{"id":25468,"date":"2021-06-23T11:12:08","date_gmt":"2021-06-23T05:42:08","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25468"},"modified":"2022-03-02T10:36:21","modified_gmt":"2022-03-02T05:06:21","slug":"ncert-solutions-for-class-9-maths-chapter-8-ex-8-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 8 Quadrilaterals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2<\/h2>\n

Question 1.
\nABCD is a quadrilateral in which P, Q, R, and S are midpoints of the sides AB, BC, CD, and DA (see fig. 8.29). AC is a diagonal. Show that
\n(i) SR || AC and SR = \\(\\frac {1}{2}\\) AC
\n(ii) PQ = SR
\n(iii) PQRS is a parallelogram.
\n\"NCERT
\nSolution:
\nGiven. ABCD is a quadrilateral in which P, Q, R, and S are the midpoints of side AB, BC, CD, and AD respectively.
\nTo prove that:
\n(i) SR || AC and SR = \\(\\frac {1}{2}\\) AC
\n(ii) PQ = SR
\n(iii) PQRS is a parallelogram.
\nProof:
\n(i) In \u0394ADC
\nS and R are the mid-point of side AD and DC respectively.
\nWe know that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of it.
\nSR || AC and SR = \\(\\frac {1}{2}\\) AC ……(i)<\/p>\n

(ii) In \u0394ABC,
\nP and Q are the mid points of side AB and BC respectively.
\nTherefore, PQ || AC and PQ = \\(\\frac {1}{2}\\) AC …….(ii)
\nFrom equation (i) and (ii),
\nPQ || SR and PQ = SR<\/p>\n

(iii) We have
\nPQ = SR (Prove above)
\nand PQ || SR (Prove above)
\nWe know that if one pair of opposite sides of a quadrilateral are parallel and equal then the quadrilateral is a parallelogram.
\n\u2234 PQRS is parallelogram.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nABCD is a rhombus and P, Q, R, and S are the midpoints of the sides AB, BC, CD, and DA respectively. Show that the quadrilateral PQRS is a rectangle.
\nSolution:
\nGiven: Arhmbus ABCD in which P, Q, Rand S are the midpoints of sides AB, BC, CD, and DA respectively. PQ, QR, RS, and SP are joined.
\nTo prove that: PQRS is a rectangle.
\nConstruction: Join AC and BD.
\n\"NCERT
\nProof: In order to prove that PQRS is a rectangle, it is sufficient to show that it is a parallelogram whose one angle is a right angle. First, we shall prove that PQRS is a parallelogram.
\nIn \u0394ABC,
\nP and Q are the mid points of AB and BC respectively.
\n\u2234 PQ || AC and PQ = \\(\\frac {1}{2}\\) AC ……(i)
\nIn \u0394ADC,
\nR and S are the mid points of CD and AD respectively.
\n\u2234 RS || AC and RS = \\(\\frac {1}{2}\\) AC ……(ii)
\nFrom equation (i) and (ii) we have
\nPQ || RS and PQ = RS
\nThus, PQRS is a quadrilateral such that one pair of opposite sides PQ and SR is equal and parallel. So, PQRS is a parallelogram.
\nPQ || AC (Opposite sides of || gm)
\n\u2234 MQ || ON
\nQR || BD (Opposite sides of || gm)
\n\u2234 QN || OM
\nTherefore, OMQN is a parallelgram.
\nand \u2220NOM = 90\u00b0
\n[Diagomals of a rhombus bisects each other at a right angle]
\nSo, \u2220MQN = 90\u00b0
\n[Opposite angles of || gm are equal]
\nTherefore, PQRS is a parallelogram in which \u2220Q is a right angle.
\n\u2234 PQRS is a rectangle.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nABCD is a rectangle and P, Q, R, and S are midpoints of the sides AB, BC, CD, and DA respectively. Show that the quadrilateral PQRS is a rhombus.
\n\"NCERT
\nSolution:
\nGiven: A rectangle ABCD in which P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA respectively. PQ, QR, RS, and SP are joined.
\nTo prove that: PQRS is a rhombus.
\nConstruction: Join AC and BD.
\nProof: In \u2206ABC,
\nP and Q are mid points of AB and BC respectively.
\n\u2234 PQ || AC and PQ = \\(\\frac {1}{2}\\) AC …..(i)
\nSimilarly in \u2206ACD,
\nSR || AC and SR = \\(\\frac {1}{2}\\) AC ……(ii)
\nFrom equation (i) and (ii) we have
\nPQ || SR and PQ = SR
\nTherefore, PQRS is a parallelogram
\n\u2234 SR || PQ (Opposite sides of || gm PQRS)
\nand SR = PQ
\nNow, in \u2206BCD,
\nQ and R are mid points of side BC & CD.
\n\u2234 QR || AC and QR = \\(\\frac {1}{2}\\) BD
\nBut, AC = BD (In rectangle diagonals are equal)
\nQR = \\(\\frac {1}{2}\\) AC ……(iii)
\nFrom equation (i) and (iii) we have
\nPQ = QR
\nbut PQ = SR and PS = QR
\nTherefore, PQ = QR = RS = PS
\n\u2234 PQRS is a rhombus.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nABCD is a trapezium in which AB || DC, BD is a diagonal, and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see fig. 8.30). Show that F is the midpoint of BC.
\n\"NCERT
\nSolution:
\nIn \u2206ABD,
\nE is the midpoints of AD and EF || AB
\n\u2234 EF || CD
\nWe know that a line through the midpoint of a side of a triangle parallel to another side bisects the third side. So, G is the midpoint of DB.
\nNow, in \u2206DBC
\nG is the midpoint of DB and GF || CD
\n\u2234 F is the midpoint of BC.<\/p>\n

Question 5.
\nIn a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see fig. 8.31). Show that the line segment AF and EC trisect the diagonal BD.
\n\"NCERT
\nSolution:
\nWe have,
\nABCD is a parallelogram
\n\u2234 AB || CD
\nSo, AE || FC ……(i)
\nAgain, AB = CD (Opposite sides of ||gm ABCD)
\n\u2234 \\(\\frac {1}{2}\\) AB = \\(\\frac {1}{2}\\) CD
\nor, AE = FC …….(ii)
\nFrom equation (i) and (ii)
\nTherefore, AFCE is a parallelogram.
\nNow, in \u0394DQC,
\nF is the mid point of CD,
\nand FP || CQ
\nWe know that a line through the midpoint of a side of a triangle parallel to another side bisects the third side.
\nSo, P is the midpoint of DQ
\nor, DP = PQ ……(i)
\nSimilarly in \u0394APB, Q is midpoint of BP
\ni.e. PQ = BQ …..(ii)
\nFrom equation (i) and (ii)
\nPQ = DP = BQ
\n\u2234 AF and CE trisect the diagonal BD.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nShow that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
\nSolution:
\nGiven: ABCD is a quadrilateral in which P, Q, R, and S are the midpoints of side AB, BC, CD, and AD respectively.
\nTo prove that: PR & QS bisect each other. Construction: Join PQ, QR, RS, and SP and join BD.
\nProof: In \u0394ABD,
\nS and P are the midpoint of side AD and AB respectively we know that the line segment joining the midpoints of any two sides of a triangle is. parallel to the third side and is half of it.
\n\"NCERT
\nTherefore, SP || BD and SP = \\(\\frac {1}{2}\\) BD …….(i)
\nSimilarly,
\nIn \u0394BCD,
\nQ and R is the mid points of side BC & CD
\n\u2234 QR || AC and QR = \\(\\frac {1}{2}\\) AC ……(ii)
\nFrom equation (i) and (ii)
\nSP || QR and SP = QR
\nWe know that if one opposite pair of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.
\n\u2234 SPQR is a parallelogram.
\nSo, PR and QS bisect each other.
\n[Because diagonal of a parallelogram bisects each other]<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that
\n(i) D is the mid point of AC
\n(ii) MD \u22a5 AC
\n(iii) CM = MA = \\(\\frac {1}{2}\\) AB
\nSolution:
\nGiven: ABC is a right-angled triangle right angle at C. M is the midpoint of side AB and DM || BC.
\nTo prove that:
\n(i) D is the mid point of AC
\n(ii) MD \u22a5 AC
\n(iii) CM = MA = \\(\\frac {1}{2}\\) AB
\n\"NCERT
\nConstruction: Join CM.
\nProof: (i) In \u0394ABC,
\nM is the mid point of side AB and MD || BC
\nWe know that a line through the midpoint of a side of a triangle parallel to another side bisects the third side.
\nSo, D is the midpoint of AC.<\/p>\n

(ii) We have DM || CB
\n\u2234 \u2220MDC + \u2220DCB = 180\u00b0
\n(Sum of interior angles of the same side of transveralsi)
\nor, \u2220MDC + 90\u00b0 = 180\u00b0
\nor, \u2220MDC = 90\u00b0
\nTherefore, MD \u22a5 AC<\/p>\n

\"NCERT<\/p>\n

(iii) In \u0394AMD and \u0394CMD
\nAD = CD (Prove above)
\n\u2220ADM = \u2220CDM, (Each 90\u00b0)
\nDM = DM (Common)
\nBy S-A-S congruency condition
\n\u0394AMD \u2245 \u0394CMD
\n\u2234 AM = CM …..(i) (By CPCT)
\nBut AM = AB ……(ii)
\n(Given M is the mid point of AB)
\nFrom (i) and (ii)
\nCM = MA = \\(\\frac {1}{2}\\) AB<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 Question 1. ABCD is a quadrilateral in which P, Q, R, and S are midpoints of the sides AB, BC, CD, …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2<\/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":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts. 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