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>\nNCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5<\/h2>\n <\/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>\nQuestion 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 <\/p>\n
<\/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 <\/p>\nQuestion 4. \nABC is an isosceles triangle right angled at C. Prove that AB2<\/sup> = 2AC2<\/sup>. \nSolution: \n \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>\nQuestion 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 <\/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 \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>\nQuestion 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 <\/p>\nQuestion 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 <\/p>\n <\/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 \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 <\/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 <\/p>\n
<\/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 \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 <\/p>\nQuestion 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 \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 <\/p>\n <\/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 <\/p>\nQuestion 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 \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
<\/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":{"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 6.5 - 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