Răspuns:
[tex]A_{\triangle MQR}~\text{este}~25\%~\text{din}~A_{\triangle MNP}[/tex]
Explicație pas cu pas:
Ipoteza:
[tex]NM\perp PM\\MR~\text{median\u{a}}\\m\sphericalangle P=30\°\\MN=10~\text{dm}\\[/tex]
Aflăm ipotenuza [tex]NP[/tex] aplicând [tex]\sin[/tex] în [tex]\sphericalangle MPN[/tex]:
[tex]\sin{30\°}=\frac{1}{2}\\\\\triangle MPN~\text{dreptunghic}\Rightarrow \sin{\sphericalangle MPN}=\frac{cat. op}{ip.}=\frac{MN}{NP}\\\Rightarrow \frac{1}{2}=\frac{MN}{NP}\\\Rightarrow 1\cdot NP=2\cdot MN\\\Rightarrow NP=2\cdot10\\\Rightarrow NP=20~\text{dm}[/tex]
Aplicăm teorema lui Pitagora în [tex]\triangle MNP[/tex] și aflăm [tex]$MP$[/tex]:
[tex]{\text{cateta}_1}^2+{\text{cateta}_2}^2=\text{ipotenuza}^2\\\\\Rightarrow MN^2+MP^2=NP^2 \Rightarrow MP=\sqrt{NP^2-MN^2}\Rightarrow\\\Rightarrow MP=\sqrt{20^2-10^2}\\\Rightarrow MP=\sqrt{400-100}\\\Rightarrow MP=\sqrt{300}\\\Rightarrow MP=10\sqrt{3}~\text{dm}[/tex]
Folosim teorema catetei pentru a afla segmentele [tex]NQ[/tex] și [tex]PQ[/tex]:
[tex]\text{cateta}^2=\text{proiectia catetei}\cdot\text{ipotenuza}\\\\\Rightarrow MN^2=NQ\cdot NP\\\Rightarrow 10^2=NQ\cdot20\\\Rightarrow NQ\cdot20=100\\\Rightarrow NQ=5~\text{dm}\\\\PQ=NP-NQ\\\Rightarrow PQ=20-5\\\Rightarrow PQ=15~\text{dm}[/tex]
Calculăm aria triunghiului:
[tex]A_{\triangle MNP}=\frac{\text{cateta}_1\cdot\text{cateta}_2}{2}\\\Rightarrow A_{\triangle MNP}=\frac{MN\cdot MP}{2}\\\Rightarrow A_{\triangle MNP}=\frac{10\cdot10\sqrt{3}}{2}\\\Rightarrow A_{\triangle MNP}=50\sqrt{3}~\text{dm}^2[/tex]
Una dintre proprietățile medianei (în orice triunghi) este că împarte aria în valori egale. Putem folosi asta pentru a afla [tex]A_{\triangle MPR}[/tex]:
[tex]MR~\text{median\u{a}}\Rightarrow A_{\triangle MPR}=\frac{A_{\triangle MNP}}{2}\\\Rightarrow A_{\triangle MPR}=\frac{50\sqrt{3}}{2}\\\\\Rightarrow A_{\triangle MPR}=25\sqrt{3}~\text{dm^2}[/tex]
Exprimăm aria [tex]\triangle MNP[/tex] prin formula [tex]\frac{\text{baza}\cdot\text{\^{i}n\u{a}ltimea}}{2}[/tex] pentru a afla înălțimea [tex]MQ[/tex]:
[tex]A_{\triangle MNP}=50\sqrt{3}~\text{dm}^2\\\\A_{\triangle MNP}=\frac{NP\cdot MQ}{2}\\\Rightarrow 50\sqrt{3}=\frac{20\cdot MQ}{2}\\\Rightarrow 100\sqrt{3}=20MQ\\\Rightarrow MQ=\frac{100\sqrt{3}}{20}\\\Rightarrow MQ=5\sqrt{3}~\text{dm}[/tex]
Aflăm [tex]A_{\triangle MPQ}[/tex] și scădem [tex]A_{\triangle MPR}[/tex] din ea pentru a determina [tex]A_{\triangle MQR}[/tex]:
[tex]A_{\triangle MPQ}=\frac{MQ\cdot QP}{2}\\\Rightarrow A_{\triangle MPQ}=\frac{5\sqrt{3}\cdot 15}{2}\\\Rightarrow A_{\triangle MPQ}=\frac{75\sqrt{3}}{2}\\\\A_{\triangle MQR}=A_{\triangle MQP}-A_{\triangle MPR}\\\Rightarrow A_{\triangle MQR}=\frac{75\sqrt{3}}{2}-25\sqrt{3}\\\Rightarrow A_{\triangle MQR}=\frac{75\sqrt{3}-2\cdot25\sqrt{3}}{2}\\\Rightarrow A_{\triangle MQR}=\frac{25\sqrt{3}}{2}~\text{dm}^2[/tex]
Comparăm ariile:
[tex]A_{\triangle MQR}=x\cdot A_{\triangle MNP}\\\frac{25\sqrt{3}}{2}=x\cdot 50\sqrt{3}\\\frac{25\sqrt{3}}{2} : 50\sqrt{3}=x\\x=\frac{25\sqrt{3}}{2}\cdot \frac{1}{50\sqrt{3}}\\x=\frac{25\sqrt{3}}{100\sqrt{3}}\\x=\frac{25}{100}\\x=25\%[/tex]
