{"id":489,"date":"2020-09-15T06:19:32","date_gmt":"2020-09-15T00:49:32","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=489"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-maths-solutions-chapter-2-ex-2-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-2-ex-2-4\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4<\/strong>\u00a0are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4<\/h2>\n

Question 1.
\nLet \u2206ABC ~ \u2206DEF and their areas be, respectively, 64 cm2<\/sup> and 121 cm2<\/sup>, If EF = 15.4 cm., find BC.
\nSolution:
\n\"KSEEB
\nIf \u2206ABC ~ \u2206DEF, then
\n\\(\\frac { { ar }(\\Delta { ABC }) }{ { ar }(\\Delta { DEF }) } =\\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \\)
\n\\(\\frac{64}{121}=\\frac{\\mathrm{BC}^{2}}{(15.4)^{2}}\\)
\n\\(\\mathrm{BC}^{2}=\\frac{64}{121}=\\frac{\\mathrm{BC}^{2}}{(15.4)^{2}}\\)
\n\\(\\mathrm{BC}^{2}=\\frac{64}{121} \\times \\frac{237.16}{1}\\)
\n\u2234 BC = 8 \u00d7 1.4
\n\u2234 BC = 11.2 cm<\/p>\n

Question 2.
\nDiagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
\n\"KSEEB
\nSolution:
\nData: Diagonals of a trapezium ABCD with AB||DC intersect each other at the point O. We have AB = 2 CD.
\nTo prove: \\(\\frac { { Ar }(\\Delta { A }{ OB }) }{ { Ar }(\\Delta { COD }) } =?\\)
\nAs per similarity criterion, \u2206AOB and \u2206COD are,
\n\u2206AOB ~ \u2206COD
\n\\(\\frac { { Ar }(\\Delta { AOB }) }{ { Ar }(\\Delta { COD }) } =\\frac { { AB }^{ 2 } }{ { DC }^{ 2 } } \\)
\n\\(=\\frac{(2 \\mathrm{DC})^{2}}{\\mathrm{DC}^{2}}(\\cdot \\mathrm{AB}=2 \\mathrm{DC})\\)
\n\\(=\\frac{4 \\mathrm{DC}^{2}}{1 \\mathrm{DC}^{2}}\\)
\n\\(\\frac { { Ar }(\\Delta { AOB }) }{ { Ar }(\\Delta { COD }) } =\\frac { 4 }{ 1 } \\)
\n= 4 : 1<\/p>\n

Question 3.
\nIn the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
\n\\(\\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { DBC }) } =\\frac { { AO } }{ { DO } } \\)
\n\"KSEEB
\nSolution:
\nData: ABC and DBC are two triangles on the same base BC. If AD intersects BC at O.
\nTo Prove: \\(\\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { DBC }) } =\\frac { { AO } }{ { DO } } \\)
\nConstruction: Draw AP \u22a5 BC and DM \u22a5 BC.
\n\"KSEEB
\nArea of \u2206 = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height
\n\\(\\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { DBC }) } =\\frac { \\frac { 1 }{ 2 } \\times BC\\times AP }{ \\frac { 1 }{ 2 } \\times BC\\times DM } \\)
\n\\(=\\frac{\\mathrm{AP}}{\\mathrm{DM}}\\)
\nIn \u2206APO and \u2206DMO,
\n\u2220APO = \u2220DMO = 90\u00b0 (construction)
\n\u2220AOP = \u2220DOM (Vertically opposite angles)
\n\u2234 \u2206APO ~ \u2206DMO
\n(\u2235 Similarity criterion is A.A.A.)
\n\\(\\frac{\\mathrm{AP}}{\\mathrm{DM}}=\\frac{\\mathrm{AO}}{\\mathrm{DO}}\\)
\n\\(\\Rightarrow \\quad \\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { DBC }) } =\\frac { { AO } }{ { DO } } \\)<\/p>\n

Question 4.
\nIf the areas of two similar triangles are equal, prove that they are congruent.
\n\"KSEEB
\nSolution:
\nIn \u2206ABC ~ \u2206PQR
\n\\(\\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { PQR }) } =\\left( \\frac { { AB } }{ { PQ } } \\right) ^{ 2 }=\\left( \\frac { { BC } }{ { QR } } \\right) ^{ 2 }=\\left( \\frac { { AC } }{ { PR } } \\right) ^{ 2 }\\) …………. (1)
\nArea of \u2206ABC = Area of \u2206PQR
\n\\(\\Rightarrow \\quad \\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { PQR }) } =1\\) …………. (2)
\nSubstituting the eqn (2) in eqn (1)
\n\\(1=\\left(\\frac{\\mathrm{AB}}{\\mathrm{PQ}}\\right)^{2}=\\left(\\frac{\\mathrm{BC}}{\\mathrm{QR}}\\right)^{2}=\\left(\\frac{\\mathrm{AC}}{\\mathrm{PR}}\\right)^{2}\\)
\n\u21d2 AB = PQ, BC = QR, AC = PR
\n\u2234 \u2206ABC \u2245 \u2206PQR (\u2235 S.S.S postulate)<\/p>\n

Question 5.
\nD, E, and F are respectively mid-points of sides AB, BC and CA of \u2206ABC. Find the ratio of the areas of \u2206DEF and \u2206ABC.
\n\"KSEEB
\nSolution:
\nData: D. E and F are respectively midpoints of sides AB, BC and CA of \u2206ABC.
\nTo Prove: Ratio of the areas of \u2206DEF and \u2206ABC.
\nIf a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side and it is half of that side
\n\\(\\mathrm{DE} \\| \\mathrm{AC} \\quad \\mathrm{DE}=\\frac{1}{2} \\mathrm{AC}\\)
\n\\(\\mathrm{EF} \\| \\mathrm{AB} \\quad \\mathrm{EF}=\\frac{1}{2} \\mathrm{AB}\\)
\n\\(\\mathrm{DF} \\| \\mathrm{BC} \\quad \\mathrm{DF}=\\frac{1}{2} \\mathrm{BC}\\)
\n\\(\\frac { { Ar }(\\Delta { DEF }) }{ { Ar }(\\Delta { ABC }) } =\\frac { { DF }^{ 2 } }{ { BC }^{ 2 } } \\quad ({ Theorem }6)\\)
\n\\(\\frac { { Ar }(\\Delta { DEF }) }{ { Ar }(\\Delta { ABC }) } =\\frac { \\left( \\frac { 1 }{ 2 } { BC } \\right) ^{ 2 } }{ { BC }^{ 2 } } \\)
\n\\(\\frac { { Ar }(\\Delta { DEF }) }{ { Ar }(\\Delta { ABC }) } =\\frac { \\frac { 1 }{ 4 } { BC }^{ 2 } }{ { BC }^{ 1 } } =\\frac { 1 }{ 4 } =1:4\\)<\/p>\n

Question 6.
\nProve that the ratio of the areas , of two similar triangles is equal to the square of the ratio of their corresponding medians.
\n\"KSEEB
\nSolution:
\nLet \u2206ABC ~ \u2206PQR
\nAD and PS are mid-point lines of \u2206ABC and \u2206PQR.
\n\u2206ABC ~ \u2206PQR
\n\\(\\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{BC}}{\\mathrm{QR}}=\\frac{\\mathrm{AC}}{\\mathrm{PR}}\\) ………………… (1)
\n\u2220A = \u2220P, \u2220R = \u2220Q, and \u2220C = \u2220R (2)
\nAD and PS are mid-point lines
\n\\(B D=D C=\\frac{B C}{2}\\)
\nand \\(\\quad \\mathrm{QS}=\\mathrm{SR}=\\frac{\\mathrm{QR}}{2}\\)
\n\\(\\frac{A B}{P Q}=\\frac{B D}{Q S}=\\frac{A C}{P R}\\) …………… (3)
\nIn \u2206ABD and \u2206PQS,
\n\u2220B = \u2220Q ………. Eqn (2)
\nand \\(\\frac{A B}{P Q}=\\frac{B D}{Q S}\\) ………. Eqn (3)
\n\u2234 \u2206ABD ~ \u2206PQS (SAS Postulate)
\n\\(\\frac{A B}{P Q}=\\frac{B D}{Q S}=\\frac{A D}{P S}\\) ……………. (4)
\n\\(\\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { PQR }) } =\\left( \\frac { { AB } }{ { PQ } } \\right) ^{ 2 }=\\left( \\frac { { BC } }{ { QR } } \\right) ^{ 2 }=\\left( \\frac { { AC } }{ { PR } } \\right) ^{ 2 }\\)
\nFrom Eqn. (1) and Eqn. (4), we have
\n\\(\\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{BC}}{\\mathrm{QR}}=\\frac{\\mathrm{AC}}{\\mathrm{PR}}=\\frac{\\mathrm{AD}}{\\mathrm{PS}}\\)
\n\\(\\Rightarrow \\quad \\frac { { Ar }(\\Delta { BC }) }{ { Ar }(\\Delta { PQR }) } =\\left( \\frac { { AD } }{ { PS } } \\right) ^{ 2 }\\)<\/p>\n

Question 7.
\nProve that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
\n\"KSEEB
\nSolution:
\nData: Equilateral triangle ABE is on AB of square ABCD. Equilateral triangle ACF is on diagonal AC.
\nTo Prove: \\((\\Delta A B E)=\\frac{1}{2} \\text { Ar }(\\Delta A C F)\\)
\n\\(\\frac { { Ar }(\\Delta { ABE }) }{ { Ar }(\\Delta { ACF }) } =\\frac { { AB }^{ 2 } }{ { AC }^{ 2 } } \\quad { Theorem }6\\)
\n\\(=\\frac{A B^{2}}{A B^{2}+B C^{2}}\\)
\n\u2235 Pythagoras theorem
\n\\(=\\frac{\\mathrm{AB}^{2}}{2 \\mathrm{AB}^{2}}\\)
\n\u2234 AB = AC sides of square
\n\\(=\\frac{1 \\times A B^{2}}{2 \\times A B^{2}}\\)
\n\\(\\frac { { Ar }(\\Delta { ABE }) }{ { Ar }(\\Delta { ACF }) } =\\frac { 1 }{ 2 } \\)<\/p>\n

Tick the correct answer and justify:<\/strong>
\nQuestion 8.
\n\u2206ABC and \u2206BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles \u2206ABC and \u2206BDE is :
\nA) 2 : 1
\nB) 1 : 2
\nC) 4 : 1
\nD) 1 : 4
\nSolution:
\nC) 4 : 1
\n\"KSEEB
\n\\(\\frac { { Ar }(\\Delta { ABC }) }{ { Ar }(\\Delta { BDE }) } =\\frac { { BC }^{ 2 } }{ { BD }^{ 2 } } =\\frac { { BC }^{ 2 } }{ \\left( \\frac { 1 }{ 2 } { BC } \\right) ^{ 2 } } \\)
\n\\(=\\quad \\frac{1}{1 \/ 4}=\\frac{4}{1}\\)
\n= 4 : 1<\/p>\n

Question 9.
\nSides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio :
\nA) 2 : 3
\nB) 4 : 9
\nC) 81 : 16
\nD) 16 : 81
\nSolution:
\nD) 16 : 81
\n\u2234 (4)2<\/sup> : (9)2<\/sup> = 16 : 81.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4\u00a0are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4. Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4 Question 1. Let \u2206ABC ~ \u2206DEF and their areas be, …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/489"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=489"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/489\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=489"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=489"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=489"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}