Question

help me pleaseeeeeeeeeee
se considera triunghiul isoscel mnp cu mn = mp = 25 cm si np = 30 cm .
calculati aria triunghiului mnp
daca prin i , centrul cercului inscris in triunghiul mnp , se duce o paralela qs la np , aflati raportul
A mqs / A mnp

Answer 1

R∈(NP),MR⊥NP
NR=RP=NP/2
NR=RP=15cm
[tex]MR^2=MP^2-PR^2\\ MR^2=625-225\\ MR^2=400 [/tex]
⇒MR=20 cm
[tex]A_{MNP}= \frac{MR*NP}{2}\\ A_{MNP}= \frac{20*30}{2}=300cm^2 [/tex]
[tex]P_{MNP}=MN+MP+NP\\ P_{MNP}=25+25+30\\ P_{MNP}=80 cm\\ p= \frac{P}{2}=40 cm [/tex]
[tex]MI= \frac{A}{p}\\ MI= \frac{300}{40}=7,5 cm [/tex]
MR⊥NP
QS||NP⇒MR⊥QS
QI||NR⇒ΔMQI≈ΔMNR si de aici vom avea fractiile:
[tex] \frac{QI}{NR}=\frac{MI}{MR}= \frac{MQ}{QN}\\ \frac{QI}{15}= \frac{7,5}{20}\\ QI= \frac{15* \frac{15}{2} }{20}\\ QI= \frac{225}{2}* \frac{1}{20}= \frac{45}{8}cm [/tex] 
QS=2*QI
QS=2*45/8=45/4 cm
[tex]A_{MQS}= \frac{MI*QS}{2}\\ A_{MQS}= \frac{ \frac{15}{2}*\frac{45}{4} }{2}\\ A_{MQS}= \frac{675}{8}* \frac{1}{2}\\ A_{MQS}= \frac{675}{16}cm^2\\ \frac{A_{MQS}}{A_{MNP}}= \frac{ \frac{675}{16} }{300}\\ \frac{A_{MQS}}{A_{MNP}}= \frac{675}{16}* \frac{1}{300}\\ \frac{A_{MQS}}{A_{MNP}}= \frac{9}{64} [/tex]