10N

Simulare Iași Evaluarea Națională Matematică 2023

18 ianuarie 2023
Subiect
Barem

Subiectul I

1.
5

Numărul 242 este multiplu al numărului:

A
7\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {7}
B
11\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {11}
C
12\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12}
D
17\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {17}
2.
5

Valoarea lui
x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x}
care verifică egalitatea
x30=715\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{x}{30} = \frac{7}{15}}
este egală cu:

A
73\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{7}{3}}
B
8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8}
C
14\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {14}
D
16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {16}
3.
5

Opusul numărului
a=(12−13):136\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = \left( \frac{1}{2} - \frac{1}{3} \right) : \frac{1}{36}}
este:

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

Cel mai mare dintre numerele raționale
2,(4)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,(4)}
;
2,4(2)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,4(2)}
;
2,22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,22}
;
2,(42)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,(42)}
este:

A
2,4(2)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,4(2)}
B
2,(4)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,(4)}
C
2,(42)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,(42)}
D
2,22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2,22}
5.
5

Dacă
a=5+3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = \sqrt{5} + \sqrt{3}}
și
b=15\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b = \sqrt{15}}
, atunci
a2−2b+1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a^2 - 2b + 1}
este egal cu:

A
15\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\sqrt{15}}
B
9−215\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {9 - 2\sqrt{15}}
C
8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8}
D
9\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {9}
6.
5

În tabelul următor sunt înregistrate temperaturile medii zilnice dintr-o săptămână.
ZiuaLMMJVSDTemperatura−4∘C−3∘C−2∘C1∘Cx∘C1∘C2∘C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\begin{array}{|c|c|c|c|c|c|c|c|} \hline \bold{Ziua} & \text{L} & \text{M} & \text{M} & \text{J} & \text{V} & \text{S} & \text{D} \\ \hline \bold{Temperatura} & -4^\circ\text{C} & -3^\circ\text{C} & -2^\circ\text{C} & 1^\circ\text{C} & x^\circ\text{C} & 1^\circ\text{C} & 2^\circ\text{C} \\ \hline \end{array}}

Dacă temperatura medie din aceea săptămână a fost
−1∘C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {-1^\circ\text{C}}
, atunci
x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x}
este egal cu:

A
−2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {-2}
B
0\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {0}
C
1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1}
D
3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {3}

Subiectul al II-lea

1.
5

În figura alăturată, punctul
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
este mijlocul segmentului
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
,
N\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {N}
este mijlocul segmentului
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
, iar
C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C}
este simetricul punctului
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
față de punctul B. Valoarea raportului
ANMN\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{AN}{MN}}
este:

A
23\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{2}{3}}
B
1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {1}
C
43\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{4}{3}}
D
32\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{3}{2}}
2.
5

În figura alăturată, unghiurile
ACD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ACD}
și
DCB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DCB}
sunt adiacente suplementare. Măsura unghiului format de bisectoarele unghiurilor
ACD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ACD}
și
DCB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DCB}
este:

A
90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {90^\circ}
B
100∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {100^\circ}
C
120∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {120^\circ}
D
150∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {150^\circ}
3.
5

Figura alăturată reprezintă schița unui teren în formă de trapez
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
cu aria de
144\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {144}
m
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
și lungimea liniei mijlocii de
36\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36}
m. Distanța dintre laturile
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
și
DC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DC}
este egală cu:

A
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2}
m
B
4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4}
m
C
8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8}
m
D
12\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {12}
m
4.
5

Dreptunghiul
EFGH\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {EFGH}
din figura alăturată are lungimea
EF=15\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {EF = 15}
cm și lățimea
FG=53\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {FG = 5\sqrt{3}}
cm. Dacă
HF∩EG={O}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {HF \cap EG = \{O\}}
, măsura unghiului
GOF\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {GOF}
este:

A
60∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {60^\circ}
B
90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {90^\circ}
C
120∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {120^\circ}
D
150∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {150^\circ}
5.
5

În figura alăturată
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
este un triunghi dreptunghic în
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
cu măsura unghiului
B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B}
de
30∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30^\circ}
. Dacă lungimea bisectoarei
CM\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CM}
este egală cu
10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {10}
cm, atunci lungimea catetei
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
este egală cu:

A
8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8}
cm
B
10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {10}
cm
C
15\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {15}
cm
D
16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {16}
cm
6.
5

În figura alăturată
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
este un tetraedru regulat cu aria feței
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
egală cu
93\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {9\sqrt{3}}
dm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
. Suma lungimilor tuturor muchiilor tetraedrului este egală cu:

A
18\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18}
dm
B
36\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {36}
dm
C
72\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {72}
dm
D
81\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {81}
dm

Subiectul al III-lea

1.
5

Radu și Tudor au împreună 800 de lei. Dacă Radu i-ar da lui Tudor o cincime din suma pe care o are, atunci cei doi copii ar avea sume egale.

a.
2
Este posibil ca Radu să aibă 600 de lei? Justifică răspunsul dat.
b.
3
Află ce sumă are Tudor.
2.
5

Se consideră numerele
a=(1820−645+3280)⋅(35)−1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = \left( \frac{18}{\sqrt{20}} - \frac{6}{\sqrt{45}} + \frac{32}{\sqrt{80}} \right) \cdot \left( \frac{3}{\sqrt{5}} \right)^{-1}}
și
b=53⋅253:1252\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b = 5^{3} \cdot 25^{3} : 125^{2}}
.

a.
2
Arată că numărul
a=5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = 5}
.
b.
3
Demonstrează 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 pătrat perfect.
3.
5

Se consideră expresia
E(x)=(3x−2)2+(1−x5)(x5+1)+4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(x) = (3x - 2)^2 + (1 - x\sqrt{5})(x\sqrt{5} + 1) + 4}
, unde
x∈R\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \in \mathbb{R}}
.

a.
2
Arată că
E(x)=4x2−12x+9\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(x) = 4x^2 - 12x + 9}
, pentru orice număr real
x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x}
.
b.
3
Determină valorile întregi ale lui
n\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {n}
pentru care
E(n)<36\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(n) < 36}
.
4.
5

Rombul
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
din figura alăturată are înălțimea
DN=8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DN = 8}
cm. Se știe că
AC∩BD={O}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC \cap BD = \{O\}}
și proiecția
AM\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AM}
a segmentului
OA\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OA}
pe dreapta
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
are lungimea de
8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8}
cm.

a.
2
Arată că
OM=4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {OM = 4}
cm.
b.
3
Dacă
AC∩DN={P}\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC \cap DN = \{P\}}
, calculează aria patrulaterului
PNMO\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {PNMO}
.
5.
5

În figura alăturată sunt reprezentate triunghiurile
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
și
BCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BCD}
dreptunghice în
B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B}
, respectiv în
C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C}
. Se știe că
∠BAC=15∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\angle BAC = 15^\circ}
,
BC=CD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC = CD}
,
BM⊥AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BM \perp AC}
și
MN⊥BD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MN \perp BD}
.

a.
2
Arată că măsura unghiului
BMN\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BMN}
este de
30∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30^\circ}
.
b.
3
Dacă
BN=2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BN = 2}
cm, află distanța de la
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
la dreapta
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
.
6.
5

În figura alăturată este desenată o prismă triunghiulară regulată
ABCA′B′C′\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC A'B'C'}
cu
AB=8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 8}
m și
AA′=6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AA' = 6}
m. Punctele
N\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {N}
,
P\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {P}
și
Q\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {Q}
sunt mijloacele segmentelor
BB′\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BB'}
,
B′C′\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B'C'}
, respectiv
A′C′\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A'C'}
.

a.
2
Arată că planele
(QPN)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {(QPN)}
și
(C′AB)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {(C'AB)}
sunt paralele.
b.
3
Calculează cosinusul unghiului determinat de dreptele
QP\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {QP}
și
C′B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C'B}
.
© 2024 ZeceLaEN.ro