{"id":26994,"date":"2024-01-19T10:28:15","date_gmt":"2024-01-19T04:58:15","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26994"},"modified":"2024-01-20T17:57:36","modified_gmt":"2024-01-20T12:27:36","slug":"ncert-solutions-for-class-8-maths-chapter-3-ex-3-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-3-ex-3-3\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3"},"content":{"rendered":"

These 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>\n

NCERT 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\"NCERT
\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

\"NCERT<\/p>\n

Question 2.
\nConsider the following parallelograms. Find the values of the unknown x, y, z.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

\"NCERT
\nSolution:
\n(i) \u2220y = 100\u00b0 (opposite angles of a parallelogram are equal)
\n\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

(iii) x = 90\u00b0 (vertically opposite angles are equal)
\n\"NCERT
\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\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT<\/p>\n

\"NCERT<\/p>\n

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

\"NCERT<\/p>\n

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\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

(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\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

Question 11.
\nFind m\u2220C in the figure given below if AB || DC.
\n\"NCERT
\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\"NCERT
\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

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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. 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