{"id":27136,"date":"2021-06-29T18:03:33","date_gmt":"2021-06-29T12:33:33","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27136"},"modified":"2021-06-29T18:03:33","modified_gmt":"2021-06-29T12:33:33","slug":"ncert-solutions-for-class-10-maths-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2"},"content":{"rendered":"

These 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>\n

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nIn the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
\n\"NCERT
\nSolution:
\n\"NCERT<\/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\"NCERT
\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\"NCERT
\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\"NCERT
\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\"NCERT
\nRatio are same, so EF || QR.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn the given figure, if LM || CB and LN || CD, Prove that \\(\\frac { AM }{ AB }\\) = \\(\\frac { AN }{ AD }\\)
\n\"NCERT
\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\"NCERT<\/p>\n

Question 4.
\nIn the adjoining figure DE || AC and DF || AE. Prove that \\(\\frac { BF }{ FE }\\) = \\(\\frac { BE }{ EC }\\)
\n\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

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\"NCERT
\nSolution:
\n\"NCERT<\/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\"NCERT
\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":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-2\/\" \/>\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 10 Maths Chapter 6 Triangles Ex 6.2 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts. 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