10N

Simulare Maramureș Evaluarea Națională Matematică 2023

16 noiembrie 2022
Subiect
Barem

Subiectul I

1.
5

Rezultatul calculului
12+13+16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{1}{2} + \frac{1}{3} + \frac{1}{6}}
este:

A
311\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{3}{11}}
B
1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1}
C
56\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{5}{6}}
D
76\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{7}{6}}
2.
5

Cel mai mare număr întreg din intervalul
[−3, 4)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {[-3,\text{ }4 )}
este:

A
4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4}
B
−3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {-3}
C
5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {5}
D
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}
3.
5

Numărul numerelor divizibile cu
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}
din mulțimea
A={1,2,3,4,5,6,7,8,9,10,11,12}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A = \{1,2,3,4,5,6,7,8,9,10,11,12\}}
este:

A
1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1}
B
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}
C
5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {5}
D
4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4}
4.
5

Media aritmetică a numerelor
4+22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4 + 2\sqrt{2}}
și
2(1−2)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2(1 - \sqrt{2})}
este:

A
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
B
3+22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3 + 2\sqrt{2}}
C
2−2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2 - \sqrt{2}}
D
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}
5.
5

Temperatura maximă măsurată este prezentată în tabelul următor:
ZiuaLuniMarțiMiercuriJoiVineriTemperatura5∘−2∘4∘−3∘2∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\begin{array}{|c|c|c|c|c|c|} \hline \text{Ziua} & \text{Luni} & \text{Marți} & \text{Miercuri} & \text{Joi} & \text{Vineri} \\ \hline \text{Temperatura} & 5^\circ & -2^\circ & 4^\circ & -3^\circ & 2^\circ \\ \hline \end{array}}

Cea mai mare diferență de temperatură este între zilele:

A
Luni și Marți
B
Miercuri și Joi
C
Luni și Joi
D
Marți și Joi
6.
5

Dacă numerele reale
a\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a}
și
b\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b}
sunt direct proporționale cu
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
și
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}
, iar
a+b=20\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a + b = 20}
, atunci egalitatea
a+2b=30\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a + 2b = 30}
este:

A
Adevărată
B
Falsă

Subiectul al II-lea

1.
5

În figura alăturată punctele
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
,
B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B}
și
C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C}
sunt coliniare, iar
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
și
N\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {N}
sunt mijloacele segmentelor
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
și
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
. Dacă
AB=4cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 4cm}
și
BC=2cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC = 2cm}
, atunci lungimea segmentului
MN\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MN}
este:

A
1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1}
cm
B
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
cm
C
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}
cm
D
1,5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1,5}
cm
2.
5

Unghiurile
∠AOB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle AOB}
și
∠COD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle COD}
din figura alăturată sunt opuse la vârf. Semidreapta
OE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OE}
este bisectoarea unghiului
∠AOC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle AOC}
, iar semidreapta
OF\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OF}
este semidreapta opusă semidreptei
OE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OE}
. Dacă
∠COD=40∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle COD = 40^\circ}
, atunci măsura unghiului
∠AOF\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle AOF}
este egală cu:

A
80∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {80^\circ}
B
70∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {70^\circ}
C
250∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {250^\circ}
D
110∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {110^\circ}
3.
5

Fie
△ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\triangle ABC}
în care
∠A=60∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle A = 60^\circ}
și
∠B=∠B+∠C2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle B = \frac{\angle B + \angle C}{2}}
. Dacă
BC=2cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC = 2cm}
, atunci perimetrul
△ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\triangle ABC}
este:

A
7\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {7}
cm
B
5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {5}
cm
C
6+23\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6 + 2\sqrt{3}}
cm
D
6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6}
cm
4.
5

Se consideră
△ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\triangle ABC}
(
∠A=90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle A = 90^\circ}
) și
BC=6cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC = 6cm}
. Dacă
G\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {G}
este centrul de greutate al
△ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\triangle ABC}
, atunci lungimea segmentului
AG\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AG}
este:

A
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
cm
B
4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4}
cm
C
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}
cm
D
6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6}
cm
5.
5

Fie
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
diametrul în cercul de centru
O\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {O}
și rază
10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {10}
cm și
CD⊥AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CD \perp AB}
. Dacă
AC=103\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC = 10\sqrt{3}}
cm, atunci aria triunghiului
 ADC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\ ADC}
este:

A
24\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {24}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
B
75\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {75}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
C
753\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {75\sqrt{3}}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
D
503\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {50\sqrt{3}}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
6.
5

Suma muchiilor unui tetraedru regulat este
36cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36cm}
. Suma ariilor tuturor fețelor tetraedrului este:

A
363\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36\sqrt{3}}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
B
30\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
C
122\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12\sqrt{2}}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
D
18\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}

Subiectul al III-lea

1.
5

Un excursionist parcurge un traseu în trei zile. În prima zi parcurge
14\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{1}{4}}
din lungimea traseului, în a doua zi
23\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{2}{3}}
din rest, iar în a treia zi, ultimii
24\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {24}
km.

a.
2
Este posibil ca lungimea traseului să fie de
100km\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {100km}
? Justificați răspunsul.
b.
3
Câți kilometri a parcurs excursionistul în a doua zi?
2.
5

Se consideră numerele
a=93−321+13+145⋅207\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = \frac{9}{\sqrt{3}} - \frac{3}{2} \sqrt{1 + \frac{1}{3}} + \sqrt{\frac{14}{5} \cdot\frac{20} {7}}}
și
b=12−8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b = \sqrt{12} - \sqrt{8}}
.

a.
2
Să se arate că
a=23+22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = 2\sqrt{3} + 2\sqrt{2}}
.
b.
3
Să se arate că media geometrică a numerelor
a\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a}
și
b\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b}
este un număr natural.
3.
5

Se consideră mulțimile
A={x∈Z∣∣2x−1∣≤3}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A = \{x \in \mathbb{Z} \mid |2x - 1| \leq 3\}}
și
B={x∈R∣x2=1}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B = \{x \in \mathbb{R} \mid x^2 = 1\}}
.

a.
3
Să se determine mulțimea
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
.
b.
2
Aflați cardinalul mulțimii
A∩B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A \cap B}
.
4.
5

În exteriorul triunghiului isoscel
 ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\ ABC}
(
AB=AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = AC}
) se construiesc triunghiurile echilaterale
 ABD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\ ABD}
și
 ACE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\ ACE}
. Notăm
BE∩CD={O}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BE \cap CD = \{O\}}
.

a.
2
Să se arate că
BE≡CD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BE \equiv CD}
.
b.
3
Demonstrați că
OA⊥DE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OA \perp DE}
.
5.
5

Se consideră paralelogramul
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
având aria
32cm2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {32cm^2}
. Fie
E\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E}
mijlocul segmentului
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
și
F\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {F}
∈\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\in}
CE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CE}
astfel încât
CF=2FE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CF = 2FE}
.

a.
2
Aflați aria triunghiului
CEB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CEB}
.
b.
3
Arătați că punctele
B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B}
,
F\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {F}
,
D\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {D}
sunt coliniare.
6.
5

În piramida patrulateră regulată
VABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {VABCD}
avem
VA=12cm\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {VA = 12cm}
și
∠VAB=70∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle VAB = 70^\circ}
. Pe muchia
VB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {VB}
se consideră punctul
E\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E}
, iar pe muchia
VC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {VC}
se consideră punctul
F\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {F}
.

a.
2
Calculați măsura unghiului
∠AVB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle AVB}
.
b.
3
Să se determine cea mai mică valoare a sumei
AE+EF+FD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AE + EF + FD}
.
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