u00b0 = 360\u00b0 \n\u21d2 m\u2220P + 310\u00b0 = 360\u00b0 \n\u21d2 m\u2220P = 360\u00b0 – 310\u00b0 \n\u21d2 m\u2220P = 50\u00b0 \nHence, m\u2220P = 50\u00b0 and m\u2220S = 90\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3 Question 1. Given a parallelogram ABCD. Complete each statement along with the definition or property used. (i) AD = …<\/p>\n
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n