In triangle DEllBC, AD=1cm and BD

1.	In triangle DEllBC, AD=1cm and BD
Answer» It is given that AD = 1 cm,BD = 2 cm and\xa0{tex}D E \\\| B C{/tex}In {tex}\\triangle{/tex}ADE and\xa0{tex}\\triangle {/tex}ABC{tex}\\angle A D E = \\angle A B C{/tex}\xa0(Corresponding angles){tex}\\angle A = \\angle A{/tex}\xa0[Common]Therefore, by A.A. similar condition{tex}\\triangle \\mathrm { ADE } \\sim \\triangle \\mathrm { ABC }{/tex}Ratio of areas of similar triangles is equal to the square of the ratio of the corresponding sides.{tex}\\therefore \\quad \\frac { \\operatorname { ar } ( \\triangle \\mathrm { ABC } ) } { \\operatorname { ar } ( \\triangle \\mathrm { ADE } ) } = \\frac { \\mathrm { AB } ^ { 2 } } { \\mathrm { AD } ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\frac { \\operatorname { ar } ( \\triangle A B C ) } { \\operatorname { ar } ( \\triangle A D E ) } = \\left( \\frac { 3 } { 1 } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad \\frac { \\operatorname { arc } \\triangle A B C ) } { \\operatorname { ar } ( \\triangle A D E ) }{/tex}{tex}= \\frac { 9 } { 1 }{/tex}

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