NCERT Solutions for Class 10 Maths<\/a> Chapter 6 Triangles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2<\/h2>\n <\/p>\n
Question 1. \nIn the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). \n \nSolution: \n <\/p>\n
Question 2. \nE and F are points on the sides PQ and PR respectively of a \u2206PQR. For each of the following cases, state whether EF || QR: \n(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm \n(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm \n(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm \n \nSolution: \nWe have to show that EF || QR. \nIf a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. \n(i) PE = 3.9cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm \n \nRatio are not same, so EF is not parallel to QR.<\/p>\n
(ii) PE = 4cm, QE = 4.5cm, PF = 8cm and RF = 9cm \n \nTherefore, the ratio are same, so EF || QR.<\/p>\n
(iii) PQ = 1.28cm, PR = 2.56cm, PE = 0.18cm and PF = 0.36cm \n \nRatio are same, so EF || QR.<\/p>\n
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Question 3. \nIn the given figure, if LM || CB and LN || CD, Prove that \\(\\frac { AM }{ AB }\\) = \\(\\frac { AN }{ AD }\\) \n \nSolution: \nGiven a quadrilateral ABCD, in which LM || BC and LN || CD \nIn \u2206ABC, LM || BC \n\\(\\frac { AM }{ MB }\\) = \\(\\frac { AL }{ LC }\\) … (i) \n[Basic proportionality theorem in fig.] \nIn \u2206ADC, LN || CD \n\\(\\frac { AN }{ AD }\\) = \\(\\frac { AL }{ LC }\\) … (2) \n[Basis proportionality theorem] \nFrom equations (1) and (2), we have \n <\/p>\n
Question 4. \nIn the adjoining figure DE || AC and DF || AE. Prove that \\(\\frac { BF }{ FE }\\) = \\(\\frac { BE }{ EC }\\) \n \nSolution: \nGiven a quadrilateral ABC, in which DF || AE and DE || AC \nIn \u2206ABE, we have \n\\(\\frac { BD }{ AD }\\) = \\(\\frac { BE }{ FE }\\) … (i) \n[Basic proportionality theorem in fig.] \nIn \u2206ABC, we have \n\\(\\frac { BD }{ AD }\\) = \\(\\frac { BE }{ EC }\\) … (2) \n[Basis proportionality theorem] \nFrom equations (1) and (2), we have \n\\(\\frac { BE }{ FE }\\) = \\(\\frac { BE }{ EC }\\)<\/p>\n
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Question 5. \nIn the given figure, DE || OQ and DF || OR. Show that EF || QR. \nSolution: \nIn the triangle PQD, ED || OQ. \nSo applying Thales theorem, we get, \n \nEquation (iii) shows that EF intersects PQ and PR at E and F respectively in equal ratio. So, according to the converse of the basic proportionality theorem (or Thales theorem), EF || QR. Flence Proved.<\/p>\n
Question 6. \nIn the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. \nSolution: \nGiven : O is any point inside \u2206PQR. A is any point PO and B is any point on OQ and C is | any point on OR. AB || PQ and AC || PR. \n \nTo Prove: BC || QR \nProof: In \u2206POQ, AB || PQ \n\u2234 \\(\\frac { OA }{ AP }\\) = \\(\\frac { OB }{ BQ }\\) …(1) [Basic proportionality theorem] \nIn \u2206POR, AC || PR \n\u2234\\(\\frac { OA }{ AP }\\) = \\(\\frac { OC }{ OR }\\) …(2) [Basic proportionality theorem] \nFrom equations (1) and (2) \n\\(\\frac { OB }{ BQ }\\) = \\(\\frac { OC }{ OR }\\) \nBut, we know that \nIn \u2206OQR,if \\(\\frac { OB }{ BQ }\\) = \\(\\frac { OC }{ OR }\\) \nthen we can say BC || QR. \n[Converse of basic proportionality theorem] \nHence Proved.<\/p>\n
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Question 7. \nUsing B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. \nSolution: \nGiven: A \u2206ABC in which D is the mid-point of AB and DE || BC \nTo Prove : E is the mid-point of AC. \n \nProof : We have to prove that E is the mid-point of AC. Suppose, let E be not the mid-point of AC. Let E’ be the mid-point of AC. Join DE’. \nNow. in \u2206ABC, D is the mid-point of AB and E’ is the mid-point of AC. Therefore, by Theorem (6.1). \nDE || BC … (1) \nAlso DE || BC … (2) \nFrom (1) and (2), we find that two intersecting lines DE and DE’ are both parallel to line BC. This is a contradiction to the parallel line axiom. \nSo, our assumption is wrong. Hence E is the mid-point of AC.<\/p>\n
Question 8. \nUsing converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. \nSolution: \nGiven: A \u0394ABC in which D and E are mid-points of sides AB and AC respectively. \nTo Prove: DE || BC \nConstruction : Produce the line segement DE to F, such that DE = EF. Join FC. \n \nProof: In \u0394AED and CEF, we have \nAE = CE \n[\u2235 E is the mid-point of AC] \n\u2220AED = \u2220CEF \n[Vertically opposite angle] \nor DE = EF \nSo by SAS congruence \n\u0394AED \u2245 \u0394CEF \n\u21d2 AD = CF … (1) \n[corresponding of congruent triangle] \nand \u2220ADE = \u2220CEF … (2) \nNow, D is the mid-point of AB, \nAD = DB \nDB = CF … (3) [From (1)] \nNow, DF intersects AD and FC at D and F such that \n\u2220ADE = \u2220CFE [From (2)] \ni.e., alternate interior angles are equal \n\u2234 AD || FC \nDB || CF … (4) \nFrom (3) and (4), DBCF is a quadrilateral such that one pair of sides are equal and parallel. \n\u2234 DBCF is a parallelogram \n\u21d2 DF || BC and DC = BC [opposite side of parallelogram] \nBut, D, E, F are collinear and DE = EF \n\u2234 DE || BC<\/p>\n
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Question 9. \nABCD is a trapezium in which AB || DC and its diagonals intersect each other at the O. Show that \\(\\frac { AO }{ BO }\\) = \\(\\frac { CO }{ DO }\\) \n \nSolution: \n <\/p>\n
Question 10. \nThe diagonals of a quadrilateral ABCD intersect each other at the point O such that \\(\\frac { AO }{ BO }\\) = \\(\\frac { CO }{ DO }\\) Show that ABCD is a trapezium. \nSolution: \nConstruction: Draw OF || AB, meeting AD in F. \nTo Prove: Quadrilateral ABCD is a trapezium. \nProof: In ABD, we have OF || AB \n \nHence Proved.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2 Question 1. In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). Solution: …<\/p>\n
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2<\/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":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 - 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