{"id":28189,"date":"2021-07-03T17:00:28","date_gmt":"2021-07-03T11:30:28","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28189"},"modified":"2021-07-03T17:00:28","modified_gmt":"2021-07-03T11:30:28","slug":"ncert-solutions-for-class-10-maths-chapter-6-ex-6-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-6-ex-6-5\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 6 Triangles Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nSides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
\n(i) 7 cm, 24 cm, 25 cm
\n(ii) 3 cm, 8 cm, 6 cm
\n(iii) 50 cm, 80 cm, 100 cm
\n(iv) 13 cm, 12 cm, 5 cm
\nSolution:
\n(i) 7 cm, 24 cm,-25 cm
\n(7)2<\/sup> + (24)2<\/sup> = 49 + 576 = 625 = (25)2<\/sup> = 25
\n\u2234 The given sides make a right angled triangle with hypotenuse 25 cm<\/p>\n

(ii) 3 cm, 8 cm, 6 cm(8)2<\/sup> = 64
\n(3)2<\/sup> + (6)2<\/sup> = 9 + 36 = 45
\n64 \u2260 45
\nThe square of larger side is not equal to the sum of squares of other two sides.
\n\u2234 The given triangle is not a right angled.<\/p>\n

(iii) 50 cm, 80 cm, 100 cm
\n(100)2<\/sup>= 10000
\n(80)2<\/sup> + (50)2<\/sup> = 6400 + 2500
\n= 8900
\nThe square of larger side is not equal to the sum of squares of other two sides.
\n\u2234The given triangle is not a right angled.<\/p>\n

(iv) 13 cm, 12 cm, 5 cm
\n(13)2<\/sup> = 169
\n(12)2<\/sup> + (5)2<\/sup>= 144 + 25 = 169
\n= (13)2<\/sup> = 13
\nSides make a right angled triangle with hypotenuse 13 cm.<\/p>\n

Question 2.
\nPQR is a triangle right angled at P and M is a point on QR such that PM \u22a5 QR. Show that PM\u00b2 = QM . MR.
\nSolution:
\nWe have PQR is a right triangle and PM \u22a5 QR.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn the given figure, ABD is a triangle right angled at A and AC \u22a5. BD. Show that
\n(i) AB2<\/sup> = BC.BD
\n(ii) AC2<\/sup> = BC.DC
\n(iii) AD2<\/sup> = BD.CD
\nSolution:
\n\"NCERT<\/p>\n

Question 4.
\nABC is an isosceles triangle right angled at C. Prove that AB2<\/sup> = 2AC2<\/sup>.
\nSolution:
\n\"NCERT
\nGiven: In \u2206ABC, \u2220C = 90\u00b0 and AC = BC
\nTo Prove: AB2<\/sup> = 2AC2<\/sup>
\nProof: In \u2206ABC,
\nAB2<\/sup>= BC2<\/sup> + AC2<\/sup>
\nAB2<\/sup> = AC2<\/sup> + AC2<\/sup> [Pythagoras theorem]
\n= 2AC2<\/sup><\/p>\n

Question 5.
\nABC is an isosceles triangle with AC = BC. If AB2<\/sup> = 2AC2<\/sup> , Prove that ABC is a right triangle.
\nSolution:
\nGiven that ABC is an isosceles triangle with AC = BC and given that AB\u00b2 = 2AC\u00b2
\nNow we have AB\u00b2 = 2AC\u00b2
\nAB\u00b2 = AC\u00b2 + AC\u00b2
\nBut AC= BC (Given)
\nAB\u00b2 = AC\u00b2 + BC\u00b2
\nHence by Pythagoras theorem \u2206ABC is a right triangle where AB is the hypotenuse of \u2206ABC.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nABC is an equilateral triangle of side la. Find each of its altitudes.
\nSolution:
\nGiven: In \u2206ABC, AB = BC = AC = 2a
\n\"NCERT
\nWe have to find length of AD
\nIn \u2206ABC,
\nAB = BC = AC = 2a
\nand AD \u22a5 BC
\nBD = \\(\\frac { 1 }{ 2 }\\) x 2 a = a
\nIn right angled triangle ADB,
\nAD2<\/sup> + BD2<\/sup> = AB2<\/sup>
\n\u21d2 AD2<\/sup> = AB2<\/sup> – BD2<\/sup>= (2a)2<\/sup> – (a)2<\/sup> = 4a2<\/sup>– a2<\/sup>= 3a2<\/sup>
\nAD = \u221a3a<\/p>\n

Question 7.
\nProve that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
\nSolution:
\nGiven: ABCD is a rhombus. Diagonals AC and BD intersect at O.
\nTo Prove: AB2<\/sup>+ BC2<\/sup>+ CD2<\/sup>+ DA2 <\/sup>= AC2<\/sup>+ BD2<\/sup>
\n\"NCERT<\/p>\n

Question 8.
\nIn the given figure, O is a point in the interior of a triangle ABC, OD \u22a5 BC, OE \u22a5 AC and OF \u22a5 AB. Show that
\n(i) OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\n(ii) AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup>.
\nSolution:
\n(i) Given: \u2206ABC, O is any point inside it,
\nOD, OE and OF are perpendiculars to BC, CA and AB respectively.
\nTo Prove:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nA ladder 10 m long reaches a window 8 m above the ground. ind the distance of the foot of the ladder from base of the wall.
\nSolution:
\nBy Pythagoras theorem
\n\"NCERT
\nHence the distance of the foot of the ladder from base of the wall in 6 m.<\/p>\n

Question 10.
\nA guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
\nSolution:
\nLet AB i.e., the vertical pole of height 18 m and AC be the guy wire of 24 m long. BC is the distance from the vertical pole to where the wire will be staked.
\nBy Pythagoras theorem
\n\"NCERT<\/p>\n

Question 11.
\nAn aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1\\(\\frac { 1 }{ 2 }\\) hours?
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nTwo poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
\nSolution:
\nWe have two poles.
\nWe have
\nBC = 12 m
\nAB = 11 – 6
\nAB = 5 m
\n\"NCERT
\nBy Pythagoras theorem in right triangle ABC
\nAC\u00b2 = AB\u00b2 + BC\u00b2
\nAC\u00b2 = (12)\u00b2 +(5)\u00b2
\nAC\u00b2 = 144 + 25
\nAC\u00b2 = 169
\nAC = 13 m
\nHence the distance between the tops is 13 m<\/p>\n

Question 13.
\nD and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2<\/sup> + BD2<\/sup> = AB2<\/sup> + DE2<\/sup>.
\nSolution:
\nGiven: \u2206ABC is a right angled at C D and E are the points on the side CA and CB.
\nTo Prove: AE\u00b2 + BD\u00b2 = AB\u00b2 + DE\u00b2
\nProof : \u2206ACE is right angled at C
\nAE\u00b2 = AC\u00b2 + CE\u00b2… (i)
\n(Pythagoras theorem)
\n\"NCERT<\/p>\n

Question 14.
\nThe perpendicular from A on side BC of a \u2206ABC intersects BC at D such that DB = 3CD (see the figure). Prove that 2AB2 <\/sup>= 2AC2<\/sup> + BC2<\/sup>.
\n\"NCERT
\nSolution:
\nWe have
\nBD = 3CD
\n\u2234 BC = BD + DC
\n\u21d2 BC = 3CD + CD
\nBC = 4CD
\n\u21d2 CD = \\(\\frac { 1 }{ 4 }\\)BC … (i)
\nAnd BD = 3CD
\n\u21d2 BD = \\(\\frac { 3 }{ 4 }\\)BC …(ii)
\nSince \u2206ABD is a right triangle, right angled at
\nAB\u00b2= AD\u00b2 + BD\u00b2 …(iii)
\nSimilarly, \u2206ACD is right angled at D.
\nAC\u00b2 = AD\u00b2 + CD\u00b2 …(iv)
\nSubstracting (iv) from (iii)
\nWe get
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 15.
\nIn an equilateral triangle ABC, D is a point on side BC, such that BD = \\(\\frac { 1 }{ 3 }\\)BC. Prove that 9AD2<\/sup> = 7AB2<\/sup>.
\nSolution:
\n\"NCERT<\/p>\n

Question 16.
\nIn an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
\nSolution:
\nLet ABC be an equailateral traingle of each side x. AD be its altitude.
\nSo, AB = BC = CA = x
\nand BD = DC = \\(\\frac { 1 }{ 2 }\\) BC = \\(\\frac { x }{ 2 }\\)
\nIn right triangle ADC in which \u2220D = 90\u00b0
\nAD = perpendicular,
\nDC = base and AC = hypotenuse.
\nApply Pythagorus theorem, we get
\n\"NCERT
\nSo, we get length of each side is x an length of altitude is \\(\\frac{\\sqrt{3} x}{2}\/latex]
\nThen, three times the square of each side = 3 x (x)\u00b2 = 3x\u00b2 … (i)
\nand four times, the square of its altitudes = 4 x [latex]\\frac { 3 }{ 4 }\\)x\u00b2 = 3x\u00b2 … (ii)
\nIt shows that equations (i) and (ii) are same. Hence times the square of one side an equilateral triangle is equal to four times the square of its altitude.<\/p>\n

\"NCERT<\/p>\n

Question 17.
\nTick the correct answer and justify : In \u2206ABC, AB = 6\\(\\sqrt{3}\\) cm, AC = 12 cm and BC = 6 cm. The angle B is:
\n(a) 120\u00b0
\n(b) 60\u00b0
\n(c) 90\u00b0
\n(d) 45
\nSolution:
\n(C) is the correct answer.
\nWhen the triangle is right angled then by Pythagoras theorem
\n(12)= (6\\(\\sqrt{3}\\))\u00b2 + (6)\u00b2
\n144 = 36 x 3 + 36
\n144 = 144
\nL.H.S. = R.H.S.
\nHence the result is (C)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5 Question 1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5<\/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 6.5 - 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-6-5\/\" \/>\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 6.5 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5 Question 1. Sides of triangles are given below. Determine which of them are right triangles. 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