{"id":5815,"date":"2021-03-06T10:22:20","date_gmt":"2021-03-06T04:52:20","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=5815"},"modified":"2021-09-15T11:39:28","modified_gmt":"2021-09-15T06:09:28","slug":"ap-ssc-10th-class-maths-solutions-chapter-8-optional-exercise","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-8-optional-exercise\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise"},"content":{"rendered":"<h2>AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Optional Exercise<\/h2>\n<p>AP State Board Syllabus\u00a0<a href=\"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions\/\">AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 8 Similar Triangles Optional Exercise Textbook Questions and Answers.<\/p>\n<h3>10th Class Maths 8th Lesson Coordinate Geometry Optional Exercise Textbook Questions and Answers<\/h3>\n<p>Question 1.<br \/>\nIn the given figure, \\(\\frac{QT}{PR}\\) = \\(\\frac{QR}{QS}\\) and \u22201 = \u22202. Prove that \u25b3PQS ~ \u25b3TQR.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5825\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-1.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 1\" width=\"179\" height=\"139\" \/><br \/>\nAnswer:<br \/>\nGiven: \\(\\frac{QT}{PR}\\) = \\(\\frac{QR}{QS}\\)<br \/>\n\u22201 = \u22202<br \/>\nR.T.P : \u25b3PQS ~ \u25b3TQR<br \/>\nProof: In \u25b3PQR; \u22201 = \u22202 Thus, PQ = PR<br \/>\n[\u2235 sides opp. to equal angles are equal]<br \/>\n\\(\\frac{QT}{PR}\\) = \\(\\frac{QR}{QS}\\) \u21d2 \\(\\frac{QT}{PQ}\\) = \\(\\frac{QR}{QS}\\)<br \/>\ni.e., the line PS divides the two sides QT and QR of \u25b3TQR in the same ratio.<br \/>\nHence, PS \/\/ TR.<br \/>\n[\u2235 If a line join of any two points on any two sides of triangle divides the two sides in the same ratio, then the line is parallel to the third side]<br \/>\nHence, PS \/\/ TR (converse of B.P.T)<br \/>\nNow in \u25b3PQS and \u25b3TQR<br \/>\n\u2220QPS = \u2220QTR<br \/>\n[\u2235 \u2220P, \u2220T are corresponding angles for PS \/\/ TR]<br \/>\n\u2220QSP = \u2220QRT<br \/>\n[\u2235 \u2220S, \u2220R are corresponding angles for PS \/\/ TR]<br \/>\n\u2220Q = \u2220Q (common)<br \/>\n\u2234 \u25b3PQS ~ \u25b3TQR (by AAA similarity)<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 2.<br \/>\nRavi is 1.82 m tall. He wants to find the height of a tree in his backyard. From the tree\u2019s base he walked 12.20 m. along the tree\u2019s shadow to a position where the end of his shadow exactly overlaps the end of the tree\u2019s shadow. He is now 6.10 m from the end of the shadow. How tall is the tree?<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5826\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-2.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 2\" width=\"285\" height=\"130\" \/><br \/>\nAnswer:<br \/>\nGiven:<br \/>\nHeight of Ravi \u2018BC\u2019 = 1.82 m.<br \/>\nDistance of Ravi from the foot of the tree BD = 12.2 m.<br \/>\nLength of the shadow of Ravi = AB = 6.10 m<br \/>\nLet DE represent the tree.<br \/>\nFrom the figure, \u25b3ABC ~ \u25b3ADE.<br \/>\nThus, \\(\\frac{AB}{AD}\\) = \\(\\frac{BC}{DE}\\) = \\(\\frac{AC}{AE}\\)<br \/>\nRatio of corresponding sides of two similar triangles are equal]<br \/>\n\\(\\frac{6.10}{6.10+12.20}\\) = \\(\\frac{1.82}{\\mathrm{DE}}\\)<br \/>\nDE = \\(\\frac{1.82 \\times 18.30}{6.10}\\) = 5.46 m<br \/>\nThus the height of the tree = 5.46 m.<\/p>\n<p>Question 3.<br \/>\nThe diagonal AC of a parallelogram ABCD intersects DP at the point Q, where \u2018P\u2019 is any point on side AB. Prove that CQ \u00d7 PQ = QA \u00d7 QD.<br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5829\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-3.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 3\" width=\"276\" height=\"117\" \/><br \/>\nGiven: \u25a1 ABCD is a parallelogram.<br \/>\nP is a point on AB.<br \/>\nDP and AC intersect at Q.<br \/>\nR.T.P.: CQ . PQ = QA . QD.<br \/>\nProof: In \u25b3CQD, \u25b3AQP<br \/>\n\u2220QCD = \u2220QAP<br \/>\n\u2220CQD = \u2220AQP<br \/>\n\u2234 \u2220QDC = \u2220QPA<br \/>\n(\u2235 Angle sum property of triangles)<br \/>\nThus, \u25b3CQD ~ \u25b3AQP by AAA similarity condition.<br \/>\n\\(\\frac{CQ}{AQ}\\) = \\(\\frac{QD}{QP}\\) = \\(\\frac{CD}{AP}\\)<br \/>\n[\u2235 Ratio of corresponding sides of similar triangles are equal]<br \/>\n\\(\\frac{CQ}{AQ}\\) = \\(\\frac{QD}{QP}\\)<br \/>\nCQ . PQ = QA . QD [Q.E.D]<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 4.<br \/>\n\u25b3ABC and \u25b3AMP are two right triangles right angled at B and M respectively. Prove that (i) \u25b3ABC ~ \u25b3AMP (ii) \\(\\frac{CA}{PA}\\) = \\(\\frac{BC}{MP}\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5830\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-4.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 4\" width=\"215\" height=\"137\" \/><br \/>\nAnswer:<br \/>\nGiven: \u25b3ABC; \u2220B = 90\u00b0<br \/>\nAAMP; \u2220M = 90\u00b0<br \/>\nR.T.P : i) \u25b3ABC ~ \u25b3AMP<br \/>\nProof: In \u25b3ABC and \u25b3AMP<br \/>\n\u2220B = \u2220M [each 90\u00b0 given]<br \/>\n\u2220A = \u2220A [common]<br \/>\nHence, \u2220C = \u2220P<br \/>\n[\u2235 Angle sum property of triangles]<br \/>\n\u2234 \u25b3ABC ~ \u25b3AMP (by A.A.A. similarity)<br \/>\nii) \u25b3ABC ~ \u25b3AMP (already proved)<br \/>\n\\(\\frac{AB}{AM}\\) = \\(\\frac{BC}{MP}\\) = \\(\\frac{CA}{PA}\\)<br \/>\n[\u2235 Ratio of corresponding sides of similar triangles are equal]<br \/>\n\\(\\frac{CA}{PA}\\) = \\(\\frac{BC}{MP}\\)<\/p>\n<p>Question 5.<br \/>\nAn aeroplane leaves an airport and flies due north at a speed of 1000 kmph. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 kmph. How far apart will the two planes be after 1\\(\\frac{1}{2}\\) hour?<br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5836\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-5.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 5\" width=\"202\" height=\"155\" \/><br \/>\nGiven: Speed of the first plane due north = 1000 kmph.<br \/>\nSpeed of the second plane due west = 1200 kmph.<br \/>\nDistance = Speed \u00d7 Time<br \/>\nDistance travelled by the first plane in<br \/>\n1\\(\\frac{1}{2}\\) hrs = 1000 \u00d7 1\\(\\frac{1}{2}\\) = 1000 \u00d7 \\(\\frac{3}{2}\\) = 1500 km.<br \/>\nDistance travelled by the second plane<br \/>\nin 1\\(\\frac{1}{2}\\) hrs = 1200 \u00d7 \\(\\frac{3}{2}\\) = 1800 km.<br \/>\nFrom the figure, \u25b3ABC is a right triangle; \u2220A = 90\u00b0.<br \/>\nAB<sup>2<\/sup> + AC<sup>2<\/sup> = BC<sup>2<\/sup><br \/>\n[Pythagoras theorem]<br \/>\n1500<sup>2<\/sup> + 1800<sup>2<\/sup> = BC<sup>2<\/sup><br \/>\n2250000 + 3240000 = BC<sup>2<\/sup><br \/>\n\u2234 BC = \u221a5490000<br \/>\n= 100 \u00d7 \u221a549 m<br \/>\n\u2245 100 \u00d7 23.43<br \/>\n\u2245 2243km.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 6.<br \/>\nIn a right triangle ABC right angled at C. P and Q are points on sides AC and CB respectively which divide these sides in the ratio of 2:1. Prove that<br \/>\ni) 9AQ<sup>2<\/sup> = 9AC<sup>2<\/sup> + 4BC<sup>2<\/sup><br \/>\nii) 9BP<sup>2<\/sup> = 9BC<sup>2<\/sup> + 4AC<sup>2<\/sup><br \/>\niii) 9(AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 13AB<sup>2<\/sup><br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5846\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-6.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 6\" width=\"182\" height=\"121\" \/><br \/>\nGiven: In \u25b3ABC; \u2220C = 90\u00b0<br \/>\nR.T.P.: i) 9AQ<sup>2<\/sup> = 9AC<sup>2<\/sup> + 4BC<sup>2<\/sup><br \/>\nProof: In \u25b3ACQ; \u2220C = 90\u00b0<br \/>\nAC<sup>2<\/sup> + CQ<sup>2<\/sup> = AQ<sup>2<\/sup><br \/>\n[side<sup>2<\/sup> + side<sup>2<\/sup> = hypotenuse<sup>2<\/sup>]<br \/>\nAQ<sup>2<\/sup> = AC<sup>2<\/sup> + \\(\\left(\\frac{2}{3} \\mathrm{BC}\\right)^{2}\\)<br \/>\n[\u2235 Q divides CB in the ratio 2 : 1<br \/>\nCQ = \\(\\frac{2}{3}\\)BC]<br \/>\nAQ<sup>2<\/sup> = AC<sup>2<\/sup> + \\(\\frac{4}{9} \\mathrm{BC}^{2}\\)<br \/>\nAQ<sup>2<\/sup> = \\(\\frac{9 \\mathrm{AC}^{2}+4 \\mathrm{BC}^{2}}{9}\\)<br \/>\n\u21d2 9AQ<sup>2<\/sup> = 9AC<sup>2<\/sup> + 4BC<sup>2<\/sup><\/p>\n<p>ii) 9BP<sup>2<\/sup> = 9BC<sup>2<\/sup> + 4AC<sup>2<\/sup><br \/>\nProof: In \u25b3PCB,<br \/>\nPB<sup>2<\/sup> = PC<sup>2<\/sup> + BC<sup>2<\/sup> [Pythagoras theorem]<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5847\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-7.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 7\" width=\"233\" height=\"168\" \/><br \/>\n\u21d2 PB<sup>2<\/sup> = AC<sup>2<\/sup> + 9BC<sup>2<\/sup><br \/>\n[!! If we take P on CA, in the ratio 2 : 1 then we get<br \/>\nBP<sup>2<\/sup> = PC<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nBP<sup>2<\/sup> = \\(\\left(\\frac{2}{3} \\mathrm{A}\\right)^{2}\\) + BC<sup>2<\/sup><br \/>\nBP<sup>2<\/sup> = \\(\\frac{4}{9} \\mathrm{AC}^{2}\\) + BC<sup>2<\/sup><br \/>\nBP<sup>2<\/sup> = \\(\\frac{4 \\mathrm{AC}^{2}+9 \\mathrm{BC}^{2}}{9}\\)<br \/>\n9BP<sup>2<\/sup> = 4AC<sup>2<\/sup> + 9BC<sup>2<\/sup><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise\" width=\"196\" height=\"17\" \/><\/p>\n<p>iii) 9 (AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 13 AB<sup>2<\/sup><br \/>\nProof: In \u25b3ABC,<br \/>\nAC<sup>2<\/sup> + BC<sup>2<\/sup> = AB<sup>2<\/sup><br \/>\n[Pythagoras theorem]<br \/>\nAlso, from (i) and (ii),<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-5848\" src=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-8.png\" alt=\"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Optional Exercise 8\" width=\"350\" height=\"170\" srcset=\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-8.png 350w, https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-8-Similar-Triangles-Optional-Exercise-8-300x146.png 300w\" sizes=\"(max-width: 350px) 100vw, 350px\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Optional Exercise AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Optional Exercise Textbook Questions and Answers. 10th Class Maths 8th Lesson Coordinate Geometry Optional Exercise Textbook Questions and Answers Question 1. In the given figure, = and &hellip;<\/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\/5815"}],"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=5815"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5815\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=5815"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=5815"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=5815"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}