{"id":46983,"date":"2023-04-25T15:20:10","date_gmt":"2023-04-25T09:50:10","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=46983"},"modified":"2023-04-27T11:05:16","modified_gmt":"2023-04-27T05:35:16","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-12-chapter-test","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-12-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 12 Pythagoras Theorem Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 12 Pythagoras Theorem Chapter Test<\/h2>\n

Question 1.
\n(a) In fig. (i) given below, AD \u22a5 BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.
\n(b) In figure (ii) given below, \u2220BAC = 90\u00b0,
\n\u2220ADC = 90\u00b0, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find :
\n(i) AC
\n(ii) AB
\n(iii) area of the shaded region.
\n(c) In figure (iii) given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate
\n(i) the length of BC (ii) the area of \u2206ADE.
\n\"ML
\nAnswer: (a) Given. In \u2206 ABC , AD \u22a5 BC, AB = 25 cm, AC = 17 cm and AD = 15 cm
\nRequired. The length of BC.
\nSolution:
\nIn right angled \u2206ABD,
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup> (By Pythagoras theorem)
\n\u2234 BD2<\/sup> = AB2<\/sup> – AD2<\/sup>
\n= (25)2<\/sup> – (15)2<\/sup>
\n= 625 – 225 = 400
\n\u21d2 BD = \\(\\sqrt{400}\\) = 20 cm.
\nNow, in right angled \u2206 ADC
\nAC2<\/sup> = AD2<\/sup> + DC2<\/sup> (By Pythagoras theorem)
\n\u2234 DC2<\/sup>= AC2<\/sup> – AD2<\/sup>
\n\u21d2 DC2<\/sup> = (17)2<\/sup> – (15)2<\/sup>
\n\u21d2 DC2<\/sup> = 289 – 225 = 64
\nDC = \\(\\sqrt{64}\\) = 8cm
\nHence, BC = BD + DC = 20 cm + 8 cm = 28 cm.<\/p>\n

(b) Given. In \u2206 ABC,
\n\u2220BAC = 90\u00b0, \u2220 ADC = 90\u00b0 AD = 6cm, CD = 8 cm and BC = 26 cm.
\nRequired. (i) AC (ii) AB
\n(iii) area of the shaded region
\nSolution:
\nIn right angled \u2206 ADC
\nAC2<\/sup> = AD2<\/sup> + DC2<\/sup> (By Pythagoras theorem)
\n= (6)2<\/sup> + (8)2<\/sup>
\n= 36 + 64 = 100
\n\u2234 = \\(\\sqrt{10}\\) = 10 cm Answer:
\nIn right angled \u2206 ABC
\nBC2<\/sup> = AB2<\/sup> + AC2<\/sup> (By Pythagoras theorem)
\n\u21d2 (26)2<\/sup> = AB2<\/sup> + (10)2<\/sup>
\n\u21d2 AB2<\/sup> = (26)2<\/sup> – (10)2<\/sup>
\n\u21d2 AB2<\/sup> = 676 – 100 = 576
\n\u21d2 AB2<\/sup> = 576
\n\u21d2 AB = \\(\\sqrt{576}\\) = 24 cm
\nNow, Area of A ABC = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 AC
\n= \\(\\frac{1}{2}\\) \u00d7 24 \u00d7 10 cm2<\/sup> = 12 \u00d7 10 cm2<\/sup> = 120 cm2<\/sup>
\nArea of \u2206 ADC = \\(\\frac{1}{2}\\) \u00d7 AD \u00d7 DE
\n= \\(\\frac{1}{2}\\) \u00d7 6 \u00d7 8 cm2<\/sup> =3 \u00d7 8 cm2<\/sup> = 24 cm2<\/sup>
\nNow, Area of \u2206 ABC = \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 AC
\n= \\(\\frac{1}{2}\\) \u00d7 24 \u00d7 10 cm2<\/sup> = 12 \u00d7 10 cm2<\/sup> = 120 cm2<\/sup>
\nArea of \u2206 ADC = \\(\\frac{1}{2}\\) \u00d7 AD \u00d7 DC
\n= \\(\\frac{1}{2}\\) \u00d7 6 \u00d7 8 cm2<\/sup> = 3 \u00d7 8 cm2<\/sup> = 24 cm2<\/sup>
\nHence, area of shaded region = Area of \u2206 ABC – Area of \u2206 ADC = 120 cm2<\/sup> – 24 cm2<\/sup> = 96 cm2<\/sup><\/p>\n

(c) Given. In right angled \u2206 ABC, AB = 9 cm, AC = 15 cm, and D, E are mid-points of the sides AB and AC respectively.
\nRequired. (i) length of BC (ii) the area of \u2206 ADE
\nSolution:
\nIn right angled \u2206 ADE, (By Pythagoras theorem)
\nAE2<\/sup> = AD2<\/sup> + DE2<\/sup>
\n\"ML
\n\"ML<\/p>\n

Question 2.
\nIf in \u2206 ABC, AB > AC and AD \u22a5 BC, prove that AB2<\/sup> – AC2<\/sup> = BD2<\/sup> – CD2<\/sup>.
\nAnswer:
\nGiven. In \u2206 ABC, AB > AC and AD \u22a5 BC
\nTo prove. AB2<\/sup> – AC2<\/sup> = BD2<\/sup> – CD2<\/sup>
\n\"ML
\nProof. In right angled \u2206 ABD
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup> …….(1) (By Pythagoras theorem
\nIn right angled \u2206ACD
\nAC2<\/sup> = AD2<\/sup> + CD2<\/sup> ……(2)
\nSubtracting (2) from (1), we get
\nAB2<\/sup> – AC2<\/sup> = (AD2<\/sup> + BD2<\/sup>) – (AD2<\/sup> + CD2<\/sup>)
\n= AD2<\/sup> + BD2<\/sup> – AD2<\/sup> – CD2<\/sup> = BD2<\/sup> – CD2<\/sup>
\n\u2234 AB2<\/sup> – AC2<\/sup> = BD2<\/sup> – CD2<\/sup> Hence, the result.<\/p>\n

Question 3.
\nIn a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that
\n(i) 9AQ2<\/sup> = 9AC2<\/sup> + 4BC2<\/sup>
\n(ii) 9BP2<\/sup> = 9BC2<\/sup> + 4AC2<\/sup>
\n(iii) 9(AQ2<\/sup> + BP2<\/sup>) = 13AB2<\/sup>.
\nSolution:
\nGiven. A right angled \u2206 ABC in which \u2220C = 90\u00b0. P and Q are points on the side CA and CB respectively such that CP : AP = 2 : 1 and CQ : BQ = 2 : 1
\nTo prove. (i) 9AQ2<\/sup> = 9AC2<\/sup> + 4BC2<\/sup>
\n(ii) 9BP2<\/sup> = 9BC2<\/sup> + 4AC2<\/sup>
\n(iii) 9 (AQ2<\/sup> + BP2<\/sup>) = 13 AB2<\/sup>
\n\"ML
\nConstruction. Join AQ and BP.
\nProof. (i) In right angled \u2206 ACQ
\nAQ2<\/sup> = AC2<\/sup> + QC2<\/sup> (By Pythagoras theorem)
\n9AQ2<\/sup> = 9AC2<\/sup> + 9QC2<\/sup> ( Multiplying both sides by 9)
\n= 9AC2<\/sup> + (3QC)2<\/sup> = 9AC2<\/sup> + (2BC)2<\/sup>
\n\"ML
\n= 9AC2<\/sup> + 4BC2<\/sup>
\n\u2234 9AQ2<\/sup> = 9AC2<\/sup> + 4BC2<\/sup> …..(1)<\/p>\n

(ii) In right angled \u2206 BPC
\nBP2<\/sup> = BC2<\/sup> + CP2<\/sup> (By Pythagoras theorem)
\n9BP2<\/sup> = 9BC2<\/sup> + 9CP2<\/sup> (\u2235 Multiplying both side by 9)
\n= 9BC2<\/sup> + (3 CP)2<\/sup> = 9BC2<\/sup> + (2AC)2<\/sup>
\n\"ML
\n= 9BC2<\/sup> + 4AC2<\/sup>
\n\u2234 9BP2<\/sup> = 9BC2<\/sup> + 4AC2<\/sup> ….(2)<\/p>\n

(iii) Adding (1) and (2),
\n9AQ2<\/sup> + 9BP2<\/sup> = 9AC2<\/sup> + 4BC2<\/sup> + 9BC2<\/sup> + 4AC2<\/sup>
\n= 13AC2<\/sup> + 13BC2<\/sup> = 13 (AC2<\/sup> + BC2<\/sup>) = 13 AB2<\/sup> [In right angled \u2206 ABC = AB2<\/sup> = AC2<\/sup> + BC2<\/sup>]
\n\u2234 9AQ + 9BP2<\/sup> = 13 AB2<\/sup>
\nHence, the result.<\/p>\n

Question 4.
\nIn the given figure, \u2206PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT2<\/sup> = 3PR2<\/sup> + 5PS2<\/sup>.
\n\"ML
\nSolution:
\nIn the \u2206PQR, \u2220Q = 90\u00b0
\nT and S are points on RQ such that these trisect it
\ni.e., RT = TS = SQ
\nTo prove : 8PT2<\/sup> = 3PR2<\/sup> + 5PS2<\/sup>
\nProof: Let RT = TS = SQ = x
\nIn right \u2206PRQ
\nPR2<\/sup> = RQ2<\/sup> + PQ2<\/sup> = (3x)2<\/sup> + PQ2<\/sup> = 9x2<\/sup> + PQ2<\/sup>
\nSimilarly in right PTS,
\nPT2<\/sup> = TQ2<\/sup> + PQ2<\/sup> = (2x)2<\/sup> + PQ2<\/sup> = 4x2<\/sup> + PQ2<\/sup>
\nand in PSQ,
\nPS2<\/sup> = SQ2<\/sup> + PQ2<\/sup> = x2<\/sup> + PQ2<\/sup>
\n8PT2<\/sup> = 8(4x2<\/sup> + PQ2<\/sup>) = 32x2<\/sup> + 8PQ2<\/sup>
\n3PR2<\/sup> = 3(9x2<\/sup> + PQ2<\/sup>) = 27x2<\/sup> + 3PQ2<\/sup>
\n5PS2<\/sup> = 5(x2<\/sup> + PQ2<\/sup>) = 5x2<\/sup> + 5PQ2<\/sup>
\nLHS = 8PT2<\/sup> = 32x2<\/sup> + 8PQ2<\/sup>
\nRHS = 3PR2<\/sup> + 5PS2<\/sup> = 27x2<\/sup> + 3PQ2<\/sup> + 5x2<\/sup> + 5PQ2<\/sup>.
\n= 32x2<\/sup> + 8PQ2<\/sup>.
\nLHS = RHS
\nHence proved.<\/p>\n

Question 5.
\nIn a quadrilateral ABCD, \u2220B = 90\u00b0. If AD2<\/sup> = AB2<\/sup> + BC2<\/sup> + CD2<\/sup>, prove that \u2220ACD = 90\u00b0.
\nSolution:
\nIn quadrilateral ABCD, \u2220B = 90\u00b0 and AD2<\/sup> = AB2<\/sup> + BC2<\/sup> + CD2<\/sup>
\nTo prove : \u2220ACD = 90\u00b0
\nProof: In \u2220ABC, \u2220B = 90\u00b0
\n\u2234 AC2<\/sup> = AB2<\/sup> + BC2<\/sup> …(i) (Pythagoras Theorem)
\nBut AD2<\/sup> = AB2<\/sup> + BC2<\/sup> + CD2<\/sup> (Given)
\n\"ML
\n\u21d2 AD2<\/sup> = AC2<\/sup> + CD2<\/sup> [From (i)]
\n\u2234 In \u2206ACD,
\n\u2220ACD = 90\u00b0 (Converse of Pythagoras Theorem)<\/p>\n

Question 6.
\nIn the given figure, find the length of AD in terms of b and c.
\n\"ML
\nSolution:
\nIn the given figure,
\nABC is a triangle, \u2220A = 90\u00b0
\nAB = c, AC = b
\nAD \u22a5 BC
\nTo find : AD in terms of b and c
\nSolution:
\n\"ML
\n\"ML<\/p>\n

Question 7.
\nABCD is a square, F is mid-point of AB and BE is one-third of BC. If area of \u2206FBE is 108 cm2<\/sup>, find the length of AC.
\nSolution:
\nGiven : In square ABCD. F is mid piont of
\nAB and BE = \\(\\frac{1}{3}\\) BC
\nArea of AFBE = 108 cm2<\/sup>
\nAC and EF are joined
\n\"ML
\nTo find: AC
\nSolution:
\nLet each side of square is = a
\nFB = \\(\\frac{1}{2}\\) AB (F is mid point of AB)
\n= \\(\\frac{1}{2}\\) a
\nand BE = \\(\\frac{1}{3}\\)BC = \\(\\frac{1}{3}\\)a
\nNow in square ABCD
\nAC = \\( \\sqrt{{2}} \\) \u00d7 Side = \\( \\sqrt{{2}} \\)a
\nand area \u2206FBE = \\(\\frac{1}{2}\\) FB \u00d7 BE
\n= \\(\\frac{1}{2}\\) \u00d7 \\(\\frac{1}{2}\\)a \u00d7 \\(\\frac{1}{2}\\)a = \\(\\frac{1}{12}\\)a2<\/sup>
\n\u2234 \\(\\frac{1}{12}\\) a2<\/sup> = 108 \u21d2 a2<\/sup> = 12 \u00d7 108 = 1296
\n\u21d2 a = \\( \\sqrt{{1296}} \\) = 36
\n\u2234 AC = \\( \\sqrt{{2}} \\) a = \\( \\sqrt{{2}} \\) \u00d7 36 = 36 \\( \\sqrt{{2}} \\) cm<\/p>\n

Question 8.
\nIn a triangle ABC, AB = AC and D is a point on side AC such that BC2<\/sup> = AC \u00d7 CD. Prove that BD = BC.
\nSolution:
\nGiven. In a triangle ABC, AB = AC and D is point on side AC such that BC2<\/sup> = AC \u00d7 CD
\nTo prove. BD = BC
\n\"ML
\nConstruction. Draw BE \u22a5 AC
\nProof. In right angled \u2206 BCE
\nBC2<\/sup> = BE2<\/sup> + EC2<\/sup> (By Pythagoras theorem)
\n= BE2<\/sup> + (AC – AE)2<\/sup>
\n= BE2<\/sup> + AC2<\/sup> + AE2<\/sup> -2AC.AE
\n= (BE2<\/sup> + AE2<\/sup>) + AC2<\/sup> – 2AC.AE
\n= AB2<\/sup> + AC2<\/sup> – 2AC.AE (In rt. \u2220 ed \u2206 ABC, AB2<\/sup> = BE2<\/sup> + AE2<\/sup>)
\n= AC2<\/sup> + AC2<\/sup> – 2AC.AE (given AB = AC)
\n= 2AC2<\/sup> – 2AC. AE = 2AC (AC – AE)
\n= 2AC.EC
\nBut BC2<\/sup> = AC \u00d7 CD (given)
\n\u21d2 AC \u00d7 CD = 2AC.EC \u21d2 CD = 2EC
\n\u2234 E is mid-points of CD \u21d2 EC = DE
\nNow, in \u2206BED and \u2206 BEC
\nEC = DE (above proved)
\nBE = BE (common)
\n\u2220 BED = \u2220 BEC (each 90\u00b0)
\n\u2234 \u2206BED \u2245 \u2206 BEC (By S. A. S. axiom of congruency)
\n\u2234 BD = BC (c.p.c.t.)
\nHence, the result.<\/p>\n

ML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 12 Pythagoras Theorem Chapter Test Question 1. (a) In fig. (i) given below, AD \u22a5 BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC. (b) In figure (ii) given below, \u2220BAC = 90\u00b0, \u2220ADC = 90\u00b0, …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/46983"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=46983"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/46983\/revisions"}],"predecessor-version":[{"id":159225,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/46983\/revisions\/159225"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=46983"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=46983"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=46983"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}