{"id":3698,"date":"2021-02-24T14:46:57","date_gmt":"2021-02-24T09:16:57","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3698"},"modified":"2021-09-15T11:40:42","modified_gmt":"2021-09-15T06:10:42","slug":"ap-ssc-10th-class-maths-solutions-chapter-8-ex-8-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-8-ex-8-4\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.4"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 8 Similar Triangles Ex 8.4 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.4<\/h2>\n

10th Class Maths 8th Lesson Similar Triangles Ex 8.4 Textbook Questions and Answers<\/h3>\n

Question 1.
\nProve that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
\nAnswer:
\n\"AP
\nGiven : \u25a1 ABCD is a rhombus.
\nLet its diagonals AC and BD bisect each other at \u2018O\u2019.
\nWe know that “the diagonals in a rhombus are perpendicular to each other\u201d.
\nIn \u25b3AOD; AD2<\/sup> = OA2<\/sup> + OD2<\/sup> ………. (1)
\n[Pythagoras theorem]
\nIn \u25b3COD; CD2<\/sup> = OC2<\/sup> + OD2<\/sup> ………. (2)
\n[Pythagoras theorem]
\nIn \u25b3AOB; AB2<\/sup> = OA2<\/sup> + OB2<\/sup> ………. (3)
\n[Pythagoras theorem]
\nIn \u25b3BOC; BC2<\/sup> = OB2<\/sup> + OC2<\/sup> ………. (4)
\n[Pythagoras theorem]
\nAdding the above equations we get AD2<\/sup> + CD2<\/sup> + AB2<\/sup> + BC2<\/sup> = 2 (OA2<\/sup> + OB2 + OC2<\/sup> + OD2<\/sup>)
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that AE2<\/sup> + CD2<\/sup> = AC2<\/sup> + DE2<\/sup>.
\n\"AP
\nAnswer:
\nGiven: In \u25b3ABC; \u2220B = 90\u00b0
\nD and E are points on AB and BC.
\nR.T.P.: AE2<\/sup> + CD2<\/sup> = AC2<\/sup> + DE2<\/sup>
\nProof: In \u25b3BCD, \u25b3BCD is a right triangle right angled at B.
\n\u2234 BD2<\/sup> + BC2<\/sup> = CD2<\/sup> ……… (1)
\n[\u2235 Pythagoras theorem states that hypotenuse2<\/sup> = side2<\/sup> + side2<\/sup>]
\nIn \u25b3ABE; \u2220B = 90\u00b0
\nAdding (1) and (2), we get
\nBD2<\/sup> + BC2<\/sup> + AB2<\/sup> + BE2<\/sup> – CD2<\/sup> + AE2<\/sup>
\n(BD2<\/sup> + BE2<\/sup>) + (AB2<\/sup> + BC2<\/sup>) = CD2<\/sup> + AE2<\/sup>
\nDE2<\/sup> + AC2<\/sup> – CD2<\/sup> + AE2<\/sup> [Q.E.D.]
\n[\u2235 (i) In \u25b3DBE, \u2220B = 90\u00b0 and DE2<\/sup> = BD2<\/sup> + BE2<\/sup>
\n(ii) In \u25b3ABC, \u2220B = 90\u00b0 and AB2<\/sup> + BC2<\/sup>]<\/p>\n

Question 3.
\nProve that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
\nAnswer:
\nGiven: \u25b3ABC, an equilateral triangle;
\nAD – altitude and the side is a units, altitude h units.
\nR.T.P: 3a2<\/sup> = 4h2<\/sup>
\n\"AP
\nProof: In \u25b3ABD, \u25b3ACD
\n\u2220B = \u2220C [\u2235 60\u00b0]
\n\u2220ADB = \u2220ADC [\u2235 90\u00b0]
\n\u2234 \u2220BAD = \u2220DAC [\u2235 Angle sum property]
\nAlso, BA = CA
\n\u2234 \u25b3ABD s \u25b3ACD (by SAS congruence condition)
\nHence, BD = CD = \\(\\frac{1}{2}\\)BC = \\(\\frac{a}{2}\\) [\u2235 c.p.c.t]
\nNow in \u25b3ABD, AB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\n[\u2235 Pythagoras theorem]
\na2<\/sup> = h2<\/sup> + \\(\\left(\\frac{a}{2}\\right)^{2}\\)
\na2<\/sup> = h2<\/sup> + \\(\\frac{a^{2}}{4}\\)
\nh2<\/sup> = \\(\\frac{4 a^{2}-a^{2}}{4}\\)
\n\u2234 h2<\/sup> = \\(\\frac{3 a^{2}}{4}\\)
\n\u21d2 4h2<\/sup> = 3a2<\/sup> (Q.E.D)<\/p>\n

\"AP<\/p>\n

Question 4.
\nPQR is a triangle right angled at P and M is a point on QR such that PM \u22a5 QR. Show that PM2<\/sup> = QM . MR.
\nAnswer:
\n\"AP
\nGiven: In \u25b3PQR, \u2220P = 90\u00b0 and PM \u22a5 QR.
\nR.T.P : PM2<\/sup> = QM . MR
\nProof: In \u25b3PQR; \u25b3MPR
\n\u2220P = \u2220M [each 90\u00b0]
\n\u2220R = \u2220R [common]
\n\u2234 \u25b3PQR ~ \u25b3MPR ……… (1)
\n[A.A. similarity]
\nIn \u25b3PQR and \u25b3MQP,
\n\u2220P = \u2220M (each 90\u00b0)
\n\u2220Q = \u2220Q (common)
\n\u2234 \u25b3PQR ~ \u25b3MQP ……… (2)
\n[A.A. similarity]
\nFrom (1) and (2),
\n\u25b3PQR ~ \u25b3MPR ~ \u25b3MQP [transitive property]
\n\u2234 \u25b3MPR ~ \u25b3MQP
\n\\(\\frac{MP}{MQ}\\) = \\(\\frac{PR}{QP}\\) = \\(\\frac{MR}{MP}\\)
\n[Ratio of corresponding sides of similar triangles are equal]
\n\\(\\frac{PM}{QM}\\) = \\(\\frac{MR}{PM}\\)
\nPM . PM = MR . QM
\nPM2<\/sup> = QM . MR [Q.E.D]<\/p>\n

Question 5.
\nABD is a triangle right angled at A and AC \u22a5 BD.
\nShow that (i) AB2<\/sup> = BC BD
\n(ii) AD2<\/sup> = BD CD
\n(iii) AC2<\/sup> = BC DC.
\n\"AP
\nAnswer:
\nGiven: In \u25b3ABD; \u2220A = 90\u00b0 AC \u22a5 BD
\nR.T.P.:
\ni) AB2<\/sup> = BC . BD
\nProof: In \u25b3ABD and \u25b3CAB,
\n\u2220BAD = \u2220ACB [each 90\u00b0]
\n\u2220B = \u2220B [common]
\n\u2234 \u25b3ABD ~ \u25b3CBA
\n[by A.A. similarity condition]
\nHence, \\(\\frac{AB}{BC}\\) = \\(\\frac{BD}{AB}\\) = \\(\\frac{AD}{AC}\\)
\n[\u2235 Ratios of corresponding sides of similar triangles are equal]
\n\\(\\frac{AB}{BD}\\) = \\(\\frac{BC}{AB}\\)
\n\u21d2 AB . AB = BC . BD
\n\u2234 AB2<\/sup> = BC . BD<\/p>\n

\"AP<\/p>\n

ii) AD2<\/sup> = BD . CD
\nProof: In \u25b3ABD and \u25b3CAD
\n\u2220BAD = \u2220ACD [each 90\u00b0]
\n\u2220D = \u2220D (common)
\n\u2234 \u25b3ABD ~ \u25b3CAD [A.A similarity]
\nHence, \\(\\frac{AB}{AC}\\) = \\(\\frac{BD}{AD}\\) = \\(\\frac{AD}{CD}\\)
\n\u21d2 \\(\\frac{BD}{AD}\\) = \\(\\frac{AD}{CD}\\)
\nAD . AD = BD . CD
\nAD2<\/sup> = BD . CD [Q.E.D]<\/p>\n

iii) AC2<\/sup> = BC . DC
\nProof: From (i) and (ii)
\n\u25b3ACB ~ \u25b3DCA
\n[\u2235 \u25b3BAD ~ \u25b3BCA ~ \u25b3ACD
\nHence, \\(\\frac{AC}{DC}\\) = \\(\\frac{BC}{AC}\\) = \\(\\frac{AB}{AD}\\)
\n\\(\\frac{AC}{DC}\\) = \\(\\frac{BC}{AC}\\)
\nAC . AC = BC . DC
\nAC2<\/sup> = BC . DC [Q.E.D]<\/p>\n

Question 6.
\nABC is an isosceles triangle right angled at C. Prove that AB2<\/sup> = 2AC2<\/sup>.
\nAnswer:
\n\"AP
\nGiven: In \u25b3ABC; \u2220C = 90\u00b0; AC = BC.
\nR.T.P.: AB2<\/sup> = 2AC2<\/sup>
\nProof: In \u25b3ACB; \u2220C = 90\u00b0
\nHence, AC2<\/sup> + BC2<\/sup> = AB2<\/sup>
\n[Square of the hypotenuse is equal to sum of the squares of the other two sides – Pythagoras theorem]
\n\u21d2 AC2<\/sup> + AC2<\/sup> = AB2<\/sup> [\u2235 AC = BC given]
\n\u21d2 AB2<\/sup> = 2AC2<\/sup> [Q.E.D.]<\/p>\n

\"AP<\/p>\n

Question 7.
\n‘O\u2019 is any point in the interior of a triangle ABC.
\nOD \u22a5 BC, OE \u22a5 AC and OF \u22a5 AB, show that
\ni) OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\nii) AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup>.
\n\"AP
\nAnswer:
\nGiven: \u25b3ABC; O\u2019 is an interior point of \u25b3ABC.
\nOD \u22a5 BC, OE \u22a5 AC, OF \u22a5 AB.
\nR.T.P.:
\ni) OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\nProof: In OAF, OA2<\/sup> = AF2<\/sup> + OF2<\/sup> [Pythagoras theorem]
\n\u21d2 OA2<\/sup> – OF2<\/sup> = AF2<\/sup> …….. (1)
\nIn \u25b3OBD,
\nOB2<\/sup> = BD2<\/sup> + OD2<\/sup>
\n\u21d2 OB2<\/sup> – OD2<\/sup> = BD2<\/sup> …….. (2)
\nIn \u25b3OCE, OC2<\/sup> = CE2<\/sup> + OE2<\/sup>
\nOC2<\/sup> – OE2<\/sup> = CE2<\/sup> ……… (3)
\nAdding (1), (2) and (3) we get,
\nOA2<\/sup> – OF2<\/sup> + OB2<\/sup> – OD2<\/sup> + OC2<\/sup> – OE2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\nOA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup> ……… (4)<\/p>\n

ii) AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup>
\nIn \u25b3OAE,
\nOA2<\/sup> = AE2<\/sup> + OF2<\/sup> ……… (1)
\n\u21d2 OA2<\/sup> – OE2<\/sup> = AE2<\/sup>
\nIn \u25b3OBF, OB2<\/sup> = BF2<\/sup> + OF2<\/sup>
\nOB2<\/sup> – OF2<\/sup> = BF2<\/sup> ……… (2)
\nIn \u25b3OCD, OC2<\/sup> = OD2<\/sup> + CD2<\/sup>
\nOC2<\/sup> – OD2<\/sup> = CD2<\/sup> ……… (3)
\nAdding (1), (2) and (3) we get
\nOA2<\/sup> – OE2<\/sup> + OB2<\/sup> – OF2<\/sup> + OC2<\/sup> – OD2<\/sup> = AE2<\/sup> + BF2<\/sup> + CD2<\/sup>
\n\u21d2 OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup>
\n\u21d2 AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup> [From problem (i)]<\/p>\n

Question 8.
\nA wire attached to vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
\nAnswer:
\n\"AP
\nHeight of the pole AB = 18 m.
\nLength of the wire AC = 24 m.
\nDistance beween the pole and the stake be \u2018d\u2019 meters.
\nBy Pythagoras theorem,
\nHypotenuse2<\/sup> = side2<\/sup> + side2<\/sup>
\n242<\/sup> = 182<\/sup> + d2<\/sup>
\nd2<\/sup> = 242<\/sup> – 182<\/sup> = 576 – 324 = 252
\n= \\(\\sqrt{36 \\times 7}\\)
\n\u2234 d = 6\u221a7 m.<\/p>\n

\"AP<\/p>\n

Question 9.
\nTwo poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
\nAnswer:
\n\"AP
\nLet the height of the first pole AB = 6 m.
\nLet the height of the second pole CD = 11 m.
\nDistance between the poles AC = 12 m.
\nFrom the figure \u25a1 ACEB is a rectangle.
\n\u2234 AB = CE = 6 m
\nED = CD – CE = 11 – 6 = 5 m
\nNow in \u25b3BED; \u2220E = 90\u00b0; DE = 5 m; BE = 12 m
\nBD2<\/sup> = BE2<\/sup> + DE2<\/sup>
\n[hypotenuse2<\/sup> = side2<\/sup> + side2<\/sup> – Pythagoras theorem]
\n= 122<\/sup> + 52<\/sup>
\n= 144 + 25
\nBD2<\/sup> = 169
\nBD = \u221al69 = 13m
\n\u2234 Distance between the tops of the poles = 13 m.<\/p>\n

Question 10.
\nIn an equilateral triangle ABC, D is a point on side BC such that BD = \\(\\frac{1}{3}\\) BC. Prove that 9AD2<\/sup> = 7AB2<\/sup>.
\nAnswer:
\n\"AP
\nIn \u25b3ABE, \u2220E = 90\u00b0
\n\u21d2 \\(\\overline{\\mathrm{AB}}\\) is hypotenuse.
\n\u2234 AB2<\/sup> = AE2<\/sup> + BE2<\/sup>
\nAE2<\/sup> = AB2<\/sup> – BE2<\/sup>
\n\u21d2 AE2<\/sup> = AB2<\/sup> – \\(\\left(\\frac{BC}{2}\\right)^{2}\\)
\n= AE2<\/sup> = AB2<\/sup> – \\(\\left(\\frac{AB}{2}\\right)^{2}\\) (\u2235 AB = BC)
\n\u21d2 AE2<\/sup> = \\(\\frac{3}{4}\\)AB2<\/sup> ……… (1)
\nIn \u25b3ADE, \u2220E = 90\u00b0
\n\u21d2 \\(\\overline{\\mathrm{AD}}\\) is hypotenuse.
\n\u21d2 AD2<\/sup> = AE2<\/sup> + DE2<\/sup>
\n\u21d2 AE2<\/sup> = AD2<\/sup> + DE2<\/sup>
\n\"AP
\n\u21d2 28 AB2<\/sup> = 36 AD2<\/sup>
\n\u21d2 7 AB2<\/sup> = 9 AD2<\/sup>
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 11.
\nIn the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it. Prove that 8 AE2<\/sup> = 3 AC2<\/sup> + 5 AD2<\/sup>.
\n\"AP
\nAnswer:
\nIn \u25b3ABC, \u2220B=90\u00b0
\n\u21d2 \\(\\overline{\\mathrm{AD}}\\) is hypotenuse.
\nAC2 = AB2 + BC2
\n3AC2<\/sup> = 3AB2<\/sup> + 3BC2<\/sup> …….. (1)
\nIn \u25b3ABD, \u2220B = 90\u00b0
\n\u21d2 AD is hypotenuse.
\n\u2234 AD2<\/sup> = AB2<\/sup> + BD2<\/sup> = AB2<\/sup> + \\(\\left(\\frac{BC}{3}\\right)^{2}\\)
\n\u21d2 AD2<\/sup> = AB2<\/sup> + \\(\\frac{\\mathrm{BC}^{2}}{9}\\)
\n\u21d2 5 AD2<\/sup> = 5 AB2<\/sup> + \\(\\frac{5 \\mathrm{BC}^{2}}{9}\\) …….. (2)
\n(1) + (2)
\n3 AC2<\/sup> + 5 AD2<\/sup> = 3 AB2<\/sup> + 3 BC2<\/sup> + 5 AB2<\/sup> + \\(\\frac{5}{9} \\mathrm{BC}^{2}\\)
\n= 8AB2<\/sup> + \\(\\frac{32}{9} \\mathrm{BC}^{2}\\) ……… (3)
\nNow in \u25b3ABE, \u2220B = 90\u00b0
\n\u21d2 \\(\\overline{\\mathrm{AE}}\\) is hypotenuse.
\n\u21d2 AE2<\/sup> = AB2<\/sup> + BE2<\/sup> = AB2<\/sup> + \\(\\left(\\frac{2}{3} BC\\right)^{2}\\)
\n= AB2<\/sup> + \\(\\frac{4}{9} \\mathrm{BC}^{2}\\)
\n\u21d2 AE2<\/sup> = 8AB2<\/sup> + \\(\\frac{32}{9} \\mathrm{BC}^{2}\\) ……… (4)
\n\u2234 RHS of (3) and (4) are equal.
\n\u2234 LHS of (3) and (4) are equal.
\n\u2234 8 AE2<\/sup> = 3 AC2<\/sup> + 5 AD2<\/sup>.
\nHence proved.<\/p>\n

Question 12.
\nABC is an isosceles triangle right angled at B. Equilateral triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of \u25b3ABE and \u25b3ACD.
\nAnswer:
\n\"AP
\nGiven: \u25b3ABC, AB = BC and \u2220B = 90\u00b0
\n\u25b3ABE on AB; \u25b3ACD on AC are equiangular triangles.
\nLet equal sides of the isosceles right triangle, AB = BC = a (say)
\nThen, in \u25b3ABC, \u2220B = 90\u00b0
\nAC2<\/sup> – AB2<\/sup> + BC2<\/sup>
\n[hypotenuse2<\/sup> = side2<\/sup> + side2<\/sup> – Pythagoras theorem] = a2<\/sup> + a2<\/sup> = 2a2<\/sup>
\nSince, \u25b3ABE ~ \u25b3ACD
\n\\(\\frac{\\Delta \\mathrm{ABE}}{\\Delta \\mathrm{ACD}}\\) = \\(\\frac{\\mathrm{AB}^{2}}{\\mathrm{AC}^{2}}\\)
\n[\u2235 Ratio of areas of two similar tri-angles is equal to the ratio of squares of their corresponding sides]
\n= \\(\\frac{a^{2}}{2 a^{2}}\\) = \\(\\frac{1}{2}\\)
\n\u25b3ABE : \u25b3ACD = 1 : 2.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.4 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.4 10th Class Maths 8th Lesson Similar Triangles Ex 8.4 Textbook Questions and Answers Question 1. Prove that the sum of the …<\/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\/3698"}],"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=3698"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3698\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3698"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3698"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3698"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}