Problema 3 de la partea a3 -a
Răspunsuri la întrebare
Răspuns:
Explicație pas cu pas:
AB=36, AM=MB=18, ND=36:3=12, NA=12·2=24,
a) MN²=AM²+AN²=18²+24²=6²·3²+6²·4²=6²·(3²+4²)=6²·5². Atunci MN=√(6²·5²)=6·5=30.
NC²=ND²+CD²=12²+36²=12²+12²·3²=12²·(1+3²)=12²·10, atunci NC=√(12²·10)=12√10.
MC²=MB²+BC²=18²+36²=18²+18²·2²=18²·(1+2²)=18²·5, deci MC=√(18²·5)=18√5.
Atunci P(ΔMNC)=MN+MC+NC=30+18√5+12√10.
Aria(ΔMNC)=Aria(ABCD)-Aria(ΔAMN)-Aria(ΔMBC)-Aria(ΔCND)=AB²-(1/2)·AM·AD-(1/2)·BM·BC-(1/2)·ND·CD=36²-(1/2)·18·24-(1/2)·18·36-(1/2)·12·36=
=36²-9·24-9·36-6·36=36(36-6-9-6)=36·16=576 cm².
b) Aria(ΔMNC)=(1/2)·MC·MN·sin(∡CMN), deci
(1/2)·18√5·30·sin(∡CMN)=576, ⇒9√5·30·sin(∡CMN)=576, ⇒
sin(∡CMN)=576/(9√5·30)=64/(30√5)=(32√5)/75.
c) d(C,MN)=CE, unde CE⊥MN.
atunci Aria(ΔMNC)=(1/2)·MN·CE, sau (1/2)·30·CE=576, ⇒15·CE=576, ⇒CE=576/15=192/5=384/10=38,4 cm.
Răspuns:
Explicație pas cu pas:
