{"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":"
Question 1. (b) Given. In \u2206 ABC, (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. Question 2. Question 3. (ii) In right angled \u2206 BPC (iii) Adding (1) and (2), Question 4. Question 5. Question 6. Question 7. Question 8.
\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
\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
\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
\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
\n<\/p>\n
\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
\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
\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
\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
\n= 9AC2<\/sup> + 4BC2<\/sup>
\n\u2234 9AQ2<\/sup> = 9AC2<\/sup> + 4BC2<\/sup> …..(1)<\/p>\n
\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
\n= 9BC2<\/sup> + 4AC2<\/sup>
\n\u2234 9BP2<\/sup> = 9BC2<\/sup> + 4AC2<\/sup> ….(2)<\/p>\n
\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
\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
\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
\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
\n\u21d2 AD2<\/sup> = AC2<\/sup> + CD2<\/sup> [From (i)]
\n\u2234 In \u2206ACD,
\n\u2220ACD = 90\u00b0 (Converse of Pythagoras Theorem)<\/p>\n
\nIn the given figure, find the length of AD in terms of b and c.
\n
\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
\n<\/p>\n
\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
\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
\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
\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>\nML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"