{"id":31856,"date":"2018-10-08T11:58:56","date_gmt":"2018-10-08T11:58:56","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=31856"},"modified":"2020-11-25T14:58:36","modified_gmt":"2020-11-25T09:28:36","slug":"ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test\/","title":{"rendered":"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test<\/h2>\n

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths<\/a>. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test.<\/p>\n

ML Aggarwal Solutions<\/a>ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n

Question 1.<\/strong><\/span>
\nIn the given figure, \u22201 = \u22202 and \u22203 = \u22204. Show that PT x QR = PR x ST.<\/strong>
\nSolution:<\/strong><\/span>
\nGiven : In the given figure,
\n\u22201 = \u22201 and \u22203 = \u22204
\n\"ML
\nTo prove : PT x QR = PR x ST
\nProof: \u22201 = \u22202
\nAdding \u22206 to both sides
\n\u22201 + \u22206 = \u22202 + \u22206
\n\u2220SPT = \u2220QPR
\nIn \u2206PQR and \u2206PST
\n\"ML<\/p>\n

Question 2.<\/strong><\/span>
\nIn the adjoining figure, AB = AC. If PM \u22a5 AB and PN \u22a5 AC, show that PM x PC = PN x PB.<\/strong>
\n\"ML
\nSolution:<\/strong><\/span>
\nGiven : In the given figure,
\nAB = AC, PM \u22a5 AB and PN \u22a5 AC
\nTo prove : PM x PC = PN x PB
\nProof: In \u2206ABC, AB = AC
\n\u2220B = \u2220C
\nNow in \u2206CPN and \u2206BPM,
\n\"ML<\/p>\n

Question 3.<\/strong><\/span>
\n(a) In the figure (1) given below. \u2220AED = \u2220ABC. Find the values of x and y.<\/strong>
\n(b) In the fig. (2) given below, CD = \\(\\\\ \\frac { 1 }{ 2 } \\) AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :<\/strong>
\n(i) CE || AG<\/strong>
\n(ii) 3 ED = GD.<\/strong>
\n\"ML
\nSolution:<\/strong><\/span>
\n(a) Given : In following figure, \u2220AED = \u2220ABC
\nRequired : The values of x and y.
\nNow, in \u2206ABC and \u2206ADE
\n\u2220AED = \u2220ABC (given)
\n\u2220A = \u2220A (common)
\n\u2234 \u2206ABC ~ \u2206ADE
\n(By A.A. axiom of similarity)
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 4.<\/strong><\/span>
\nIn the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:<\/strong>
\n(i) DF || BH<\/strong>
\n(ii) AH = 3 AF.<\/strong>
\n\"ML
\nSolution:<\/strong><\/span>
\nGiven : E is the mid-point of BD and F is mid-point of AC also 2 AD = BD and EC || BH
\nTo Prove : (i) DF || BH
\n(ii) AH = 3 AF
\n\"ML
\n\"ML<\/p>\n

Question 5.<\/strong><\/span>
\nIn a \u2206ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.<\/strong>
\nSolution:<\/strong><\/span>
\nGiven : In \u2206ABC, D and E are the points on the sides AB and AC respectively
\nDE || BC
\nAD = 2.4 cm, AE = 3.2 cm, DE = 2 cm, BC = 5 cm
\n\"ML
\n\"ML<\/p>\n

Question 6.<\/strong><\/span>
\nIn a \u2206ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.<\/strong>
\nSolution:<\/strong><\/span>
\nIn \u2206ABC, D and E are points on the sides AB and AC respectively
\nAD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm
\n\"ML
\n\"ML<\/p>\n

Question 7.<\/strong><\/span>
\nIn a \u2206ABC, DE is parallel to the base BC, with D on AB and E on AC. If \\(\\frac { AD }{ DB } =\\frac { 2 }{ 3 } ,\\frac { BC }{ DE } \\)<\/strong>
\nSolution:<\/strong><\/span>
\nIn \u2206ABC, DE || BC
\nD is on AB and E is on AC
\n\"ML
\n\"ML<\/p>\n

Question 8.<\/strong><\/span>
\nIf the area of two similar triangles are 360 cm\u00b2 and 250 cm\u00b2 and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.<\/strong>
\nSolution:<\/strong><\/span>
\nLet \u2206ABC and \u2206DEF are similar and area of
\n\u2206ABC = 360 cm\u00b2
\n\"ML
\n\"ML<\/p>\n

Question 9.<\/strong><\/span>
\nIn the adjoining figure, D is a point on BC such that \u2220ABD = \u2220CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find<\/strong>
\n(i) BC<\/strong>
\n(ii) DC<\/strong>
\n(iii) area of \u2206ACD : area of \u2206BCA.<\/strong>
\n\"ML
\nSolution:<\/strong><\/span>
\nIn \u2206ABC and \u2206ACD
\n\u2220C = \u2220C (Common)
\n\u2220ABC = \u2220CAD (given)
\n\u2234 \u2206ABC ~ \u2206ACD
\n\"ML
\n\"ML<\/p>\n

Question 10.<\/strong><\/span>
\nIn the adjoining figure the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of DAOE : area of ||gm ABCD.<\/strong>
\n\"ML
\nSolution:<\/strong><\/span>
\n(a) In the figure
\nDiagonals of parallelogram ABCD are AC and BD which intersect each other at O.
\nOE is drawn parallel to CB to meet AB in E.
\n\"ML
\n\"ML<\/p>\n

Question 11.<\/strong><\/span>
\nIn the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of \u2206AOB and \u2206COD.<\/strong>
\n\"ML
\nSolution:<\/strong><\/span>
\nIn the given figure, ABCD is trapezium in
\nwhich AB || DC, 2AB = 3DC
\n\"ML<\/p>\n

Question 12.<\/strong><\/span>
\nIn the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .<\/strong>
\n\"ML
\n(i) DO : OE = 2 : 1<\/strong>
\n(ii) area of \u2206OEC : area of \u2206OAD = 1 : 4.<\/strong>
\nSolution:<\/strong><\/span>
\nGiven : In || gm ABCD,
\nE is mid point of BC and DE meets the diagonal AC at O and meet AB produced at F.
\nTo prove : (i) DO : OE = 2 : 1
\n(ii) area of \u2206OEC : area of \u2206OAD = 1 : 4
\nProof: In \u2206AOD and \u2206EDC
\n\u2220AOD = \u2220EOC (vertically opposite angle)
\n\u2220OAD = \u2220OCB (alt. angles)
\n\u2206AOD ~ \u2206EOC (AA postulate)
\n\"ML<\/p>\n

Question 13.<\/strong><\/span>
\nA model of a ship is made to a scale of 1 : 250. Calculate :<\/strong>
\n(i) the length of the ship, if the length of model is 1.6 m.<\/strong>
\n(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m\u00b2.<\/strong>
\n(iii) the volume of the model, if the volume of the ship is 1 km\u00b3.<\/strong>
\nSolution:<\/strong><\/span>
\nScale factor (k) of the model of the ship = \\(\\\\ \\frac { 1 }{ 250 } \\)
\n(i) Length of model = 1.6 m
\n\"ML<\/p>\n

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test are helpful to complete your math homework.<\/p>\n

If you have any doubts, please comment below. APlusTopper <\/a>try to provide online math tutoring for you.<\/p>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test. ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions Question 1. In the given figure, […]<\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[6768],"tags":[],"yoast_head":"\nML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test - A Plus Topper<\/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:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. 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Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test. ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions Question 1. In the given figure, […]","og_url":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test\/","og_site_name":"A Plus Topper","article_publisher":"https:\/\/www.facebook.com\/aplustopper\/","article_published_time":"2018-10-08T11:58:56+00:00","article_modified_time":"2020-11-25T09:28:36+00:00","og_image":[{"url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/10\/ML-Aggarwal-Class-10-Solutions-for-ICSE-Maths-Chapter-14-Similarity-Chapter-Test-Q1.1.png"}],"twitter_card":"summary","twitter_misc":{"Written by":"Nirmala","Est. reading time":"4 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/www.aplustopper.com\/#organization","name":"Aplus Topper","url":"https:\/\/www.aplustopper.com\/","sameAs":["https:\/\/www.facebook.com\/aplustopper\/"],"logo":{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/#logo","inLanguage":"en-US","url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","contentUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","width":1585,"height":375,"caption":"Aplus Topper"},"image":{"@id":"https:\/\/www.aplustopper.com\/#logo"}},{"@type":"WebSite","@id":"https:\/\/www.aplustopper.com\/#website","url":"https:\/\/www.aplustopper.com\/","name":"A Plus Topper","description":"Improve your Grades","publisher":{"@id":"https:\/\/www.aplustopper.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.aplustopper.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test\/#primaryimage","inLanguage":"en-US","url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/10\/ML-Aggarwal-Class-10-Solutions-for-ICSE-Maths-Chapter-14-Similarity-Chapter-Test-Q1.1.png","contentUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/10\/ML-Aggarwal-Class-10-Solutions-for-ICSE-Maths-Chapter-14-Similarity-Chapter-Test-Q1.1.png","width":223,"height":215,"caption":"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q1.1"},{"@type":"WebPage","@id":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test\/#webpage","url":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-14-chapter-test\/","name":"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test - 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