10N

Simulare Evaluarea Națională Matematică 2023

21 martie 2023
Subiect
Barem

Subiectul I

1.
5

Numărul natural scris în baza zece, de forma
17x‾\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\overline{17x}}
, divizibil cu 10, este egal cu:

A
17\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {17}
B
70\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {70}
C
100\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {100}
D
170\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {170}
2.
5

Numărul care reprezintă
20%\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {20\%}
din
50\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {50}
este egal cu:

A
10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {10}
B
20\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {20}
C
25\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {25}
D
100\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {100}
3.
5

Suma numerelor întregi din intervalul
[−2, 3]\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {[-2,\text{ }3]}
este egală cu:

A
−9\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {-9}
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

Inversul numărului
23\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\frac{2}{3}}
este numărul:

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

Patru elevi, Elena, Maria, George și Mihai, au calculat media geometrică a numerelor
x=3−22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x = 3 - 2\sqrt{2}}
și
y=3+22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {y = 3 + 2\sqrt{2}}
și au obținut următoarele rezultate:
ElenaMariaGeorgeMihai17213\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\begin{array}{|c|c|c|c|} \hline \text{Elena} & \text{Maria} & \text{George} & \text{Mihai} \\ \hline \sqrt{17} & \sqrt{2} & 1 & 3 \\ \hline \end{array}}

Dintre cei patru elevi, cel care a calculat corect media geometrică este:

A
Elena
B
Maria
C
George
D
Mihai
6.
5

Afirmația „Numărul
4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4}
este mai mare decât numărul
25\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {2\sqrt{5}}
.” este:

A
adevărată
B
falsă

Subiectul al II-lea

1.
5

În figura alăturată este reprezentat segmentul
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
cu lungimea de
5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {5}
cm. Punctul
C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C}
este simetricul punctului
B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B}
față de punctul
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
, iar punctul
D\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {D}
este simetricul punctului
C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C}
față de punctul
B\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B}
. Lungimea segmentului
CD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CD}
este egală cu:

A
5\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {5}
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
20\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {20}
cm
2.
5

În figura alăturată, unghiurile
AOC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AOC}
și
BOD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BOD}
sunt opuse la vârf. Măsura unghiului
AOC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AOC}
este egală cu
30∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30^\circ}
, iar semidreapta
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}
. Măsura unghiului
DOE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DOE}
este egală cu:

A
75∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {75^\circ}
B
90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {90^\circ}
C
105∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {105^\circ}
D
150∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {150^\circ}
3.
5

În figura alăturată este reprezentat triunghiul
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
cu
AB=12\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB=12}
cm,
BC=13\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC=13}
cm și
AC=7\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC=7}
cm. Punctele
M, N\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M,\text{ }N}
și
P\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {P}
sunt mijloacele segmentelor
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
,
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
, respectiv
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
. Perimetrul triunghiului
MNP\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {MNP}
este egal cu:

A
8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8}
cm
B
16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {16}
cm
C
18\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {18}
cm
D
32\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {32}
cm
4.
5

În figura alăturată este reprezentat paralelogramul
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
. Dreapta
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
este perpendiculară pe dreapta
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
și dreapta
AD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD}
este perpendiculară pe dreapta
BD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BD}
. 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}
și măsura unghiului
DCM\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DCM}
este egală cu
40∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {40^\circ}
. Măsura unghiului
CMD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CMD}
este egală cu:

A
80∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {80^\circ}
B
90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {90^\circ}
C
100∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {100^\circ}
D
120∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {120^\circ}
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 diametru
BC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC}
. Punctul
A\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {A}
aparține cercului, astfel încât măsura arcului mic
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
este egală cu
120∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {120^\circ}
. Măsura unghiului
ACB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ACB}
este egală cu:

A
30∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30^\circ}
B
60∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {60^\circ}
C
90∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {90^\circ}
D
120∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {120^\circ}
6.
5

În figura alăturată este reprezentat un con circular drept cu secțiunea axială triunghiul dreptunghic
VAB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {VAB}
și raza bazei conului
AO=4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AO=4}
cm. Generatoarea acestui con are lungimea egală cu:

A
4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4}
cm
B
42\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {4\sqrt{2}}
cm
C
8\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8}
cm
D
82\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {8\sqrt{2}}
cm

Subiectul al III-lea

1.
5

Într-un bloc de locuințe sunt 22 de apartamente cu două, respectiv cu patru camere, în total fiind
60\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {60}
de camere.

a.
2
Este posibil ca în acest bloc să fie
16\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {16}
apartamente cu patru camere? Justifică răspunsul dat.
b.
3
Determină numărul de apartamente cu două camere din acest bloc.
2.
5

Se consideră expresia
E(x)=(x2−9x2−16−1):(1x+4+1x−4−3x2−16)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(x) = \left(\frac{ x^2 - 9 }{ x^2 - 16 } - 1 \right) : \left( \frac{1}{ x + 4 } + \frac{1}{ x - 4 } - \frac{3}{ x^2 - 16 } \right)}
, unde
x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x}
este număr real,
x≠−4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \neq -4}
,
x≠4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \neq 4}
și
x≠32\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \neq \frac{3}{2}}
.

a.
2
Arată că
E(x)=72x−3\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(x) = \frac{7}{2x - 3}}
, unde
x\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x}
este număr real,
x≠−4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \neq -4}
,
x≠4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \neq 4}
și
x≠32\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {x \neq \frac{3}{2}}
.
b.
3
Determină numerele naturale
n\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {n}
pentru care
E(n)\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E(n)}
este număr natural.
3.
5

Se consideră numerele
a=(−13)32:(−13)30⋅(−6)2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = \left(-\frac{1}{3}\right)^{32} : \left(-\frac{1}{3}\right)^{30} \cdot (-6)^2}
și
b=(11+2+11+2+3)⋅(0,5)−2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {b = \left(\frac{1}{1+2} + \frac{1}{1+2+3}\right) \cdot (0,5)^{-2}}
.

a.
2
Arată că
a=4\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {a = 4}
.
b.
3
Calculează media aritmetică 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}
.
4.
5

În figura alăturată este reprezentat triunghiul
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
cu
BC=10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BC=10}
cm,
AC=20\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC=20}
cm și măsura unghiului
ACB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ACB}
este egală cu
30∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {30^\circ}
. Punctul
D\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {D}
aparține segmentului
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
, astfel încât unghiul
DBC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DBC}
este congruent cu
BAC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BAC}
.

a.
2
Arată că aria triunghiului
ABC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABC}
este egală cu
50\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {50}
cm
2\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {^2}
.
b.
3
Calculează lungimea segmentului
CD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CD}
.
5.
5

În figura alăturată sunt reprezentate rombul
ABCD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD}
cu măsura unghiului
BAD\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BAD}
egală cu
45∘\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {45^\circ}
și triunghiul dreptunghic isoscel
ABE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABE}
cu
AB=BE=10\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = BE = 10}
cm. Punctele
C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {C}
și
E\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {E}
sunt de o parte și de alta a dreptei
AB\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB}
.

a.
2
Arată că dreapta
DA\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {DA}
este perpendiculară pe dreapta
AE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AE}
.
b.
3
Arată că tangenta unghiului
CAE\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CAE}
este egală cu
2+1\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {\sqrt{2} + 1}
.
6.
5

În figura alăturată este reprezentat cubul
ABCDA′B′C′D′\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {ABCD A'B'C'D'}
cu
AB=6\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AB = 6}
cm. Punctul
M\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {M}
este mijlocul segmentului
B′C′\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B'C'}
și dreptele
BM\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {BM}
și
B′C\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {B'C}
se intersectează în punctul
R\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {R}
. Punctul
P\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {P}
aparține segmentului
AC\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AC}
, astfel încât
AP=22\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AP = 2\sqrt{2}}
cm.

a.
2
Arată că
CP=2⋅AP\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {CP = 2 \cdot AP}
.
b.
3
Determină măsura unghiului dreptelor
PR\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {PR}
și
AD′\def\arraystretch{1.5} \def,{{\char`,}} \def\Div{{\space\raisebox{-0.1em}{$\vdots$}\space}} {AD'}
.
© 2024 ZeceLaEN.ro