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>\nNCERT 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 \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
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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 \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
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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 \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
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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 \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 \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
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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 \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
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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 \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
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(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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n