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| 1. |
In a triangle and,DE parallel BC.if AD:DB=3:5 then area of triangle DEC by area of triangle CFB |
| Answer» In \u200b\u200b{tex}\\triangle {/tex}\u200bADE and \u200b\u200b{tex}\\triangle {/tex}\u200bABC\u200b{tex}\\angle{/tex}\u200b 1 = \u200b{tex}\\angle{/tex}\u200b 1 [Common]\u200b{tex}\\angle{/tex}\u200b 2 = \u200b{tex}\\angle{/tex}\u200b ACB [Corresponding \u200b{tex}\\angle{/tex}\u200bs]\u200b\u200b{tex}\\therefore {/tex}\u200b\u200b\u200b{tex}\\triangle {/tex}\u200b\u200bADE \u200b\u200b{tex}\\sim{/tex}\u200b\u200b{tex}\\triangle {/tex}\u200b ABC [By A.A Rule]{tex}\\therefore {/tex}{tex}\\frac{{DE}}{{BC}} = \\frac{{AD}}{{AB}}{/tex}.....(i)Again in \u200b\u200b{tex}\\triangle {/tex}\u200bDEF and \u200b\u200b{tex}\\triangle {/tex}\u200bCFB\u200b{tex}\\angle{/tex}\u200b 3 = \u200b{tex}\\angle{/tex}\u200b 6 [Alternate \u200b{tex}\\angle{/tex}\u200b s]\u200b{tex}\\angle{/tex}\u200b 4 = \u200b{tex}\\angle{/tex}\u200b 5 [Vertically opposite \u200b{tex}\\angle{/tex}\u200b s]\u200b{tex}\\therefore {/tex}\u200b\u200b\u200b{tex}\\triangle {/tex}\u200b\u200b DFE \u200b{tex} \\sim {/tex}\u200b\u200b{tex}\\triangle {/tex}\u200b CFB [By A.A Rule]\u200b{tex}\\therefore {/tex}\u200b{tex}\\frac{{area\\left( {\\triangle DFE} \\right)}}{{area\\left( {\\triangle CFB} \\right)}} = \\frac{{D{E^2}}}{{B{C^2}}} = {\\left( {\\frac{{AD}}{{AB}}} \\right)^2}{/tex}[From (i)]{tex}= {\\left( {\\frac{5}{9}} \\right)^2}\\left[ {\\because \\frac{{AD}}{{DB}} = \\frac{5}{4} \\Rightarrow \\frac{{AD}}{{AD + DB}} = \\frac{5}{{5 + 4}} \\Rightarrow \\frac{{AD}}{{DB}} = \\frac{5}{9}} \\right]{/tex}\u200b{tex}\\therefore {/tex}\u200b{tex}\\frac{{area\\left( {\\triangle DFE} \\right)}}{{area\\left( {\\triangle CFB} \\right)}} = \\frac{{25}}{{81}}{/tex} | |