a) Aria=AD*AB=30√3*30=900√3 m²
b)construiesti diagonala AC si in triunghiul ABC aplici teorema lui pitagora AC²=AB²+BC²
si vei obtine AC=60 m
apoi folosesti teorema asemanarii: triunghiul BAC este asemenea cu triunghiul BMN
[tex] \frac{BM}{BA}= \frac{BN}{BC} = \frac{MN}{AC} [/tex]
BM=10,BA=30, AC=60, BC=30√3
[tex] \frac{10}{30} = \frac{BN}{30 \sqrt{3} } [/tex]⇒BN=10√3
[tex] \frac{10}{30} = \frac{MN}{60} [/tex]⇒MN=20
NC=BC-BN=30√3-10√3=20√3
AM=AB-MB=30-10=20
CP=AM=20
NP²=NC²+CP² ⇒NP=40=MQ
cos BMN=[tex] \frac{MB}{MN} [/tex]=[tex] \frac{10}{20}= \frac{1}{2} [/tex]
⇒m(BMN)=60 grade
cos AMQ=[tex] \frac{AM}{MQ}= \frac{20}{40} = \frac{1}{2} [/tex]
⇒ m(AMQ)=60 grade
m(NMQ)=180-m(AMQ)-m(BMN)=180-60-60=60 grade
c) MN=QP=20
NP=MQ=40
Perimetrul= 2*20+2*40=40+80=120
