10N

Simulare Sibiu Evaluarea Națională Matematică 2024

15 decembrie 2023
Subiect
Barem

Subiectul I

1.
5

Se dă mulțimea
A={(−3)2;−14;(−2)3;(12)−1;−6−2;7548;5;0,09}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A = \left\{ (-3)^2; -\sqrt{\frac{1}{4}}; (-2)^3; \left(\frac{1}{2}\right)^{-1}; \frac{-6}{-2}; \sqrt{\frac{75}{48}}; \sqrt{5}; \sqrt{0,09} \right\}}
.
Mulțimea
A∩N\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A \cap \mathbb{N}}
este egală cu:

A
{(−3)2;(−2)3;−6−2}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\tiny \left\{ (-3)^2; (-2)^3; \frac{-6}{-2} \right\}}
B
{(−3)2;(−2)3;(12)−1;−6−2}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\tiny \left\{ (-3)^2; (-2)^3; \left(\frac{1}{2}\right)^ {-1}; \frac{-6}{-2} \right\}}
C
{(−3)2;(12)−1;−6−2}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\tiny \left\{ (-3)^2; \left(\frac{1}{2}\right)^{-1}; \frac{-6}{-2} \right\}}
D
{(−3)2;−6−2}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\tiny \left\{ (-3)^2; \frac{-6}{-2} \right\}}
2.
5

Prețul unui obiect este 1200 lei. Dacă prețul obiectului se mărește cu 20%, noul preț este egal cu:

A
240 lei
B
960 lei
C
1224 lei
D
1440 lei
3.
5

Dacă
a9=4b\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{a}{9} = \frac{4}{b}}
, atunci 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 egală cu:

A
13\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\sqrt{13}}
B
6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6}
C
6,5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6,5}
D
36\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36}
4.
5

Rezultatul calculului
2(3x+1)−3(−2x−1)−5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2(3x+1)-3(-2x-1)-5}
este egal cu:

A
12x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12x}
B
12x−6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12x-6}
C
0\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {0}
D
−6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {-6}
5.
5

Probabilitatea ca aruncând un zar să se obțină un număr par este egală cu:

A
16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{1}{6}}
B
13\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{1}{3}}
C
12\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{1}{2}}
D
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
6.
5

Suma numerelor prime de o cifră este egală cu:

A
17\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {17}
B
18\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18}
C
26\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {26}
D
27\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {27}

Subiectul al II-lea

1.
5

În figura alăturată sunt reprezentate punctele
A, B, C, D\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A,\text{ }B,\text{ }C,\text{ }D}
și
E\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E}
astfel încât
B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B}
este mijlocul segmentului
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
și
D\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {D}
este mijlocul segmentului
CE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CE}
. Dacă
BD=6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BD = 6}
cm, atunci lungimea segmentului
AE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AE}
este egală cu:

A
9\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {9}
cm
B
10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {10}
cm
C
12\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12}
cm
D
18\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18}
cm
2.
5

În figura alăturată, unghiurile
∠AOB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle AOB}
și
∠BOC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle BOC}
sunt adiacente și complementare, iar
OE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OE}
este bisectoarea unghiului
BOC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BOC}
. Știind că măsura unghiului
AOB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AOB}
este egală cu
40∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {40^\circ}
, atunci măsura unghiului
AOE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AOE}
este egală cu:

A
25∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {25^\circ}
B
50∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {50^\circ}
C
60∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {60^\circ}
D
65∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {65^\circ}
3.
5

În figura alăturată este reprezentat un trapez
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
, cu
AB∥CD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB \parallel CD}
, iar punctele
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
AD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD}
și
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
. Se știe că
CD=4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CD = 4}
cm și
MN=8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MN = 8}
cm. Lungimea segmentului
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
este egală cu:

A
6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6}
cm
B
12\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12}
cm
C
16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {16}
cm
D
24\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {24}
cm
4.
5

În figura alăturată este reprezentat triunghiul
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
isoscel, de bază
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
, în care s-au construit înălțimea
AD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD}
și mediana
BE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BE}
. Dacă
AB=10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 10}
cm,
BC=16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC = 16}
cm, iar
AD∩BE={M}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD \cap BE = \{M\}}
, atunci lungimea segmentului
MD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MD}
este egală cu:

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

În figura alăturată este reprezentat cercul de centru
O\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {O}
și rază
6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {6}
cm. Fie punctul
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
exterior cercului, iar
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}
tangente cercului în punctele
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}
. Știind că
AO=10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AO = 10}
cm, atunci perimetrul patrulaterului
ABOC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABOC}
este egal cu:

A
22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {22}
cm
B
26\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {26}
cm
C
28\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {28}
cm
D
48\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {48}
cm
6.
5

În figura alăturată este reprezentat cubul
ALGEBRIC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ALGEBRIC}
. Un exemplu de două muchii necoplanare este:

A
LG\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {LG}
și
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
B
CE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CE}
și
LR\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {LR}
C
BR\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BR}
și
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
D
LR\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {LR}
și
EG\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {EG}

Subiectul al III-lea

1.
5

Suma a două numere naturale este egală cu 150, iar raportul dintre primul număr micșorat cu 8 și al doilea număr mărit cu 12 are valoarea 1.

a.
2
Este posibil ca cele două numere să fie egale? Justifică răspunsul dat.
b.
3
Determină cele două numere.
2.
5

Fie mulțimile
A={x∈R∣∣2x−1∣≤7}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A = \left\{ x \in \mathbb{R} | |2x - 1| \leq 7 \right\}}
și
B={x∈R∣1≤3x+25<4}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B = \left\{ x \in \mathbb{R} | 1\leq \frac {3x + 2}{5}< 4 \right\}}
.

a.
2
Arată că mulțimea
A=[−3;4]\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A = [-3;4]}
.
b.
3
Calculează mulțimea
C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C}
, unde
C=A∩B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C = A \cap B}
.
3.
5

Fie numerele
a=(1850−1018)⋅30\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = (\frac{18}{\sqrt{50}} - \frac{10}{\sqrt{18}})\cdot 30}
și
b=202−162⋅143\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b = \sqrt{20^2 - 16^2} \cdot \frac{1}{4\sqrt{3}}}
.

a.
2
Arată că
a=42\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = 4\sqrt{2}}
.
b.
3
Arată că
a2−b3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a\sqrt{2} - b\sqrt{3}}
este număr prim.
4.
5

În figura alăturată este reprezentat dreptunghiul
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
, cu
AD=15\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD = 15}
cm și
AC=25\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC = 25}
cm și
DM\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DM}
perpendiculară pe
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
.

a.
2
Arată că perimetrul dreptunghiului
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
este egal cu 70 cm.
b.
3
Arată că tangenta unghiului
MDC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MDC}
este egală cu
43\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{4}{3}}
.
5.
5

În figura alăturată este reprezentat trapezul isoscel
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
, cu
AB∥CD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB \parallel CD}
,
AB>CD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB > CD}
,
AB=24\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 24}
cm,
CD=8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CD = 8}
cm și înălțimea
CE=82\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CE = 8\sqrt{2}}
cm. Fie
AD∩CB={M}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD \cap CB = \{M\}}
.

a.
2
Arată că aria trapezului
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
este
1282\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {128\sqrt{2}}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
.
b.
3
Calculează perimetrul triunghiului
AMB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AMB}
.
6.
5

În figura alăturată este reprezentat tetraedrul regulat
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
cu
AB=63\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 6\sqrt{3}}
cm. Se notează cu
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
mijlocul segmentului
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
.

a.
2
Calculează suma tuturor muchiilor tetraedrului
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
.
b.
3
Calculează aria triunghiului
AMD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AMD}
.
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