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| 1. |
In a tri ABC , DE||BC .if AD is 2.4 AE is 3.2DE is 2 Bc is 5Find BD nd CE |
| Answer» We have,DE || BCNow, In {tex}\\triangle{/tex}ADE and {tex}\\triangle {/tex}ABC{tex}\\angle A = \\angle A{/tex}\xa0[common]{tex}\\angle A D E = \\angle A B C{/tex}\xa0[{tex}\\because{/tex}\xa0DE || BC {tex}\\Rightarrow{/tex}\xa0Corresponding angles are equal]{tex}\\Rightarrow \\triangle A D E= \\triangle A B C{/tex}\xa0[By AA criteria]{tex}\\Rightarrow \\frac { A B } { B C } = \\frac { A D } { D E }{/tex}\xa0[{tex}\\because{/tex}\xa0Corresponding sides of similar triangles are proportional]{tex}\\Rightarrow \\frac { A B } { 5 } = \\frac { 2.4 } { 2 }{/tex}{tex}\\Rightarrow A B = \\frac { 2.4 \\times 5 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0AB = 1.2 {tex}\\times{/tex}\xa05= 6.0 cm{tex}\\Rightarrow{/tex}\xa0AB = 6 cm{tex}\\therefore{/tex}\xa0BD = AB - AD= 6 - 2.4= 3.6 cm{tex}\\Rightarrow{/tex}\xa0DB = 3.6 cmNow,{tex}\\frac { A C } { B C } = \\frac { A E } { D E }{/tex}\xa0[{tex}\\because{/tex}\xa0Corresponding sides of similar triangles are equal]{tex}\\Rightarrow \\frac { A C } { 5 } = \\frac { 3.2 } { 2 }{/tex}{tex}\\Rightarrow A C = \\frac { 3.2 \\times 5 } { 2 }{/tex}= 1.6 {tex}\\times{/tex}\xa05= 8.0 cm{tex}\\Rightarrow{/tex}\xa0AC = 8 cm{tex}\\therefore{/tex}\xa0CE = AC - AE= 8 - 3.2= 4.8 cmHence, BD = 3.6 cm and CE = 4.8 cm | |