{"id":15142,"date":"2020-09-09T15:25:17","date_gmt":"2020-09-09T09:55:17","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=15142"},"modified":"2021-07-02T15:01:54","modified_gmt":"2021-07-02T09:31:54","slug":"kseeb-solutions-for-class-7-maths-chapter-6-ex-6-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-6-ex-6-5\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5"},"content":{"rendered":"

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5<\/h2>\n

Question 1.
\nPQR is a triangle, right-angled at P. If PQ = 10cm and PR = 24 cm, find QR.
\nSolution:
\n\"KSEEB
\nPQR is a right angled triangle. \u2220RPQ = 90\u00b0
\n\u2234 QR2<\/sup>\u00a0= PQ2<\/sup>\u00a0+ PR2<\/sup>\u00a0(By pythagoras property)
\n= 102<\/sup>\u00a0+ 242<\/sup>
\n= 100 + 576
\n\u2234 QR =\u00a0\"\\sqrt{676}\"\u00a0= 26 cms<\/p>\n

Question 2.
\nABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
\nSolution:
\n\"KSEEB
\nABC is a right-angled triangle.
\n\u2220ACB = 90\u00b0
\n\u2234 AB2<\/sup> = AC2<\/sup> + BC2<\/sup> (By pythagoras property)
\n252<\/sup> = 72<\/sup> + BC2<\/sup>
\nBC2<\/sup> = 252<\/sup> – 72 <\/sup>= 625 – 49
\n\u2234 BC = \\(\\sqrt{576}\\) = 24
\n\u2234 BC = 24 cms<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nA 15m 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.
\nSolution:
\n\"KSEEB
\nThe distance of the foot of the ladder from the wall is \u2018a\u2019 m.
\nThen
\na2<\/sup>\u00a0+ 122<\/sup>\u00a0= 122<\/sup>\u00a0(pythagoras property)
\na2<\/sup>\u00a0= 152<\/sup>\u00a0\u2013 122<\/sup>
\na2<\/sup>\u00a0= 225 \u2013 144
\na2<\/sup>\u00a0= 81
\n\u2234 a =\u00a0\"\\sqrt{81}=9
\n\u2234 The distance of the foot of the ladder from the wall be 9m.<\/p>\n

Question 4.
\nWhich of the following can be the sides of a right triangle?
\nSolution:
\nIn the case of right-angled triangles, identify the right angles.
\ni) 2.5 cm, 6.5 cm, 6 cm
\nSolution:
\nThe greatest side is hypotenuse = 6.5 cm
\n(2.5)2<\/sup> + (6)2<\/sup> = (6.5)2<\/sup> (According to Pythagoras property)
\n6.25 + 36 = 42.25
\n42.25 = 42.25
\n\u2234 This is a right-angled triangle. The given length can be the sides of a right-angled triangle. The right angle between the lengths of 2.5 cm and 6 cm.<\/p>\n

\"KSEEB<\/p>\n

ii) 2 cm, 2 cm, 5 cm
\nThe greatest side is 5 cm
\n(2)2<\/sup> + (2)2<\/sup> = (5)2<\/sup> (According to pythagoras property)
\n4 + 4 \u2260 25
\n\u2234 It is not the lengths of right angled triangle.<\/p>\n

iii) 1.5 cm, 2 cm, 2.5 cms
\nThe greatest side is 2.5 cms.
\nAccording to pythagoras property
\n(2.5)2<\/sup> = (1.5)2<\/sup> + (2)2<\/sup>
\n6.25 = 2.25 + 4
\n6.25 = 6.25
\nIt is a right angled trianges, the given lengths are sides of the right angled triangles.
\n\u2234 The right angle between the lengths of 1.5 cms and 2 cms.<\/p>\n

Question 5.
\nA Tree is broken at a height of 5m from the ground and its top touches the ground at a distance of 12 m from the base tree. Find the original height of the tree.
\nSolution:
\n\"KSEEB
\nAccording to the given statement the sketch we get is right angled triangle
\nRQ2<\/sup>\u00a0= PR2<\/sup>\u00a0+ PQ2<\/sup>
\n= 52<\/sup>\u00a0+ 122<\/sup>
\n= 25 + 144 = 169
\n\u2234 RQ =\u00a0\"\\sqrt{169}\"\u00a0= 13m
\n\u2234 The original height of the tree is PR + RS = 5 + 13 = 18 m<\/p>\n

Question 6.
\nAngles Q and R of a \u2206 PQR are 25\u00b0 and 65\u00b0.
\nWrite which of the following is true :
\nSolution:
\n\"KSEEB
\n\u2220p = 90\u00b0
\n\u2234 According to Pythagoras property<\/p>\n

ii) PQ2<\/sup> + RP2<\/sup> = QR2<\/sup>
\nPQ2<\/sup> + RP2<\/sup> = QR2<\/sup> is true<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFind the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
\nSolution:
\n\"KSEEB
\nPQRS is a rectangle.
\nAccording to the figure, PQS is a right-angled triangle.
\n\u2234 QS2<\/sup>\u00a0= PQ2<\/sup>\u00a0+ PS2<\/sup>
\n412<\/sup>\u00a0= 402<\/sup>\u00a0+ PS2<\/sup>
\n\u2234 PS2<\/sup>\u00a0= 422<\/sup>\u00a0+ 402<\/sup>
\n= 1681 \u2013 1600
\nPS =\u00a0\"\\sqrt{81}\"\u00a0= 9 cms
\nThe perimeter of the rectangle = 2 (PQ + PS)
\n= 2 (40 + 9) = 2 (49)
\n= 98 cms<\/p>\n

Question 8.
\nThe diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
\nSolution:
\n\"KSEEB
\nPQRS is a rhombus. The diagonals are PQ & RS.
\nPQ = 30 cms RS = 16 cms
\nAccording to fig In Rhombus the diagonals bisect
\nEach other. \u2234 PQ & RS bisect at O.
\nPO = OQ SO = OR
\n\"KSEEB
\nConsidering the \u2206 POR
\nPR2<\/sup> = PO2<\/sup> + OR2<\/sup>
\n= 152<\/sup> + 82<\/sup>
\n= 225 + 64 = 289
\n\u2234 PR = \\(\\sqrt{289}\\) = 17cms
\n\u2234 The perimeter of the rhombus = 4 \u00d7 side PR
\n= 4 \u00d7 17 = 68 cms<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15142"}],"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=15142"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15142\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=15142"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=15142"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=15142"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}