{"id":3609,"date":"2021-02-23T11:08:06","date_gmt":"2021-02-23T05:38:06","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3609"},"modified":"2021-09-15T11:40:42","modified_gmt":"2021-09-15T06:10:42","slug":"ap-ssc-10th-class-maths-solutions-chapter-8-ex-8-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-8-ex-8-1\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1<\/h2>\n

10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers<\/h3>\n

Question 1.
\nIn \u25b3PQR, ST is a line such that \\(\\frac{PS}{SQ}\\) = \\(\\frac{PT}{TR}\\) and also \u2220PST = \u2220PRQ.
\nProve that \u25b3PQR is an isosceles triangle.
\nAnswer:
\nGiven : In \u25b3PQR,
\n\\(\\frac{PS}{SQ}\\) = \\(\\frac{PT}{TR}\\) and \u2220PST= \u2220PRQ.
\n\"AP
\nR.T.P: \u25b3PQR is isosceles.
\nProof: \\(\\frac{PS}{SQ}\\) = \\(\\frac{PT}{TR}\\)
\nHence, ST || QR (Converse of Basic proportionality theorem)
\n\u2220PST = \u2220PQR …….. (1)
\n(Corresponding angles for the lines ST || QR)
\nAlso, \u2220PST = \u2220PRQ ……… (2) given
\nFrom (1) and (2),
\n\u2220PQR = \u2220PRQ
\ni.e., PR = PQ
\n[\u2235 In a triangle sides opposite to equal angles are equal]
\nHence, APQR is an isosceles triangle.<\/p>\n

\"AP<\/p>\n

Question 2.
\nIn the given figure, LM || CM and LN || CD. Prove that \\(\\frac{AM}{AB}\\) = \\(\\frac{AN}{AD}\\).
\n\"AP
\nAnswer:
\nGiven : LM || CB and LN || CD In \u25b3ABC, LM || BC (given) Hence,
\n\"AP
\nAdding ‘1’ on both sides.
\n\"AP
\nFrom (1) and (2)
\n\u2234 \\(\\frac{AM}{AB}\\) = \\(\\frac{AN}{AD}\\).<\/p>\n

Question 3.
\nIn the given figure, DE || AC and DF || AE. Prove that \\(\\frac{BF}{FE}\\) = \\(\\frac{BE}{AC}\\).
\n\"AP
\nAnswer:
\nIn \u25b3ABC, DE || AC
\nHence \\(\\frac{BE}{EC}\\) = \\(\\frac{BD}{DA}\\) ………. (1)
\n[\u2235 A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]
\nAlso in \u25b3ABE, DF || AE
\nHence \\(\\frac{BF}{FE}\\) = \\(\\frac{BD}{DA}\\) ………. (2)
\nFrom (1) and (2), \\(\\frac{BF}{FE}\\) = \\(\\frac{BE}{AC}\\) Hence proved.<\/p>\n

\"AP<\/p>\n

Question 4.
\nProve that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).
\nAnswer:
\n\"AP
\nGiven: In \u25b3ABC; D is the mid-point of AB.
\nA line ‘l’ through D, parallel to BC, meeting AC at E.
\nR.T.P: E is the midpoint of AC.
\nProof:
\nDE || BC (Given)
\nthen
\n\\(\\frac{AD}{DB}\\) = \\(\\frac{AE}{EC}\\)(From Basic Proportional theorem)
\nAlso given \u2018D\u2019 is mid point of AB.
\nThen AD = DB.
\n\u21d2 \\(\\frac{AD}{DB}\\) = \\(\\frac{DB}{DB}\\) = \\(\\frac{AE}{EC}\\) = 1
\n\u21d2 AE = EC
\n\u2234 \u2018E\u2019 is mid point of AC
\n\u2234 The line bisects the third side \\(\\overline{\\mathrm{AC}}\\).
\nHence proved.<\/p>\n

Question 5.
\nProve that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)
\nAnswer:
\nGiven: \u25b3ABC, D is the midpoint of AB and E is the midpoint of AC.
\nR.T.P : DE || BC.
\nProof:
\n\"AP
\nSince D is the midpoint of AB, we have AD = DB \u21d2 \\(\\frac{AD}{DB}\\) = 1 ……. (1)
\nalso \u2018E’ is the midpoint of AC, we have AE = EC \u21d2 \\(\\frac{AE}{EC}\\) = 1 ……. (2)
\nFrom (1) and (2)
\n\\(\\frac{AD}{DB}\\) = \\(\\frac{AE}{EC}\\)
\nIf a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.
\n\u2234 DE || BC by Basic proportionality theorem.
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 6.
\nIn the given figure, DE || OQ and DF || OR. Show that EF || QR.
\n\"AP
\nAnswer:
\nGiven: \u25b3PQR, DE || OQ; DF || OR
\nR.T.P: EF || QR
\nProof:
\nIn \u25b3POQ;
\n\\(\\frac{PE}{EQ}\\) = \\(\\frac{PD}{DO}\\) ……. (1)
\n[\u2235 ED || QO, Basic proportionality theorem]
\nIn \u25b3POR; \\(\\frac{PF}{FR}\\) = \\(\\frac{PD}{DO}\\) ……. (2) [\u2235 DF || OR, Basic Proportionality Theorem]
\nFrom (1) and (2),
\n\\(\\frac{PE}{EQ}\\) = \\(\\frac{PF}{FR}\\)
\nThus the line \\(\\overline{\\mathrm{EF}}\\) divides the two sides PQ and PR of \u25b3PQR in the same ratio.
\nHence, EF || QR. [\u2235 Converse of Basic proportionality theorem]<\/p>\n

Question 7.
\nIn the given figure A, B and C are points on OP, OQ and OR respec\u00actively such that AB || PQ and AC || PR. Show that BC || QR.
\n\"AP
\nAnswer:
\nGiven:
\nIn \u25b3PQR, AB || PQ; AC || PR
\nR.T.P : BC || QR
\nProof: In \u25b3POQ; AB || PQ
\n\\(\\frac{OA}{AP}\\) = \\(\\frac{OB}{BQ}\\) ……… (1)
\n(\u2235 Basic Proportional theorem)
\nand in \u25b3OPR, Proof: In \u25b3POQ; AB || PQ
\n\\(\\frac{OA}{AP}\\) = \\(\\frac{OC}{CR}\\) ……… (2)
\nFrom (1) and (2), we can write
\n\\(\\frac{OB}{BQ}\\) = \\(\\frac{OC}{CR}\\)
\nThen consider above condition in \u25b3OQR then from (3) it is clear.
\n\u2234 BC || QR [\u2235 from converse of Basic Proportionality Theorem]
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 8.
\nABCD is a trapezium in which AB || DC and its diagonals intersect each other at point \u2018O’. Show that\\(\\frac{AO}{BO}\\) = \\(\\frac{CO}{DO}\\).
\nAnswer:
\nGiven: In trapezium \u25a1 ABCD, AB || CD. Diagonals AC, BD intersect at O.
\nR.T.P: \\(\\frac{AO}{BO}\\) = \\(\\frac{CO}{DO}\\)
\n\"AP
\nConstruction:
\nDraw a line EF passing through the point \u2018O\u2019 and parallel to CD and AB.
\nProof: In \u25b3ACD, EO || CD
\n\u2234 \\(\\frac{AO}{CO}\\) = \\(\\frac{AE}{DE}\\) …….. (1)
\n[\u2235 line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
\nIn \u25b3ABD, EO || AB
\nHence, \\(\\frac{DE}{AE}\\) = \\(\\frac{DO}{BO}\\)
\n[\u2235 Basic proportionality theorem]
\n\\(\\frac{BO}{DO}\\) = \\(\\frac{AE}{ED}\\) …….. (2) [\u2235 Invertendo]
\nFrom (1) and (2),
\n\\(\\frac{AO}{CO}\\) = \\(\\frac{BO}{DO}\\)
\n\u21d2 \\(\\frac{AO}{BO}\\) = \\(\\frac{CO}{DO}\\) [\u2235 Alternendo]<\/p>\n

\"AP<\/p>\n

Question 9.
\nDraw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
\nAnswer:
\n\"AP
\nSteps of construction:<\/p>\n

    \n
  1. Draw a line segment \\(\\overline{\\mathrm{AB}}\\) of length 7.2 cm.<\/li>\n
  2. Construct an acute angle \u2220BAX at A.<\/li>\n
  3. Mark off 5 + 3 = 8 equal parts (A1<\/sub>, A2<\/sub>, …., A8<\/sub>) on \\(\\stackrel{\\leftrightarrow}{\\mathrm{AX}}\\) with same radius.<\/li>\n
  4. Join A8<\/sub> and B.<\/li>\n
  5. Draw a line parallel to \\(\\stackrel{\\leftrightarrow}{\\mathrm{A}_{8} \\mathrm{~B}}\\) at A5<\/sub> meeting AB at C.<\/li>\n
  6. Now the point C divides AB in the ratio 5:3.<\/li>\n
  7. Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.<\/li>\n<\/ol>\n

     <\/p>\n","protected":false},"excerpt":{"rendered":"

    AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1 10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers Question 1. In \u25b3PQR, ST is a line …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3609"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=3609"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3609\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3609"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3609"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3609"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}