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Modellina
Pe: 2008-07-01T18:13:38+03:00 In: In:

! a7a

in trapezul ABCD, AB||CD, AB=36cm, BC=18, AD=15 , DC=18 se noteaza AD intersectat cu BC={M}.
a)perimetrul tr MAB
b)daca AC intersectat cu BD={O} si MO intersectat cu AB={E}, stabiliti natura patrulaterului BCDE
c)aratati ca BD perpendicular pe CE
d)aratati ca MD/MA+BE/BA=1

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5 raspunsuri

  1. omegatheo

        \[ \begin{array}{l} {\rm a)in }\Delta {\rm MAB}{\rm aplicam TFA si avem }\frac{{{\rm ME}}}{{{\rm MA}}} = \frac{{MC}}{{{\rm MB}}} = \frac{{{\rm DC}}}{{{\rm AB }}}{\rm si MA = 30 si MB = 36 } \\ {\rm deci P[MAB] = 102cm}{\rm .} \\ {\rm b)}\angle {\rm DBA = < CBD = < CDB deci BO e bisectoare in }\Delta {\rm ABC si }\frac{{{\rm CO}}}{{{\rm AO}}} = \frac{{CB}}{{AB}} = \frac{1}{2}( \\ {\rm din teorema bisectoarei)}{\rm .} \\ {\rm De asemenea in ABC cu transversala M - O - E aplicam teorema lui Menelaos:} \\ \frac{{{\rm AE}}}{{{\rm EB}}} \cdot \frac{{BM}}{{{\rm CM}}} \cdot \frac{{CO}}{{AO}} = 1 \Rightarrow AE = BE{\rm deci EB = DC = 18 cm si EB||DC}{\rm deci DCBE} \\ {\rm e paralelogram}{\rm .} \\ {\rm DE e linie mijlocie si DE = }\frac{{\rm 1}}{{\rm 2}}*MB{\rm = 18 cm deci paralelogramul e romb si DB} \bot {\rm CE} \\ {\rm d)reiese imediat prin calcule} \\ \end{array} \]

  2. Modellina

    sunt numai coduri…nu se intelege nimic

    multumesc

  4. omegatheo

    Modellina wrote: sunt numai coduri…nu se intelege nimic

    multumesc


    Imi cer scuze.Multumesc!

  5. dan
    expert (VI)

    omegatheo wrote:

        \[ \begin{array}{l} {\rm a)in }\Delta {\rm MAB}{\rm aplicam TFA si avem }\frac{{{\rm ME}}}{{{\rm MA}}} = \frac{{MC}}{{{\rm MB}}} = \frac{{{\rm DC}}}{{{\rm AB }}}{\rm si MA = 30 si MB = 36 } \\ {\rm deci P[MAB] = 102cm}{\rm .} \\ {\rm b)}\angle {\rm DBA = < CBD = < CDB deci BO e bisectoare in }\Delta {\rm ABC si }\frac{{{\rm CO}}}{{{\rm AO}}} = \frac{{CB}}{{AB}} = \frac{1}{2}( \\ {\rm din teorema bisectoarei)}{\rm .} \\ {\rm De asemenea in ABC cu transversala M - O - E aplicam teorema lui Menelaos:} \\ \frac{{{\rm AE}}}{{{\rm EB}}} \cdot \frac{{BM}}{{{\rm CM}}} \cdot \frac{{CO}}{{AO}} = 1 \Rightarrow AE = BE{\rm deci EB = DC = 18 cm si EB||DC}{\rm deci DCBE} \\ {\rm e paralelogram}{\rm .} \\ {\rm DE e linie mijlocie si DE = }\frac{{\rm 1}}{{\rm 2}}*MB{\rm = 18 cm deci paralelogramul e romb si DB} \bot {\rm CE} \\ {\rm d)reiese imediat prin calcule} \\ \end{array} \]

    \rm{ Scuze ! In loc de" \frac{ME}{MA}=" cred ce trebuia ...\frac{MD}{MA}=...

  6. omegatheo

    posibil…imi cer scuze pentru eventuale greseli 😀

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