{"id":27569,"date":"2021-07-01T11:24:57","date_gmt":"2021-07-01T05:54:57","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=27569"},"modified":"2022-03-02T10:31:12","modified_gmt":"2022-03-02T05:01:12","slug":"ncert-solutions-for-class-7-maths-chapter-6-ex-6-5","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-7-maths-chapter-6-ex-6-5\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 6 The Triangles and Its Properties Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.5<\/h2>\n

Question 1.
\nPQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR. Ans: In the right APQR
\nAnswer:
\nIn the right \u0394PQR
\n\"NCERT
\nQR2<\/sup> = PQ2<\/sup> + PR2<\/sup>
\n(pythagoras property)
\n= 102<\/sup> + 242<\/sup>
\n= 100 + 576
\n= 676
\nQR2<\/sup> = 262<\/sup>
\n\u2234 QR = 26 cm<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
\nAnswer:
\nIn the right \u0394ABC
\n\"NCERT
\nAB2<\/sup> = AC2<\/sup> + BC2<\/sup>
\n(using pythagoras property)
\n252<\/sup> = 72<\/sup> + BC2<\/sup>
\n625 = 49 + BC2<\/sup>
\n625 – 49 = BC2<\/sup>
\n576 = BC2<\/sup>
\n242<\/sup> = BC2<\/sup>
\n\u2234 BC = 24 cm<\/p>\n

Question 3.
\nA 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
\nAnswer:
\nLet the distance of the foot of a ladder from the wall be \u2018a\u2019m
\n\"NCERT
\na22<\/sup> + 122<\/sup> = 152<\/sup>
\n(using pythagoras property)
\na2<\/sup> + 144 = 225
\na2<\/sup> = 225 – 144
\na2<\/sup> = 81
\na2<\/sup> = 92<\/sup>
\na = 9 m
\nThe distance of the foot of the ladder from the wall = 9 m.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nWhich of the following can be the sides of a right triangle?
\n(i) 2.5 cm, 6.5 cm, 6 cm.
\n(ii) 2 cm, 2 cm, 5 cm.
\n(iii) 1.5 cm, 2 cm, 2.5 cm.
\nIn the case of right-angled triangles, identify the right angles.
\nAnswer:
\n(i) 2.5 cm, 6.5 cm, 6 cm2<\/sup>.
\nThe longest side is 6.5 cm
\n2.52<\/sup> + 62<\/sup> = 6.25 + 36
\n= 42.25
\n6.52<\/sup> = 42.25
\n\u2234 6.52<\/sup> = 2.52<\/sup> + 62<\/sup>
\n(pythagoras property)
\nThe given lengths can be the sides of a right triangle.
\nThe right angle is the angle between the sides 2.5 cm and 6 cm<\/p>\n

(ii) 2 cm, 2 cm, 5 cm
\nThe longest side is 5 cm
\n22<\/sup> + 22<\/sup> = 4 + 4 = 8
\n52<\/sup> = 25
\n52<\/sup> \u2260 22<\/sup> + 22<\/sup>
\n\u2234 The given length cannot be the sides of a right triangle.<\/p>\n

(iii) 1.5 cm, 2 cm, 2.5 cm
\nThe longest side is 2.5 cm
\n1.52<\/sup> + 22<\/sup> = 2.25 + 4
\n= 6.25
\n2.52<\/sup> = 6.25
\n1.52<\/sup> + 22<\/sup> = 2.52<\/sup>
\n(pythagoras property)
\nThe given lengths can be sides of a right triangle.
\nThe right angle is the angle between the sides 1.5 cm and 2 cm.<\/p>\n

Question 5.
\nA tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
\n\"NCERT
\nAnswer:
\nLet the tree BD be broken at the point C, such that CD = CA
\nIn the right triangle ABC, using the pythagoras property; we get
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\nAC2<\/sup> = 122<\/sup> + 52<\/sup>
\n= 144 + 25
\nAC2<\/sup> = 169
\nAC2<\/sup> = 132<\/sup>
\n\u2234 AC = 13
\nNow, height of the tree
\nBD = BC + CD (CD = AC)
\n= 5 m + 13 m
\n= 18 m
\nThe height of the tree is 18 m.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nAngles Q and R of PQR are 25\u00b0 and 65\u00b0. Write which of the following is true:
\n(i) PQ2<\/sup> + QR2<\/sup> = RP2<\/sup>
\n(ii) PQ2<\/sup> + RP2<\/sup> = QR2<\/sup>
\n(iii) RP2<\/sup> + QR2<\/sup> = PQ2<\/sup>
\nAnswer:
\nIn \u0394PQR
\n\u2220P + \u2220Q +\u2220R = 180\u00b0
\n\u2220P + 25\u00b0 + 65\u00b0 = 180\u00b0
\n\"NCERT
\n\u2220P + 90\u00b0 = 180\u00b0
\n\u2220P = 180\u00b0 – 90\u00b0
\n= 90\u00b0
\nSo, \u0394PQR is a triangle right angled at P.
\n\u2234 QR is the hypotenuse.
\n\u2234 using the pythagoras property
\nQR2<\/sup> = PQ2<\/sup> + PR2<\/sup> So,<\/p>\n

(ii) PQ2<\/sup> + RP2<\/sup> = QR2<\/sup> is true<\/p>\n

Question 7.
\nFind the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
\nAnswer:
\nLet the breadth AD be x cm.
\nIn the right triangle ABD,
\n\"NCERT
\nBD2<\/sup> = AD2<\/sup> + AB2<\/sup>
\n412<\/sup>= x2<\/sup> + 402<\/sup>
\nx2<\/sup> = 412<\/sup> – 402<\/sup>
\n= 1681 – 1600
\nx2<\/sup> = 81
\nx2<\/sup> = 92<\/sup>
\nx = 9 cm
\nPerimeter of the rectangle = 2 (l + b)
\n= 2 (40 + 9)
\n= 2 \u00d7 49
\n= 98 cm
\nThus, the perimeter of the rectangle
\n= 98 cm<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nThe diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
\nAnswer:
\nSince the diagonals of a rhombus bisect each other at right angles.
\n\"NCERT
\n\u2220AOB = \u2220BOC = \u2220COD = \u2220AOD = 90\u00b0
\nAO = \\(\\frac { 1 }{ 2 }\\) AC; AO = \\(\\frac { 1 }{ 2 }\\) \u00d7 30 = 15
\n\u2235 AO = OC = 15 cm and BO = OD = 8 cm
\nIn the right \u0394AOB,
\nAB2<\/sup> = AO2<\/sup> + BO2<\/sup>
\nAB2<\/sup> = 152<\/sup> + 82<\/sup>
\nAB2<\/sup> = 225 + 64
\nAB2<\/sup> = 289
\nAB2<\/sup> = 172<\/sup>
\nAB = 17 cm
\n\u2235 Side of the rhombus is 17 cm.
\n\u2234 Perimeter of the rhombus ABCD = 4 \u00d7 17
\n(all the four sides are equal) = 68 cm<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.5 Question 1. PQR is a triangle, right-angled at P. If PQ = 10 …<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties 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":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-7-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 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. 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