Question

[tex]Fie\ triunghiul\ oarecare\ ABC\, in\ care\ AB=30\ cm,BC=51\ cm,\\
AC=63\ cm.Stiind\ ca\ M \in(AB)\ astfel\ incat\ BM=\frac{1}{4}AB,N \in(BC)\\
astfel\ incat\ 3BN=2BC\ si\ P\in(AC)\ astfel\ incat\ AC=7AP,\\
calculati\ aria\ triunghiului\ MNP.\\
\\
Am\ gasit\ problema\ asta\ in\ culegerea\ de\ clasa\ a\ saptea\ si\ nu\ stiu\\
cum\ s-o\ rezolv.Poate\ cineva\ sa\ ma\ ajute?Oricum,\ m-am\ uitat\\
la\ sfarsit\ si\ raspunsul\ era\ 333\ cm^2,\ dar\ vreau\ si\ rezolvarea.[/tex]

Answer 1

$\displaystyle Ok...deci~cu~Heron~am~gasit~A_{ABC}=756~cm^2. ~Avem:\\ \\ \frac{AP}{AC}= \frac{1}{7}~;~ \frac{PC}{AC}= \frac{6}{7}~; ~ \frac{BN}{BC}= \frac{2}{3}~;~ \frac{NC}{BC}= \frac{1}{3}~;~ \frac{BM}{AB}= \frac{1}{4}~;~ \frac{AM}{AB} = \frac{3}{4}. \\ \\ \frac{A_{AMP}}{A_{ABC}}= \frac{ \frac{1}{2} \Big(AM \cdot AP \cdot sinA \Big) }{ \frac{1}{2} \Big(AB \cdot AC \cdot sinA \Big) } = \frac{AM}{AB} \cdot \frac{AP}{AC}= \frac{3}{4} \cdot \frac{1}{7}= \frac{3}{28}. \\ \\ Gasim~analog:$

$\displaystyle \frac{A_{PNC}}{A_{ABC}}= \frac{PC}{AC} \cdot \frac{NC}{BC}= \frac{6}{7} \cdot \frac{1}{3}= \frac{2}{7}. \\ \\ \frac{A_{BMN}}{A_{ABC}}= \frac{BM}{AB} \cdot \frac{BN}{BC}= \frac{1}{4} \cdot \frac{2}{3}= \frac{1}{6}. \\ \\ \\ A_{MNP}=A_{ABC}- \Big( A_{AMP}+A_{PNC}+A_{BMN} \Big) = A_{ABC}- A_{ABC}\Big( \frac{3}{28}+ \\ \\ + \frac{2}{7}+ \frac{1}{6} \Big)=calcule...$