Question

ajutor repede ex 3 dau coroana

Answer 1

Răspuns:

Explicație pas cu pas:

MP⊥(ABC), PA=PD, deci oblicele MA=MD, deci ΔMAD isoscel.

ABCD patrat, AB=MP=2a.

P(ΔMAD)=MA+AD+MD

Din ΔMAP, dreptunghic in P, MA²=MP²+AP²=(2a)²+a²=5a². deci MA=a√5.

P(ΔMAD)=MA+AD+MD=a√5+2a+a√5=2a√5+2a=2a·(√5 +1)

b) (NEF)║(MBC)   ???

EF║BC, in ΔMPG, NO este linie mijlocie, deci NO║MG. Atunci avem doua drepte (EF si NO) concurente dintr-un plan (NEF), paralele cu doua drepte concurente (BC si MG) din planul MBC, deci (NEF)║(MBC).

c) sin(∠(NE,NF))=???

PO⊥EF, deci NO⊥EF. PO=a=PN. Atunci NO=a√2. Atunci NE²=NO²+EO²=(a√2)²+a²=3a². Deci NE=a√3. Aria(ΔNEF)=(1/2)·EF·NO=(1/2)·2a·a√2=a²√2.

Aria(ΔNEF)=(1/2)·NE·NF·sin(∠(NE,NF))=a²√2 |·2, ⇒NE·NF·sin(∠(NE,NF))=2·a²√2, ⇒3a²·sin(∠(NE,NF))=2·a²√2, ⇒sin(∠(NE,NF))=(2·a²√2)/(3a²)=(2√2)/3.

d) d(B,MC)=???

ΔMBC isoscel, MG⊥BC. Din ΔMPG, MP=2a=PG, deci MG=2a√2.

Aria(ΔMBC)=(1/2)·BC·MG=(1/2)·2a·2a√2=2a²√2.

Din ΔMGC, MC²=MG²+GC²=(2a√2)²+a²=8a²+a²=9a², deci MC=3a.

Aria(ΔMBC)=(1/2)·MC·d(B,MC)=2a²√2 |·2 ⇒MC·d(B,MC)=4a²√2, ⇒d(B,MC)=(4a²√2):(3a)=(4/3)·a√2.