{"id":25107,"date":"2020-09-15T14:14:36","date_gmt":"2020-09-15T08:44:36","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25107"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-solutions-for-class-10-maths-chapter-2-ex-2-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-2-ex-2-6\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 2 Triangles Ex 2.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.6<\/h2>\n

Question 1.
\nIn Fig. 2.56, PS is the bisector of \u2220QPR of \u2206 PQR, Prove that \\(\\frac{\\mathbf{Q S}}{\\mathbf{S R}}=\\frac{\\mathbf{P Q}}{\\mathbf{P R}}\\)
\n\"KSEEB
\nAnswer:
\n\"KSEEB
\nGiven: In the figure, PS is the bisector of<\/p>\n

\"KSEEBKSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nIn Fig. 2.57, D is a point on hypotenuse AD of \u2206 ABC, such that BD \u22a5 AC, DM \u22a5 BC and DN \u22a5 AB. Prove that:
\ni) DM2<\/sup> = DN.MC ii) DN2<\/sup> = DM.AN
\n\"KSEEB<\/p>\n

Answer:
\nGiven: D is a point on hypotenuse AC of \u2206 ABC DM \u22a5 BC, DN \u22a5AB and BD \u22a5 AC
\n\"KSEEB<\/p>\n

ii) Consider \u2206le <\/sup>ADN And DBN
\n\"KSEEB<\/p>\n

Question 3.
\nIn Fig. 2.58, ABC is a triangle in which \u2220ABC > 90\u00b0 and AD \u22a5 CB produced. Prove that AC2<\/sup> = AB2<\/sup> + BC2<\/sup> + 2BC. BD
\n\"KSEEB
\nAnswer:
\nGiven: In figure, ABC is a triangle in which \u2220 ABC > 90\u00b0 and AD \u22a5 BC.
\nTo prove: AC2<\/sup> = AB2<\/sup> + BC2<\/sup> + 2BC. BD.
\nProof: In right triangle ABC
\n\u2220D = 90\u00b0
\nAC2<\/sup> = AD2<\/sup> + CD2<\/sup> [Pythagoras theorem]
\nAC2<\/sup> = AD2<\/sup> + (BD + CB)2<\/sup> [DC = BD + BC]
\nAC2<\/sup> = AD2<\/sup> + BD2<\/sup>+ BC2 +2.BD. BC
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> + 2BD.BC [In right angle \u2206le<\/sup> ADB AB2<\/sup> = AD2<\/sup> + BD2<\/sup>]<\/p>\n

Question 4.
\nIn Fig. 2.59, ABC is a triangle in which \u2220 ABC < 90\u00b0 and AD \u22a5 BC. Prove that AC2<\/sup> = AB2 <\/sup>+ BC2<\/sup> – 2BC. BD.
\n\"KSEEB
\nAnswer:
\nGiven: In figure, ABC is a triangle in which \u2220ABC < 90\u00b0 and AD \u22a5 BC
\nTo prove: AC2<\/sup> = AB2<\/sup> + BC2<\/sup> – 2BC.BD
\nProof: In right triangle ABC \u2220 D = 90\u00b0
\nAC2<\/sup> = AD2<\/sup> + DC2<\/sup> [By pythagoras theorem]
\n= AD2<\/sup> + (BC – BD)2<\/sup> [CD = BC – BD]
\n= AD2<\/sup> \u00b1 BC2<\/sup> + BD2<\/sup> – 2BC. BD
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> – 2BC.BD
\n[In right angle ADB with
\n\u2220D = 90\u00b0 AB2<\/sup> = AD2<\/sup> + BD2<\/sup> By Pythagoras theorem]<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIn Fig. 2.60,AD is a median of a triangle ABC and AM I BC. Prove that:
\n\"KSEEB
\nAnswer:
\nGiven: In figure, AD is a median of a triangle ABC and AM \u22a5 BC.
\n\u2234 D is mid-point of BC (\u2235 AD is median)
\n\"KSEEB
\nProof:
\ni) In right triangle AMC \u2220M = 90\u00b0
\n\u2234 AC2<\/sup> = AM2<\/sup> + MC2<\/sup> [By Pythagoras theorem]
\n= AM2<\/sup> +(MD + DC)2<\/sup>
\n= AM2<\/sup> + MD2<\/sup> + DC2<\/sup> + 2MD. DC
\n\"KSEEB
\n[In right angle \u2206le<\/sup>\u00a0AMD with \u2220M = 90\u00b0
\nAM2<\/sup> + MD2<\/sup> = AD2<\/sup> (By pythagoras theorem)]
\n\"KSEEB
\nii) In right triangle AMB \u2220M = 90\u00b0
\nAB2 = AM2 + MB2 [Pythagoras
\ntheorem]
\n= AM2<\/sup> + (BD – MD)2<\/sup>
\n= AM2<\/sup> + BD2<\/sup> + MD2<\/sup> – 2BD . MD
\n= AM2<\/sup> + MD2<\/sup> + BD2<\/sup> – (2 BD)MD
\n\"KSEEB<\/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.
\nAnswer:
\n\"KSEEB<\/p>\n

Given: In parallelogram ABCD
\nAB = CD & AD = BC
\nConstruction: convert parallelogram into a rectangle and Draw AG \u22a5 CD
\nTo prove:
\nAC2<\/sup> + BD2<\/sup> = AB2<\/sup> + BC2<\/sup>2 + CD2<\/sup> – AD2<\/sup>
\nProof: consider \u2206le<\/sup> BDF \u2220D = 90\u00b0
\nBD2<\/sup> = BF2<\/sup> + FD2<\/sup> = h2<\/sup> + (x+d)2<\/sup> \u2014 (1)
\nConsider \u2206le<\/sup> AGC \u2220G = 90\u00b0
\nAC2<\/sup> = AG22<\/sup> + GC2 <\/sup>= h2<\/sup> + (x – d)2<\/sup> \u2014 (2)
\nAdding (I) and (2)
\nAC2<\/sup>+ BD2<\/sup> = h2<\/sup> + (x + d)2 <\/sup>+ h2<\/sup> + (x – d)2<\/sup>
\n2h2 <\/sup>+ x2<\/sup> + 2xd + d2<\/sup> + x2 <\/sup>– 2xd + d2
\n2h2<\/sup> + 2x2<\/sup> + 2d2<\/sup>
\n= 2x2<\/sup> + 2(h2 + d2)
\n= 2x2<\/sup> + 2y2<\/sup>
\n= x2<\/sup> + y2<\/sup> + x2<\/sup> + AC2<\/sup> + BD2<\/sup> =AB2<\/sup> +BC2<\/sup> i-CD2<\/sup> + AD2<\/sup>.<\/p>\n

Question 7.
\nIn Fig. 2.61, two chords AB and CD intersect each other at the point P.
\nProve that:
\ni) \u2206 APC ~ \u2206 DPB
\nii) AP. PB = CP.DP
\n\"KSEEB
\nAnswer:
\nGiven: In figure, two chords AB and CD intersect each other at the point P.
\nTo prove: i) \u2206 APC ~ \u2206 DPB
\nii) AP. PB = CP . DP
\nProof: i) In \u2206 APC and \u2206 DPB
\n\"KSEEB<\/p>\n

Question 8.
\nIn Fig. 2.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
\n\"KSEEB
\nAnswer:
\nGiven: In figure, two chords AB and CD of a circle intersect each other at the point P. (when produced) outside the circle.
\nTo prove: i) \u2206 PAC ~ \u2206 PDB
\nii )PA. PB = PC . PD
\nProof: i) we know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
\nTherefore, for cyclic quadrilateral ABCD.
\nconsider ABCD \u2206 PAC and \u2206 PBD.
\n\u2220PAC = \u2220PDB \u2192 (i)
\n\u2220PCA = \u2220PBD \u2192 (2)
\n\u2206 PAC ~ \u2206 PDB [A A similarity criterion]<\/p>\n

ii) A PAC ~ \u2206 PDB (Proved)
\n\\(\\frac{P A}{P D}=\\frac{P C}{P B}\\) [Corresponding sides of the similar \u2206le<\/sup> are proportional]
\nPA . PB = PC . PD<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nIn Fig. 2.63, D is a point on side BC of \u2206 ABC such that \\(\\frac{B D}{C D}=\\frac{A B}{A C}\\) Prove that AD is the bisector of \u2220BAC.<\/p>\n

\"KSEEB
\nAnswer:
\n\"KSEEB
\nGiven: In figure, D is a point on side BC. of \u2206 ABC such that \\(\\frac{B D}{C D}=\\frac{A B}{A C}\\)
\n\"KSEEB<\/p>\n

Question 10.
\nNazima is fly fishing in a stream. The tip of her fishing rod \u00a1s 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 nad 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 2.64)?
\n\"KSEEB
\nIf she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
\nAnswer:
\n\"KSEEB
\nAC = 3m
\nHence, she has 3 m string out.
\nLength of the string pulled in 12 seconds
\nat the rate of5 cm\/ sec 5 \u00d7 12cm = 60 cm = O.6 m.
\n\u2234 Length of remaining string left out
\n= AD = 3.0 – 0.6 = 2.4m
\n\"KSEEB<\/p>\n

In right angled \u2206 ABD \u2220B = 90\u00b0
\nAD2<\/sup>\u00a0= AB2<\/sup> + BD2<\/sup>
\nBD2<\/sup> = AD2<\/sup> – AB2<\/sup>
\n[pythagoras theorem]
\n= (2.4)2<\/sup> – (1.8)2<\/sup>
\n= 5.76 – 3.24 = 2.52
\nBD = 12.52 = 1.59 m (approx)
\nHence, the horizontal distance of the fly form Nazirna after 12 seconds = 1.2 + 1.59 = 2.79 m.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 2 Triangles Ex 2.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.6 Question 1. In Fig. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25107"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=25107"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25107\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25107"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25107"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25107"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}