{"id":21737,"date":"2022-01-12T17:19:27","date_gmt":"2022-01-12T11:49:27","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=21737"},"modified":"2022-01-12T17:19:27","modified_gmt":"2022-01-12T11:49:27","slug":"ap-board-7th-class-maths-solutions-chapter-5-exercise-5-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-5-exercise-5-4\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 5 Triangles Exercise 5.4"},"content":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf<\/a> Chapter 5 Triangles Ex 5.4 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.4<\/h2>\n

Question 1.
\nTwo sides of a triangle are 5cm and 4cm respectively. Write any three possible measurements that suit for the third side.
\nAnswer:
\nGiven two sides of a triangle are 5 cm and 4 cm.
\nWe know that, in triangle sum of lengths of any two sides is always greater than the third side and in the triangle difference of lengths of any two sides is always less than the third side.
\nSo, 5 cm, 4 cm, 2 cm
\n5 cm, 4 cm, 3 cm
\n5 cm, 4 cm, 4 cm
\n5 cm, 4 cm, 5 cm
\n5 cm, 4 cm, 6 cm
\n5 cm, 4 cm, 7 cm
\nMeasurements that suit for third side are 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 7 cm.<\/p>\n

Question 2.
\nThe lengths of line segments are 3 cm, 5 cm, 6 cm and 9 cm.
\n(i) From the above measurements which group of the line segments can form a triangle.
\nAnswer:
\n3 cm, 5 cm, 6 cm and 5 cm, 6 cm, 9 cm groups form triangle.
\nBecause, sum of any two sides is greater than the third side in this group.<\/p>\n

(ii) Which group of line segments cannot form a triangle, give the reason?
\nAnswer:
\nWe know that, sum of the lengths of any two sides is always greater than the third side.
\n3 cm, 6 cm, 9 cm does not form a triangle.
\nBecause sum of 3 cm and 6 cm is not greater than 9 cm.<\/p>\n

\"AP<\/p>\n

Question 3.
\nFind the value of \u2018x\u2019 in the following triangles :
\n\"AP
\nAnswer:
\nGiven, in \u2206PEN, PE = 4 cm, PN = 4 cm \u2220E = 65\u00b0, \u2220N = x\u00b0
\nTwo sides are equal in length.
\nSo, \u2206PEN is an isosceles triangle.<\/p>\n

In isosceles triangle, the angles which are opposite to equal sides are also equal.
\nSo, \u2220N = \u2220E
\n\u2234 x = 65\u00b0<\/p>\n

\"AP
\nAnswer:
\nGiven, in \u2206BGA,
\n\u2220B = 56\u00b0, \u2220G = x, \u2220A = 56\u00b0 and BG = x cm, GA = 4.3 cm
\nWe know that, sum of angles of a triangle is 180\u00b0.
\nIn \u2206BGA, \u2220B + \u2220G + \u2220A = 180\u00b0
\n\u21d2 56\u00b0 + x + 56\u00b0 = 180\u00b0
\n\u21d2 112\u00b0 + x = 180\u00b0 :
\n\u21d2 112\u00b0 + x – 112\u00b0 =180\u00b0- 112\u00b0
\n\u2234 x = 68\u00b0<\/p>\n

Two angles are equal in \u2206BGA.
\nSo, \u2206BGA is an isosceles triangle.
\nIn isosceles triangle the sides which are opposite to equal angles are also equal.
\nSo, BG = GA
\n\u2234 x = 4.3 cm<\/p>\n

Question 4.
\n\u2206ABC is an isosceles triangle in which AB = AC. If \u2220A = 80\u00b0, then find \u2220B and \u2220C.
\nAnswer:
\n\"AP
\nIn isosceles triangle the angles which are opposite to equal sides are also equal.
\nSo, \u2220B = \u2220C = x\u00b0
\nIn \u2206ABC, we know
\n\u2220A + \u2220B + \u2220C \u2014 180\u00b0
\n\u21d2 80\u00b0 + x + x = 180\u00b0
\n\u21d2 80\u00b0 + 2x – 80\u00b0 = 180\u00b0 – 80\u00b0
\n\u21d2 2x = 100\u00b0
\n\u21d2 \\(\\frac{2 x}{2}=\\frac{100^{\\circ}}{2}\\)
\n\u2234 x = 50\u00b0
\n\u2234 \u2220B = \u2220C = 50\u00b0<\/p>\n

\"AP<\/p>\n

Question 5.
\nFind the values of ‘x’ in each of the below triangles.
\n\"AP
\nAnswer:
\nIn \u2206PQR, PQ = QR and \u2220P = 53\u00b0
\n\u2206PQR is an isosceles triangle.<\/p>\n

In isosceles triangle, the angles which are opposite to equal sides are also equal.
\nSo, \u2220P = \u2220R = 53\u00b0
\nIn \u2206PQR, we know
\n\u2220P + \u2220Q + \u2220R = 180\u00b0
\n\u21d2 53\u00b0 + x + 53\u00b0 = 180\u00b0
\n\u21d2 106\u00b0 + x = 180\u00b0
\n\u21d2 106\u00b0 + x- 106\u00b0 = 180\u00b0- 106\u00b0
\n\u2234 x = 74\u00b0<\/p>\n

\"AP
\nAnswer:
\nIn \u2206LMN, LM = LN and exterior angle at L = 110\u00b0 .
\n\u2206LMN is an isosceles triangle.
\nIn isosceles triangle, the angles which are opposite to equal sides are also equal.<\/p>\n

So, \u2220M = \u2220N = x\u00b0 .
\nWe know that an exterior angle of a triangle is equal to the sum of its opposite interior angles.
\nExterior angle at
\nL = \u2220M + \u2220N = 110\u00b0
\n\u21d2 x + x = 110\u00b0
\n\u21d2 2x = 110\u00b0
\n\u21d2 \\(\\frac{2 x}{2}=\\frac{110}{2}\\)
\n\u2234 x = 55\u00b0<\/p>\n

Question 6.
\nWhich of the following statements are true in the following diagram.
\n\"AP
\n(i) OY < OT
\nAnswer:
\n\"AP
\nIn \u2206TOY, \u2220T + \u2220O + \u2220Y = 180\u00b0
\n\u21d2 \u2220T + 90\u00b0 + 50\u00b0 = 180\u00b0
\n\u21d2 \u2220T + 140\u00b0 = 180\u00b0
\n\u21d2 \u2220T + 140\u00b0- 140\u00b0 = 180\u00b0 – 140\u00b0
\n\u2234 \u2220T – 40\u00b0
\nWe know that side opposite to the largest angle is longest.
\nOY< OT – True<\/p>\n

(ii) TY < TO
\nAnswer:
\nTY < TO False<\/p>\n

(iii) \u2220Y < \u2220T
\nAnswer:
\n\u2220Y < \u2220T
\n50\u00b0 < 40\u00b0 – False<\/p>\n

(iv) TY < OY
\nAnswer:
\nTY < OY – False<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.4 Textbook Exercise Questions and Answers. AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.4 Question 1. Two sides of a triangle are 5cm and 4cm respectively. Write any three possible measurements that suit for the third side. Answer: Given two sides …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/21737"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=21737"}],"version-history":[{"count":4,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/21737\/revisions"}],"predecessor-version":[{"id":21830,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/21737\/revisions\/21830"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=21737"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=21737"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=21737"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}