{"id":28295,"date":"2021-07-05T13:35:08","date_gmt":"2021-07-05T08:05:08","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28295"},"modified":"2021-07-05T17:27:56","modified_gmt":"2021-07-05T11:57:56","slug":"ncert-solutions-for-class-10-maths-chapter-6-ex-6-6","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-6\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 6 Triangles Ex 6.6 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nIn the given figure, PS is the bisector of \u2220QPR of \u2206PQR. Prove that \\(\\frac { QS }{ SR } =\\frac { PQ }{ PR } \\)
\n\"NCERT
\nSolution:
\n\"NCERT<\/p>\n

Question 2.
\nIn the given figure, D is a point on hypotenuse AC of \u2206ABC, DM \u22a5 BC and DN \u22a5 AB. Prove that:
\n(i) DM2<\/sup> = DN . MC
\n(ii) DN2<\/sup> = DM . AN
\nSolution:
\nGiven \u2206ABC is right angled at B, DM \u22a5 BC and DN \u22a5 AB.
\nConstrucation : Join BD.
\nTo Prove: (i) DM\u00b2 = DN. MC
\n(ii) DN\u00b2 = DM. AN.
\nProof:
\n(i) Consider \u2206BDC
\n\u2220BDC = 90\u00b0
\n\u21d2 BDM + MDC = 90\u00b0 … (i)
\nAlso in AMCD
\n\u2220MCD + \u2220MDC = 90\u00b0 … (ii)
\n(\u2235 DMB = 90\u00b0 by exterior theorem)
\nFrom (i) and (ii) \u2220MCD = \u2220BDM … (iii)
\nIn \u2206s BMD and CMD
\n\u2220CMD = \u2220BMD (90\u00b0 each)
\n\u2220MCD = \u2220MDB (from (iii)
\n\u2234 BMD ~ DMC
\n\"NCERT
\nProof (ii): Now consider similar triangles BND and AND
\nWe
\n\\(\\frac { DN }{ DM }\\) = \\(\\frac { AN }{ DN }\\)
\nDN\u00b2 = DM.MN (Using BN = DM)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn the given figure, ABc is triangle in which\u00a0\u2220ABC > 90\u00b0 and AD \u22a5 CB produced. Prove that AC2<\/sup> = AB2<\/sup> + BC2<\/sup>\u00a0 + 2BC . BD
\n\"NCERT
\nSolution:
\nIn right angle triangle ADC. \u2220D = 90\u00b0
\nBy Pythagoras theorem.
\nAC\u00b2 = AD\u00b2 + DC\u00b2
\nor AC\u00b2 = AD\u00b2 + (BD + BC)\u00b2 (DC – BD + BC)
\nor AC\u00b2 = AD\u00b2 + BD\u00b2 + BC\u00b2 + 2BD.BC
\n(a + b)\u00b2 = a\u00b2 + b\u00b2 + 2ab)]
\nNow in \u2206ADB \u2220D produced right angle.
\nAB\u00b2 = AD\u00b2 + BD\u00b2
\nAD\u00b2 = AB\u00b2 – BD\u00b2 … (ii)
\nFrom (i) and (ii)
\nAC\u00b2 = AB\u00b2 – BD\u00b2 + BD\u00b2 + BC\u00b2 + \u00b2BC. BD
\n\u21d2 AC\u00b2 = AB\u00b2 + BC\u00b2 + 2BC. BD<\/p>\n

Question 4.
\nIn the given figure, ABC is atriangle in which \u2220ABC 90\u00b0 and AD\u00a0\u22a5 CB. Prove that AC2<\/sup> = AB2<\/sup> + BC2<\/sup> – 2BC . BD
\n\"NCERT
\nSolution:
\n\"NCERT<\/p>\n

Question 5.
\nIn the given figure, Ad is a median of a triangle ABC and AM \u22a5\u00a0BC. Prove that.
\n\"NCERT
\nSolution:
\n\"NCERT<\/p>\n

Question 6.
\nProve that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
\nSolution:
\nABCD is a parallelogram and AC and BD are its diagonals.
\nBy the property of parallelogram
\n\u2206ODC ~ \u2206OAB
\n\u2206OAD ~ \u2206OCB
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nIn the given figure, two chords AB and CD intersect each other at the point P. Prove that:
\n(i) \u2206APC ~ \u2206DPB
\n(ii) AP . PB = CP . DP
\n\"NCERT
\nGiven : Let AB and CD be the chords intersecting at P.
\nTo Prove: (i) \u2206AFC ~ \u2206DPB
\n(ii) AP. PB = CP. DP
\nProof:
\n(i) In \u2206PAC and \u2206PDB. We have
\n\u2220PAC = \u2220PDB (\u2220s in the same segement)
\n\u2220APC = \u2220BPD (Vertically opp. \u2220s)
\n\u2234 \u2206APC ~\u2206DPB (Proved)<\/p>\n

(ii) \u2206APC ~\u2206DPB
\n\\(\\frac { AP }{ DP }\\) = \\(\\frac { CP }{ PB }\\)
\n\u21d2 AP. PB = CP. DP (Proved)<\/p>\n

Question 8.
\nIn the given figure, two chords Ab and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
\n(i) \u2206PAC ~ \u2206PDB
\n(ii) PA . PB = PC . PD
\n\"NCERT
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nIn the given figure, D is a point on side BC of \u2206ABC, such that \\(\\frac{B D}{C D}=\\frac{A B}{A C}\\) Prove that AD is the bisector of \u2206BAC.
\n\"NCERT
\nSolution:
\nGiven: A \u2206ABC, in which D is a point on BC such that
\n\\(\\frac{B D}{C D}=\\frac{A B}{A C}\\)
\nTo Prove: AD is the bisector of \u2220BAC.
\nConstruction: Produce BA to E such that AE = AC. Join EC.
\nProof: In \u2206ACE, we have
\n\"NCERT<\/p>\n

Question 10.
\nNazima is fly fishing in a stream. The trip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the trip of the rod. Assuming that her string (from the trip of the rod to the fly) is that, how much string does she have out (see the figure)? If she pills in the string at the rate of 5 cm per second, what will be the\u00a0 horizontal distance of the fly from her after 12 seconds?
\n\"NCERT
\nSolution:
\nABC is a right triangle right angled at C by Pythagoras theorem
\nAB\u00b2 = AC\u00b2 + BC\u00b2
\nAB\u00b2 = (1.8)\u00b2 + (2.4)\u00b2
\nAB\u00b2 = 9
\nAB = 3 m.
\nHence 3 m the distance is
\nNow after the pull the vertical distance is 1.8 – 0.6 = 1.2 m
\nNow
\n\u2220ABC= \u2220PRQ = 90\u00b0
\n\u2220B = \u2220Q (The angle from water)
\n\u2206ABC ~ \u2206PQR
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6 Question 1. In the given figure, PS is the bisector of \u2220QPR of \u2206PQR. Prove that Solution: Question 2. In the …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6<\/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 6.6 - 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-6-6\/\" \/>\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 6.6 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 Questions and Answers are prepared by our highly skilled subject experts. 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