{"id":693,"date":"2025-01-16T10:07:00","date_gmt":"2025-01-16T04:37:00","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=693"},"modified":"2025-01-17T10:18:40","modified_gmt":"2025-01-17T04:48:40","slug":"samacheer-kalvi-10th-maths-guide-chapter-4-ex-4-3","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-10th-maths-guide-chapter-4-ex-4-3\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.3"},"content":{"rendered":"

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.3<\/h2>\n

Question 1.
\nA man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
\nSolution:
\nLet the initial position of the man be “O” and his final
\nposition be “B”.
\nBy Pythagoras theorem
\nIn the right \u2206 OAB,
\n\"Samacheer
\nOB2<\/sup> = OA2<\/sup> + AB2<\/sup>
\n= 182<\/sup> + 242<\/sup>
\n= 324 + 576 = 900
\nOB = \\(\\sqrt { 900 }\\) = 30
\nThe distance of his current position is 30 m<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nThere are two paths that one can choose to go from Sarah\u2019s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).
\n\"Samacheer
\nSolution:
\nDistance between Sarah House and James House using “C street”.
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n= 22<\/sup> + 1.52<\/sup>
\n= 4 + 2.25 = 6.25
\nAC = \\(\\sqrt { 6.25 }\\)
\nAC = 2.5 miles
\n\"Samacheer
\nDistance covered by using “A Street” and “B Street”
\n= (2 + 1.5) miles = 3.5 miles
\nDifference in distance = 3.5 miles – 2.5 miles = 1 mile<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nTo get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
\nSolution:
\nIn the right \u2206ABC,
\nBy Pythagoras theorem
\nAC2<\/sup>= AB2<\/sup> + BC2<\/sup> = 342<\/sup> + 412<\/sup>
\n= 1156 + 1681 = 2837
\nAC = \\(\\sqrt { 2837 }\\)
\n= 53.26 m
\n\"Samacheer
\nThrough A one must walk (34m + 41m) 75 m to reach C.
\nThe difference in Distance = 75 – 53.26
\n= 21.74 m<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nIn the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
\nCalculate the length and breadth of the rectangle?
\n\"Samacheer
\nSolution:
\nLet the length of the rectangle be “a” and the breadth of the rectangle be “b”.
\nXY + YZ = 17 cm
\nb + a = 17 …….. (1)
\nIn the right \u2206 WXZ,
\nXZ2<\/sup> = WX2<\/sup> + WZ2<\/sup>
\n(XZ)2<\/sup> = a2<\/sup> + b2<\/sup>
\n\"Samacheer
\nXZ = \\(\\sqrt{a^{2}+b^{2}}\\)
\nSimilarly WY = \\(\\sqrt{a^{2}+b^{2}}\\) \u21d2 XZ + WY = 26
\n2 \\(\\sqrt{a^{2}+b^{2}}\\) = 26 \u21d2 \\(\\sqrt{a^{2}+b^{2}}\\) = 13
\nSquaring on both sides
\na2<\/sup> + b2<\/sup> = 169
\n(a + b)2<\/sup> – 2ab = 169
\n172<\/sup> – 2ab = 169 \u21d2 289 – 169 = 2 ab
\n120 = 2 ab \u21d2 \u2234 ab = 60
\na = \\(\\frac { 60 }{ b } \\) ….. (2)
\nSubstituting the value of a = \\(\\frac { 60 }{ b } \\) in (1)
\n\\(\\frac { 60 }{ b } \\) + b = 17
\nb2<\/sup> – 17b + 60 = 0
\n(b – 2) (b – 5) = 0
\nb = 12 or b = 5
\n\"Samacheer
\nIf b = 12 \u21d2 a = 5
\nIf b = 6 \u21d2 a = 12
\nLenght = 12 m and breadth = 5 m<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nThe hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
\nSolution:
\nLet the shortest side of the right \u2206 be x.
\n\u2234 Hypotenuse = 6 + 2x
\nThird side = 2x + 6 – 2
\n= 2x + 4
\n\"Samacheer
\nIn the right triangle ABC,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n(2x + 6)2<\/sup> = x2<\/sup> + (2x + 4)2<\/sup>
\n4x2<\/sup> + 36 + 24x = x2<\/sup> + 4x2<\/sup> + 16 + 16x
\n0 = x2<\/sup> – 24x + 16x – 36 + 16
\n\u2234 x2<\/sup> – 8x – 20 = 0
\n(x – 10) (x + 2) = 0
\nx – 10 = 0 or x + 2 = 0
\n\"Samacheer
\nx = 10 or x = -2 (Negative value will be omitted)
\nThe side AB = 10 m
\nThe side BC = 2 (10) + 4 = 24 m
\nHypotenuse AC = 2(10) + 6 = 26 m<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\n5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
\nSolution:
\n“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.
\nIn the right \u2206ABC,
\nBC2<\/sup> = AC2<\/sup> – AB2<\/sup> = 52<\/sup> – 42<\/sup>
\n= 25 – 16 = 9
\nBC = \\(\\sqrt { 9 }\\) = 3m.
\nWhen the foot of the ladder moved 1.6 m toward the wall.
\nThe distance between the foot of the ladder to the ground is
\nBE = 3 – 1.6 m
\n= 1.4 m
\n\"Samacheer
\nLet the distance moved upward on the wall be “h\u201d m
\nThe ladder touch the wall at (4 + h) M
\nIn the right triangle BED,
\nED2<\/sup> = AB2<\/sup> + BE2<\/sup>
\n52<\/sup> = (4 + h)2<\/sup> + (1.4)2<\/sup>
\n25 – 1.96= (4 + h)2<\/sup>
\n\u2234 4 + h = \\(\\sqrt { 23.04 }\\)
\n4 + h = 4. 8 m
\nh = 4.8 – 4
\n= 0.8 m
\nDistance moved upward on the wall = 0.8 m<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nThe perpendicular PS on the base QR of a \u2206PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2<\/sup> = 2PR2<\/sup> + QR2<\/sup>.
\nSolution:
\nGiven QS = 3SR
\nQR = QS + SR
\n= 3SR + SR = 4SR
\nSR = \\(\\frac { 1 }{ 4 } \\) QR …..(1)
\nQS = 3SR
\nSR = \\(\\frac { QS }{ 3 } \\) ……..(2)
\n\"Samacheer
\nFrom (1) and (2) we get
\n\\(\\frac { 1 }{ 4 } \\) QR = \\(\\frac { QS }{ 3 } \\)
\n\u2234 QS = \\(\\frac { 3 }{ 4 } \\) QR ………(3)
\nIn the right \u2206 PQS,
\nPQ2<\/sup> = PS2<\/sup> + QS2<\/sup> ……….(4)
\nSimilarly in \u2206 PSR
\nPR2<\/sup> = PS2<\/sup> + SR2<\/sup> ………..(5)
\nSubtract (4) and (5)
\nPQ2<\/sup> – PR2<\/sup> = PS2<\/sup> + QS2<\/sup> – PS2<\/sup> – SR2<\/sup>
\n= QS2<\/sup> – SR2<\/sup>
\n\"Samacheer
\nPQ2<\/sup> – PR2<\/sup> = \\(\\frac { 1 }{ 2 } \\) QR2<\/sup>
\n2PQ2<\/sup> – 2PR2<\/sup> = QR2<\/sup>
\n2PQ2<\/sup> = 2PR2<\/sup> + QR2<\/sup>
\nHence the proved.<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nIn the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2<\/sup> = 3AC2<\/sup> + 5AD2<\/sup>.
\nSolution:
\nSince the Points D, E trisect BC.
\nBD = DE = CE
\nLet BD = DE = CE = x
\nBE = 2x and BC = 3x
\n\"Samacheer
\nIn the right \u2206ABD,
\nAD2<\/sup> = AB2<\/sup> + BD2<\/sup>
\nAD2 <\/sup>= AB2<\/sup> + x2<\/sup> ……….(1)
\nIn the right \u2206ABE,
\nAE2<\/sup> = AB2<\/sup> + 2BE2<\/sup>
\nAE2<\/sup> = AB2<\/sup> + 4X2<\/sup> ………..(2) (BE = 2x)
\nIn the right \u2206ABC
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\nAC2<\/sup> = AB2<\/sup> + 9x2<\/sup> …………… (3) (BC = 3x)
\nR.H.S = 3AC2<\/sup> + 5AD2<\/sup>
\n= 3[AB2<\/sup> + 9x2<\/sup>] + 5 [AB2<\/sup> + x2<\/sup>] [From (1) and (3)]
\n= 3AB2<\/sup> + 27x2<\/sup> + 5AB2<\/sup> + 5x2<\/sup>
\n= 8AB2<\/sup> + 32x2<\/sup>
\n= 8 (AB2<\/sup> + 4 x2<\/sup>)
\n= 8AE2<\/sup> [From (2)]
\n= R.H.S.
\n\u2234 8AE2<\/sup> = 3AC2<\/sup> + 5AD2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.3 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[2],"tags":[],"class_list":["post-693","post","type-post","status-publish","format-standard","hentry","category-class-10"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/693"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=693"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/693\/revisions"}],"predecessor-version":[{"id":39872,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/693\/revisions\/39872"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=693"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=693"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=693"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}