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| 1. |
In figure xp/py=xq/qz=3if the area of triangle xyz is32cm, then find area of qadrilateral pyzq |
| Answer» Given {tex}\\frac{{XP}}{{PY}} = \\frac{{XQ}}{{QZ}}{/tex}\xa0\u200b{tex}\\Rightarrow {/tex}\u200b PQ {tex}\\parallel{/tex} YZ .....(i) [By converse of B.P.T]In \u200b{tex}\\triangle {/tex}\u200b XPQ and \u200b{tex}\\triangle {/tex}\u200b XYZ we have[ \u200b{tex}\\angle{/tex}\u200b XPQ = \u200b{tex}\\angle{/tex}\u200b Y [From (i) corresponding angles]\u200b{tex}\\angle{/tex}\u200b X = \u200b{tex}\\angle{/tex}\u200bX [common]\u200b{tex}\\therefore {/tex}\u200b\u200b{tex}\\triangle {/tex}\u200b XPQ \u200b{tex}\\cong{/tex}\u200b{tex}\\triangle {/tex}XYZ [By AA similarity]\u200b{tex}\\therefore {/tex}\u200b{tex}\\frac{{ar\\left( {\\triangle XYZ} \\right)}}{{ar\\left( {\\triangle XPQ} \\right)}}=\\frac{{X{Y^2}}}{{X{P^2}}}{/tex}{tex}\\frac { P Y } { X P } = \\frac { 1 } { 3 }{/tex}We have\xa0{tex}\\Rightarrow \\frac { P Y } { X P } + 1 = \\frac { 1 } { 3 } + 1{/tex}{tex}\\Rightarrow \\frac { P Y + X P } { X P } = \\frac { 4 } { 3 }{/tex}\xa0\u200b\u200b\u200b{tex}\\Rightarrow {/tex}\u200b{tex}\\frac{{XY}}{{XP}} = \\frac{4}{3}{/tex}Substituting in (i), we get{tex}\\frac{{ar\\left( {\\triangle XYZ} \\right)}}{{ar\\left( {\\triangle XPQ} \\right)}} = {\\left( {\\frac{4}{3}} \\right)^2} = \\frac{{16}}{9}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b{tex}\\frac{{32}}{{ar\\left( {XPQ} \\right)}} = \\frac{{16}}{9}{/tex}{tex}ar\\left( {XPQ} \\right) = \\frac{{32 \\times 9}}{{16}} = 18c{m^2}{/tex}Area of quadrilateral PYZQ = 32 - 18 = 14 cm2 | |