Unitatea 1 - SĂ NE AMINTIM!

Recapitulare

1. a) $A=\{-5,\,-4,\,-3,\,-2,\,-1,\,0,\,1\}$; b) $B=\{-3\}$; c) $C=\{-3,\,-2,\,-1,\,0,\,1,\,2,\,3\}$; d) $D=\{-3,\,-2,\,2,\,3\}$.

2. 120, 180, 240, 300, 360.

3. $(a,\,b,\,c)=(24,\,16,\,12)$ sau $(a,\,b,\,c)=(-24,\,-16,\,-12)$.

4. a) -4; b) 0; c) -6; d) -12; e) 5; f) -2; g) -4; h) 2; i) 26; j) -1.

5. 450 minute.

6. $\cfrac{29}{13}$.

7. 5600 kg.

8. $ \sphericalangle EOC=\cfrac{1}{2}\sphericalangle AOC$ ($OE$ bisectoarea, $\sphericalangle AOC$) $=\cfrac{1}{2}\sphericalangle BOC $ ($OC$ bisectoarea $\sphericalangle AOB$) $=\sphericalangle FOC$ ($OF$ bisectoarea $\sphericalangle BOC$) $\Rightarrow OC$ bisectoarea $\sphericalangle EOF$.

9. $33^{\circ}$.

10. a) corespondente; b) opuse la vârf; c) alterne externe; d) alterne interne; e) suplementare; f) suplementare.

11. $f\parallel g$ (secanta $d$) $\Rightarrow$ unghiurile corespondente formate sunt congruente $\Rightarrow d\perp g$.

$d\parallel e$ (secanta $f$) $\Rightarrow$ unghiurile corespondente formate sunt congruente $\Rightarrow e\perp f$.

$d\parallel e$ (secanta $g$) $\Rightarrow$ unghiurile corespondente formate sunt congruente $\Rightarrow e\perp g$.

12. a) conciclice, 0 puncte comune; b) interioare, 0 puncte comune; c) tangente interioare, 1 punct comun; d) secante, 2 puncte comune; e) secante, 2 puncte comune; f) secante, 2 puncte comune; g) tangente exterior, 1 punct comun; h) cercuri exterioare, 0 puncte comune.

13. $30^{\circ}$, $60^{\circ}$, $90^{\circ}$.

14. $\sphericalangle ABC=60^{\circ}$ ,$\sphericalangle BAC=40^{\circ}$ ,$\sphericalangle ACB=80^{\circ}$, $\sphericalangle BEA=125^{\circ}$, $DNM=120^{\circ}$.

15. Fie $M'$ mijlocul laturii $LN$, $Y'$ mijlocul laturii $XZ$. $\triangle LM{{G}_{1}}\equiv \triangle XY{{G}_{2}}$ $\Rightarrow \left[ M{{G}_{1}} \right]\equiv \left[ Y{{G}_{2}} \right]\Rightarrow \left[ MM' \right]\equiv \left[ YY' \right]\left( 1 \right)$ centrele de greutate împart medianele $MM'$ și $YY'$ în raportul $\cfrac{2}{3}$, $\cfrac{1}{3}$.

$\triangle LM{{G}_{1}}\equiv \triangle XY{{G}_{2}} \Rightarrow \left\{ \begin{array}{l} \sphericalangle LMM'\equiv \sphericalangle XYY' \\ \left[ LM \right]\equiv \left[ XY \right] \\ \end{array} \right\}\Rightarrow \triangle LMM'\equiv \triangle XYY$ $\Rightarrow \left\{ \begin{array}{l} \left[ LM' \right]\equiv \left[ XY' \right]\Rightarrow \left[ LN \right]\equiv \left[ XZ \right]\ \left( M'Y'\text{ mijloace }\right) \\ \sphericalangle MLN\equiv \sphericalangle YXZ \\ \left[ LM \right]\equiv \left[ XY \right] \\ \end{array} \right\}\xRightarrow[]{LUL}\triangle LMN\equiv \triangle XYZ$

16. $\sphericalangle ABC=\sphericalangle ACB=30^{\circ}$, $\sphericalangle BAC=120^{\circ}$.

Unitatea 2 - Mulțimea numerelor reale

Rădăcina pătrată a pătratului unui număr natural

6. a) F; b) A; c) F; d) A.

7. 9, 16, 100, 121, 400.

9. a) $5^2\cdot 3^4$; b) $7^3\cdot 2$; c) $5^2\cdot 17^5$; d) $2020\cdot2^{1010}$; e) $22\cdot 3^2 \cdot 2^3$; f) $5^3\cdot 7^2 \cdot 10^4$.

10. a) 45; b) 16; c) 32; d) 81; e) 625; f) 128; g) 27; h) 648.

11. a) 4; b) 10; c) 5; d) 10; e) 6; f) 2; g) 5; h) 3; i) 8.

12. a) 3, 4, 5; b) 5, 12, 13.

Estimarea rădăcinii pătrate dintr-un număr rațional

3. 1-e; 2-a; 3-c; 4-b.

4. a) $\cfrac{2}{5}$; b) $\cfrac{1}{11}$; c) $\cfrac{8}{13}$; d) $\cfrac{10}{7}$; e) $\cfrac{12}{19}$; f) $\cfrac{1}{20}$; g) $\cfrac{15}{4}$; h) $\cfrac{7}{30}$; i) $\cfrac{40}{9}$; j) $\cfrac{3}{14}$.

8. a) 1,5; b) 0,4; c) 1,2; d) 0,7; e) 0,3; f) 0,25.

9. a) $\cfrac{25}{7}$; b) $\cfrac{1}{81}$; c) $\cfrac{1000}{32}$; d) $\cfrac{7}{27}$; e) $\cfrac{64}{125}$; f) $\cfrac{81}{243}$.

10. a) A; b) A; c) F; d) A; e) F; f) F.

Numere iraționale, exemple; mulțimea numerelor reale; incluziunea $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$

5. a-4, b-1, c-2, d-3.

Compararea și ordonarea numerelor reale

4. a) 1 este egal cu $\sqrt{1}$; $\sqrt{2}$; $\sqrt{3}$; 2; 3. b) $\sqrt{5}$; 2,3; $\sqrt{6}$; 2,5; $\sqrt{7}$2,7. c) $\sqrt{2,25}$ este egal cu $\cfrac{3}{2}$; 1,55; $\sqrt{2,56}$; $\cfrac{81}{50}$.

Reprezentarea numerelor reale pe axa numerelor prin aproximări. Modulul unui număr real (definiție, proprietăți)

9. a) $\sqrt{7}$; b) $5$ si $-5$; c) $-4$, $-3$, $-2$, $-1$, $0$ , $1$, $2$, $3$, $4$; d) $-9$; e) $-2$, $-1$, $0$, $1$, $2$.

Unitatea 3 - Operații cu numere reale

Adunarea și scăderea numerelor reale

3. a) F; b) F; c) A; d) A; e) F; f) F; g) A; h) F; i) F; j) A; k) F; l) F.

4. a) $12\sqrt{3}$; b) $7\sqrt{5}$; c) $-7\sqrt{2}$; d) $5\sqrt{5}$; e) $-12\sqrt{11}$; f) $5\sqrt{7}$; g) $6\sqrt{5}$; h) $-3\sqrt{7}$.

5. a) $7\sqrt{13}-8\sqrt{5}$; b) $9\sqrt{3}+3\sqrt{5}$; c) $-4\sqrt{2}+3\sqrt{11}$; d) $3\sqrt{7}$; e) $-3\sqrt{7}+2\sqrt{3}$; f) 0.

Înmulțirea și împărțirea numerelor reale

2. a) $\sqrt{15}$; b) 2; c) $\sqrt{3}$; d) $\sqrt{15}$; e) $15\sqrt{14}$; f) $7\sqrt{6}$; g) $2\sqrt{3}$; h) $21\sqrt{2}$; i) $4\sqrt{3}$; j) $32\sqrt{3}$; k) $6\sqrt{6}$; l) 1.

3. a) $2\sqrt{15}$; b) $12\sqrt{15}$; c) $30\sqrt{15}$; d) $72\sqrt{5}$; e) $2\sqrt{14}$; f) $200\sqrt{2}$; g) $135\sqrt{2}$; h) $88$; i) $270\sqrt{5}$; j) $16\sqrt{2}$; k) $286$; l) 306.

4. a) $5\sqrt{2}$; b) $3\sqrt{5}-10\sqrt{3}$; c) $\sqrt{3}$; d) $32$; e) $2\sqrt{5}$; f) $6\sqrt{15}+8\sqrt{5}$; g) $11\sqrt{10}$; h) $-2\sqrt{5}$; i) $3\sqrt{2}$.

6. $P=12\sqrt{5}$ cm.

7. $P=48\sqrt{2}$ cm.

Puteri cu exponent număr întreg. Raționalizarea numerelor de forma $a\sqrt{b}$

2. a) $9\sqrt{3}$; b) $7\sqrt{7}$; c) $6\sqrt{6}$; d) 6; e) 11; f) 15; g) 61.

3. a) ${{(\sqrt{3}})^{8}}$; b) ${{(\sqrt{7}})^{15}}$; c) $\sqrt{6}$; d) $\sqrt{6}$; e) ${{(\sqrt{11}})^{-3}}$; f) ${{(\sqrt{15}})^{-1}}$; g) ${{(\sqrt{61}})^{-3}}$; h) ${{(\sqrt{15}})^{5}}$; i) ${{(\sqrt{14}})^{3}}$; j) ${{(\sqrt{3}})^{3}}$; k) ${{(\sqrt{3}})^{5}}$; l)${{(\sqrt{2}})^{-7}}$; m) ${{(\sqrt{3}})^{15}}$; n) ${{(\sqrt{2}})^{-9}}$; o) ${{(\sqrt{3}})^{27}}$.

4. a) $\cfrac{1}{9\sqrt{3}}$; b)$7\sqrt{7}$; c) $8\sqrt{2}$; d) 45; e)$40\sqrt{5}$; f)$147$; g)$-250\sqrt{2}$; h) $-9\sqrt{3}$; i) $\cfrac{1}{7}$; j) ${{5}^{2}}$; k) $-\cfrac{1}{9\sqrt{3}}$; l) $\cfrac{1}{18}$.

6. a) $2\sqrt{5}$; b) $\cfrac{2\sqrt{5}}{5}$; c) $\cfrac{\sqrt{3}}{3}$; d) $\cfrac{4\sqrt{6}}{3}$; e) $-\cfrac{3\sqrt{5}}{5}$; f) $-2\sqrt{7}$; g) $\cfrac{5\sqrt{3}}{3}$; h)$-3\sqrt{7}$; i) $-\cfrac{5\sqrt{11}}{11}$; j) $\cfrac{3\sqrt{2}}{2}$; k)$-\cfrac{11\sqrt{5}}{5}$; l) $-4\sqrt{3}$.

7. a)$\cfrac{3\sqrt{2}}{2}$; b) $\cfrac{2\sqrt{3}}{3}$; c) $-\sqrt{2}$; d) $\cfrac{2\sqrt{3}}{3}$; e) $\cfrac{19\sqrt{2}}{4}$; f) $-\cfrac{4\sqrt{5}}{5}$; g) $\cfrac{4\sqrt{6}}{9}$; h)$-\cfrac{\sqrt{2}}{4}$; i) $-\cfrac{7\sqrt{2}}{20}$; j) $\cfrac{2\sqrt{3}}{3}$; k)$-\cfrac{9\sqrt{2}}{40}$; l)$\cfrac{4\sqrt{3}}{9}$.

Media aritmetică ponderată a $n$ numere reale, $n \geq 2$. Media geometrică a două numere reale pozitive

3. a) $\cfrac{54}{5}$; b)$\cfrac{86}{23}$; c)$\cfrac{65}{9}$.

4. $\cfrac{321}{26}$.

5. a)20; b)$4\sqrt{5}$; c)$8\sqrt{2}$; d) $9\sqrt{5}$; e)$10\sqrt{3}$; f)$\cfrac{4}{5}$; g)$\cfrac{2}{3}$; h)$\cfrac{\sqrt{2}}{2}$; i) $\cfrac{\sqrt{3}}{2}$; j)$\cfrac{3\sqrt{7}}{2}$.

6. 104.

7. 224.

8. a) 15; b) 20; c) $3\sqrt{14}$ d) 12; e) $3\sqrt{5}$; f) $6\sqrt{10}$.

Ecuații de forma $x^2=a$, unde $a\in \mathbb{R}$

3. a) C; b) A; c) B; d) D.

4. a) $x\in \left\{ -6,\,6 \right\}$; b) $x\in \varnothing $; c) $x\in \left\{ 0 \right\}$; d) $x\in \left\{ -2\sqrt{2},\,2\sqrt{2} \right\}$; e) $x\in \left\{ -9\sqrt{2},\,9\sqrt{2} \right\}$; f) $\left\{ -\sqrt{95},\,\sqrt{95} \right\}$; g) $x\in \left\{ -3\sqrt{2},\,3\sqrt{2} \right\}$; h) $x\in \varnothing $; i) $x\in \left\{ -\sqrt{215},\,\sqrt{215} \right\}$; j) $x\in \left\{ -5\sqrt{2},\,5\sqrt{2} \right\}$; k) $x\in \left\{ -7\sqrt{7},\,7\sqrt{7} \right\}$; l) $x\in \left\{ -3\sqrt{6},\,3\sqrt{6} \right\}$.

5. a) $\sqrt{35}$ cm; b) $5\sqrt{2}$ cm; c) $3\sqrt{7}$ cm; d) 10 cm.

6. Dacă latura se măreşte de două ori, atunci aria se măreşte de patru ori. Dacă latura se măreşte de trei ori,atunci aria se măreşte de nouă ori. Observăm că dacă latura se măreşte de $k$ ori ($k>0$), aria se măreşte de $k^2$ ori.

Recapitulare

1. a) 6; b) 5; c) 81; d) $6\sqrt{2}$; e) 9; f) $9\sqrt{3}$; g) $168\sqrt{14}$; h) 49; i) $70\sqrt{2}$; j) $45\sqrt{5}$; k) $30\sqrt{7}$; l) $15\sqrt{14}$.

2. a) $4\sqrt{5}$; b) $2\sqrt{5}$; c) 15; d) 3; e) $-5\sqrt{5}$; f)30; g) $2\sqrt{5}$; h) $3\sqrt{3}$.

3. a) 9; b) 2; c) $-2\sqrt{2}$; d) 175; e) $-24\sqrt{3}$; f) 1; g) 32; h) 324; i) 75; j) 1331; k) $-250\sqrt{2}$; l) $-49\sqrt{7}$.

4. a) $\cfrac{3\sqrt{5}}{5}$; b) $\cfrac{7\sqrt{3}}{3}$; c) $-\cfrac{5\sqrt{2}}{6}$; d) $-\cfrac{\sqrt{11}}{3}$; e) $-\cfrac{3\sqrt{5}}{2}$; f) $-15\sqrt{6}$.

5. a) $11\sqrt{2}$; b) $15\sqrt{5}$; c) $-\sqrt{7}$.

7. a) $4\sqrt{3}$; b) $5\sqrt{6}$; c)3; d) $6\sqrt{5}$; e) $5\sqrt{7}$.

8. a) $x\in \left\{ -5,\,5 \right\}$; b) $x\in \varnothing $; c) $x\in \left\{ 0 \right\}$; d) $x\in \left\{ -\sqrt{10},\,\sqrt{10} \right\}$; e) $x\in \left\{ -3\sqrt{5},\,3\sqrt{5} \right\}$.

Exersezi și progresezi

1. a) $10\sqrt{3}$; b) 0; c) $-2\sqrt{7}$; d) $16\sqrt{6}$; e) -1; f) $10\sqrt{2}$; g) $\cfrac{7\sqrt{3}}{3}$; h) $\cfrac{49\sqrt{5}}{30}$; i) $\cfrac{103\sqrt{7}}{420}$.

3. a) $P=16\sqrt{5}+8\sqrt{3}$m; b) $16\sqrt{5}+8\sqrt{3}<16\cdot 3+8\cdot 2=48+16=64$.

4.

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5. Media ponderată a celor două temperaturi este:

$\cfrac{180\cdot 5+270\cdot 4}{5+4}=\cfrac{900+1080}{9}=\cfrac{1980}{9}={220}^{\circ}C$.

6. Media ponderată a celor două preţuri este: $\cfrac{80\cdot 2+120\cdot 5}{80+120}=\cfrac{160+600}{200}=\cfrac{760}{200}=3,8$ lei.

Unitatea 4 - Ecuații și sisteme de ecuații liniare

Transformarea unei egalități într-o egalitate echivalentă; identităţi

3. a) $\triangle$; b) $\square$; c) $\square$; d) $\square$; e) $\square$; f) $\triangle$; g) $\square$ sau $\triangle$; h) $\square$.

4. a) $x+y+9=3x-3$; b) $3{{x}^{2}}+5{{y}^{2}}=7{{x}^{2}}+{{y}^{2}}$; c) $xy-2{{x}^{2}}=0$;

d) $5{{a}^{2}}+3ab-8{{b}^{2}}=-8$; e) $2ab+5b=a+b$; f) $-7ab+5{{a}^{2}}=13$.

5. a) $\cfrac{x}{3}=\cfrac{y}{7}\Rightarrow 7x=3y$. Ridicând la pătrat membrii egalităţii, rezultă $49{{x}^{2}}=9{{y}^{2}}$. b) Nu. Din egalitatea dată, extragând radical, se obţine $5\left| x \right|=4\left| y \right|$ care nu implică $5x=4y$. Contraexemplu: $x=-4$, $y=5$.

6. La penultimul pas, Petra aîmpărţit prin $x-2$ care este $0$.

Ecuaţii de forma $ax + b = 0$, unde $a$, $b\in\mathbb{R}$; mulţimea soluţiilor unei ecuaţii; ecuaţii echivalente

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3. a) A; b) F; c) A; d) A; e) F; f) A; g) F; h) F.

4. a) $x=-5$; b) $x=\cfrac{9}{2}$; c) $x=2$; d) $x=\cfrac{17}{3}$; e) $x=-\cfrac{110}{3}$; f) $x=\cfrac{4}{5}$; g) $x=-2$; h) $x=36$;

i) $x=-\cfrac{11}{2}$; j) $x=-2$; k) $x=-\cfrac{175}{18}$; l) $x=\cfrac{185}{18}$.

5. a) $x=\sqrt{\cfrac{5}{2}}-\sqrt{\cfrac{2}{2}}$; b) $x=2$; c) $x=\cfrac{\sqrt{42}}{3}$; d) $x=\cfrac{7\sqrt{3}-1}{2}$; e) $x=\cfrac{16}{9}$; g) $x=3+\cfrac{2\sqrt{3}}{3}$; h) $x=\cfrac{3}{2}$.

7. Vom nota în fiecare caz cu $S$ mulţimea soluţiilor:
i) $\left\{ \begin{array}{l} a=-2\Rightarrow S=\varnothing \\ a\ne -2\Rightarrow S=\left\{ \cfrac{a\sqrt{3}-2}{a+2} \right\} \\ \end{array} \right.$
ii) $\left\{ \begin{array}{l} a=\cfrac{\sqrt{11}}{3}\Rightarrow S=\mathbb{R} \rule[-4mm]{0mm}{2mm}\\ a\ne \cfrac{\sqrt{11}}{3}\Rightarrow S=\left\{ 1 \right\} \\ \end{array} \right.$
iii) $\left\{ \begin{array}{l} a=\cfrac{5\sqrt{3}}{2}\Rightarrow S=\varnothing \rule[-4mm]{0mm}{2mm}\\ a\ne \cfrac{5\sqrt{3}}{2}\Rightarrow S=\left\{ \cfrac{5a+2\sqrt{3}}{2a-5\sqrt{3}} \right\} \\ \end{array} \right.\rule[-12mm]{0mm}{2mm}$
iv) $\left\{ \begin{array}{l} a\in \left\{ \sqrt{2},\,2\sqrt{3} \right\}\Rightarrow S=\varnothing \\ a\notin \left\{ \sqrt{2},\,2\sqrt{3} \right\}\Rightarrow S=\left\{ -\cfrac{\sqrt{6}}{{{a}^{2}}-1} \right\} \\ \end{array} \right.$
v) $\left\{ \begin{array}{l} a\in \left\{ -1,\,1 \right\}\Rightarrow S=\varnothing \\ a\notin \left\{ -1,\,1 \right\}\Rightarrow S=\left\{ -\cfrac{{{a}^{2}}+4}{{{a}^{2}}-1} \right\} \\ \end{array} \right.$
vi) $S=\left\{ \cfrac{{{a}^{2}}-1}{{{a}^{2}}+4} \right\}$

Sisteme de două ecuaţii liniare cu două necunoscute

2. a) A; b) A; c) F; d) A.

3. a) A; b) F; c) F; d) A.

4. a) $\left\{ \begin{array}{l} 6x-4y=18 \\ 3x+3y=24 \\ \end{array} \right.$ b) $\left\{ \begin{array}{l} 8x+2y=22 \\ 3x+12y=52 \\ \end{array} \right.$ c) $\left\{ \begin{array}{l} -20x+30y=20 \\ 35x-7y=7 \\ \end{array} \right.$ d) $\left\{ \begin{array}{l} 12x+2y=14 \\ 12x-15y=-3 \\ \end{array} \right.$

5. a) $y=0$; b) $y=\sqrt{2}$; c) $x=\sqrt{3}$; d) $x=0$.

6. a) $\left( x,\,y \right)=\left( 7,\,0 \right)$; b) $\left( x,\,y \right)=\left( 7-\sqrt{7},\,\sqrt{5} \right)$.

Rezolvare prin metoda substituţiei

2. a) $y=4-3x$; b) $y=9+2x$; c) $y=5x-14$; d) $y=\cfrac{10-4x}{3}$; e) $y=\cfrac{11+x}{-2}$; f) $y=\cfrac{24-5x}{7}$; g) $y=\cfrac{15-x}{7}$; h) $y=\cfrac{5x-1}{3}$.

3. a) $(x,\,y)=(1,\,3)$; b) $(x,\,y)=(1,\,-2)$; c) $(x,\,y)=(-2,\,-1)$; d) $(x,\,y)=(1,\,2)$; e) $(x,\,y)=(-1,\,5)$; f) $(x,\,y)=(1,\,1)$; g) $(x,\,y)=(1,\,3)$; h) $(x,\,y)=(1,\,-2)$.

4. a) $\left( x,\,y \right)=\left( \cfrac{2}{7},\,\cfrac{1}{3} \right)$; b) $\left( x,\,y \right)=\left( \cfrac{2}{15},\,0 \right)$; c) $\left( x,\,y \right)=\left( \cfrac{31}{27},\,-\cfrac{13}{45} \right)$, d) $\left( x,\,y \right)=\left( \cfrac{2}{7},\,-\cfrac{1}{3} \right)$.

Rezolvare prin metoda reducerii

1. a) $\left\{ \begin{array}{l} 3\cdot 1+2\cdot 1=5 \\ 4\cdot 1-2\cdot 1=2 \\ \end{array} \right.$ b) A doua ecuaţie din al doilea sistem este obţinută prin adunarea ecuaţiilor din primul sistem.

2. a) $\left\{ \begin{array}{l} 2x+6y=20 \\ 10x-6y=-28 \\ \end{array} \right.$ b) $\left\{ \begin{array}{l} 5x-7y=11 \\ -7x+7y=7 \\ \end{array} \right.$ c) $\left\{ \begin{array}{l} x-9y=-7 \\ 3x+9y=15 \\ \end{array} \right.$ d) $\left\{ \begin{array}{l} 5x-22y=7 \\ -3x+22y=-11 \\ \end{array} \right.$

3. a) $\left\{ \begin{array}{l} 12x-2y=8 \\ 12x-5y=-18 \\ \end{array}\right.$ b) $\left\{ \begin{array}{l} 15x-21y=48 \\ 15x+5y=-30 \\ \end{array} \right.$ c) $\left\{ \begin{array}{l} 24x-36y=60 \\ 24x+15y=9 \\ \end{array} \right.$ d) $\left\{ \begin{array}{l} -30x+65y=-5 \\ 30x-18y=-42 \\ \end{array} \right.$

4. a) $(x,\,y)=(1,\,3)$; b) $(x,\,y)=(1,\,-2)$; c) $(x,\,y)=(-2,\,-1)$; d) $(x,\,y)=(1,\,2)$; e) $(x,\,y)=(-1,\,5)$; f) $(x,\,y)=(1,\,1)$; g) $(x,\,y)=(1,\,3)$; h) $(x,\,y)=(1,\,-2)$.

5. a) $\left( x,\,y \right)=\left( \cfrac{1}{7},\,\cfrac{1}{11} \right)$; b) $\left( x,\,y \right)=\left( \cfrac{1}{13},\,\cfrac{2}{5} \right)$; c) $\left( x,\,y \right)=\left( \cfrac{2}{7},\,\cfrac{3}{5} \right)$; d) $\left( x,\,y \right)=\left( \cfrac{1}{7},\,\cfrac{1}{11} \right)$; $\left( x,\,y \right)=\left( -\cfrac{5}{36},\,-\cfrac{9}{28} \right)$.

Probleme care se rezolvă cu ajutorul ecuaţiilor sau a sistemelor de ecuaţii liniare

2. 22 persoane.

3. 108 lei.

4. 20 kg.

5. lungimea $=\cfrac{44}{3}$ lățimea $=\cfrac{22}{3}$.

6. 7 km.

7. 13 caiete de 3 lei, 7 caiete de 5 lei.

8. 65.

9. 75 de monede de cincizeci de bani, 105 de monede de zece bani.

10. Constructorul nu va avea suficient material. Sunt 16 bare albastre de 1 m, deci muncitorul rămâne cu 14 m pentru diagonale. Sunt 10 diagonale de lungime $\sqrt{2}\sim 1\cdot 1,414=14,14>14$ m.

11. a) $AB=CD=EF=FG=HI=\sqrt{5}$, $BC=DE=GH=\sqrt{13}$; b) Pe porţiunile de tip AB: $4\sqrt{5}$ l/100 km, $BC:2\sqrt{13}$ l/100 km.

12. dist $=\sqrt{{{h}^{2}}+{{d}^{2}}}=\sqrt{{{h}^{2}}+{{19}^{2}}{{h}^{2}}}=h\sqrt{362}=8,75\sqrt{362}\sim 166,4$ km. Carburant utilizat $=2000$ l. Carburant/km $=12$ l.

Recapitulare

1. a) A; b) F; c) A; d) A; e) F; f) A; g) A; h) A.

2. a) $x=3$; b) $x=-\cfrac{17}{3}$; c) $x=-4$; d) $x=5$; e) $x=\cfrac{19}{9}$; f) $x=\cfrac{2}{5}$; g) $x=-\cfrac{9}{8}$; h) $x=-12$.

4. a) $(x,\,y)=(2,\,-3)$; $(x,\,y)=(5,\,-10)$; b) $(x,\,y)=\left(2\sqrt{\cfrac{7}{5}},\,-3\sqrt{3}\right)$.

5. a) $(x,\,y)=(-2,\,1)$; b) $(x,\,y)=(-2,\,-1)$; c) $(x,\,y)=(1,\,3)$; d) $(x,\,y)=(-3,\,1)$.

6. a) $\left( x,\,y \right)=\left( \sqrt{5},\,-\sqrt{11} \right)$; b) $\left( x,\,y \right)=\left( \sqrt{7},\,\sqrt{6} \right)$.

7. 306 km.

8. $\cfrac{1040}{17}$.

9. a) 1000 ori; b) i) $100{N}_{0}$, ii) $10000{N}_{0}$.

10. 10 km.

Unitatea 5 - Elemente de organizare a datelor

Produsul cartezian a două mulţimi nevide; sistem de axe ortogonale în plan; reprezentarea într-un sistem de axe

1. a) (Ana, Dan), (Ana, Radu), (Ana, Alin), (Corina, Dan), (Corina, Radu), (Corina, Alin); b) 6 perechi, numărul de perechi este numărul de fete înmulţit cu numărul de băieţi.

2. C2, B6.

3. De a nota cu uşurinţă mutările pieselor pe tablă.

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6. A(2, 3); B(0, 0); C(-3, 1); D(-4, -3); E(-1, -1); F(1, 0); G(2, -2); H(4, -2); I(1, 4); J(-2, 3); K(-5, -1); L(4, 2); M(3, -3); N(-2, -4).

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10. a) 15; b) 60; c) 81; d) 300.

11. Mălina şi Cezara – 3 unităţi pe verticală, 1 pe orizontală $=6$ unităţi verticală, Ave, Delia şi Ilinca $=6$ unităţi pe verticală, Alex şi Teo $=3$ unităţi pe verticală, 1 pe orizontală $=6$ unităţi verticală.

Reprezentarea punctelor într-un sistem de axe ortogonale; distanţa dintre două puncte din plan

1. $OB=5$ unităţi.

4. a) 5; b) 13; c) $\sqrt{34}$; d) $\sqrt{13}$; e) 2; f) $\sqrt{17}$.

5. a) 18; b) 30; c) triunghi degenerat $P=12\sqrt{2}$.

6. $AB=5$, $AC=\sqrt{20}$, $AD=\sqrt{34}$, $BC=\sqrt{53}$, $BD=\sqrt{117}$, $CD=\sqrt{50}$. $AC$ are lungimea cea mai mică, $BD$ are lungimea cea mai mare.

7. a) punctele sunt coliniare; b) punctele sunt coliniare; c) punctele formează un triunghi.

8. a) $ABCD$ pătrat; b) $ABCD$ dreptunghi; c) $ABCD$ paralelogram; d) $ABCD$ romb.

9. $ AB=\sqrt{45}=3\sqrt{5}$ unități, $BC=6$ unități, $AC=\sqrt{117}$ unități, $BD= \sqrt{65}$ unități, $CD=\sqrt{17}$ unități $AD= \sqrt{200}$ unități. $A-B-D$ distanța minimă $= \sqrt{45}+\sqrt{65}\sim 147,70$ kilometri, $AD\sim 141,42$ kilometri. Distana s-ar scurta cu aproximativ $6,2$ km.

10. a) $AU=100\sqrt{2}$ m, $AV=100\sqrt{5}$ m, $AW=200\sqrt{5}$ m, $AX=100\sqrt{10}$ m, $AZ=100\sqrt{13}$ m. b) Cel mai apropiat este $U$, cel mai depărtat este $X$. c) cu cât sunt mai puţine cercuri, între $A$ şi un submarin, cu atât acesta este mai aproape: între $AU$ şi este doar un cerc; cu cât sunt mai multe cercuri, între $A$ şi un submarin, cu atât acesta este mai departe: între $AX$ sunt $5$ cercuri. d) $XY=100\sqrt{5}$ m, cercurile nu mai sunt utile.

11. $Z$.

Reprezentarea şi interpretarea unor dependenţe funcţionale prin tabele, diagrame şi grafice; poligonul frecvenţelor

5. a) Ianuarie, februarie, decembrie; b) 7500; c) 2083 (3); d) În histogramă de pot observa mai uşor informaţiile cerute.

r006

6.

r007 a) a VIII-a B; b) 231; c) 132.

7.

r008

Acelaşi timp, viteza lui C e dublă faţă de viteza lui A, deci distanţa parcursă de C e dublă faţă de distanţa parcursă de A. Aceeaşi viteză, timpul lui D e dublu faţă detimpul lui F, deci distanţa parcursă de D e dublă faţă de distanţa parcursă de F.

8. a) $8100$ euro, 5184 euro; c) preţ $2016=0,81$ preţ initial, preţ $2017=0,64$ preţ $2016$. Media geometrică $=0,72$.

9. a) $\sqrt{330}\sim 18$, $t_f=18$ s.

r010

10. a) $d=12,4968\times 11\times \sqrt{3} \sim 238$ km.

r012

11. d).

Recapitulare

1. a) – 7; b) – 6; c) – 2; d) - 3; e) – 5.

2. $A\times A=\left\{ \left( 3,\,3 \right),\,\left( 3,\,5 \right),\,\left( 3,\,7 \right),\,\left( 5,\,3 \right),\,\left( 5,\,5 \right),\,\left( 5,\,7 \right),\,\left( 7,\,3 \right),\,\left( 7,\,5 \right),\,\left( 7,\,7 \right) \right\}$, card $A\times A=9$
$A\times B=\left\{ \left( 3,\,1 \right),\,\left( 3,\,2 \right),\,\left( 3,\,3 \right),\,\left( 3,\,7 \right),\,\left( 5,\,1 \right),\,\left( 5,\,2 \right),\,\left( 5,\,3 \right),\,\left( 5,\,7 \right),\,\left( 7,\,1 \right),\,\left( 7,\,2 \right),\,\left( 7,\,3 \right),\,\left( 7,\,7 \right) \right\}$, card $A\times B=12$
$A\times C=\left\{ \left( 3,\,1 \right),\,\left( 3,\,7 \right),\,\left( 5,\,1 \right),\,\left( 5,\,7 \right),\,\left( 7,\,1 \right),\,\left( 7,\,7 \right) \right\}$, card $A\times C=6$
$A\times D=\left\{ \left( 3,\,5 \right),\,\left( 3,\,6 \right),\,\left( 3,\,8 \right),\,\left( 5,\,5 \right),\,\left( 5,\,6 \right),\,\left( 5,\,8 \right),\,\left( 7,\,5 \right),\,\left( 7,\,6 \right),\,\left( 7,\,8 \right) \right\}$, card $A\times D=9$
$B\times A=\left\{ \left( 1,\,3 \right),\,\left( 1,\,5 \right),\,\left( 1,\,7 \right),\,\left( 2,\,3 \right),\,\left( 2,\,5 \right),\,\left( 2,\,7 \right),\,\left( 3,\,3 \right),\,\left( 3,\,5 \right),\,\left( 3,\,7 \right),\,\left( 7,\,3 \right),\,\left( 7,\,5 \right),\,\left( 7,\,7 \right) \right\}$, card $B\times A=12$
$B\times B=\left\{ \left( 1,\,1 \right),\,\left( 1,\,2 \right),\,\left( 1,\,3 \right),\,\left( 1,\,7 \right),\,\left( 2,\,1 \right),\,\left( 2,\,2 \right),\,\left( 2,\,3 \right),\,\left( 2,\,7 \right),\,\left( 3,\,1 \right),\,\left( 3,\,2 \right),\,\left( 3,\,3 \right),\,\left( 3,\,7 \right),\,\left( 7,\,1 \right),\right.$ $\left. \left( 7,\,2 \right),\,\left( 7,\,3 \right),\,\left( 7,\,7 \right) \right\}$, card $B\times B=16$
$B\times C=\left\{ \left( 1,\,1 \right),\,\left( 1,\,7 \right),\,\left( 2,\,1 \right),\,\left( 2,\,7 \right),\,\left( 3,\,1 \right),\,\left( 3,\,7 \right),\,\left( 7,\,1 \right),\,\left( 7,\,7 \right) \right\}$, card $B\times C=8$
$B\times D=\left\{ \left( 1,\,5 \right),\,\left( 1,\,6 \right),\,\left( 1,\,8 \right),\,\left( 2,\,5 \right),\,\left( 2,\,6 \right),\,\left( 2,\,8 \right),\,\left( 3,\,5 \right),\,\left( 3,\,6 \right),\,\left( 3,\,8 \right),\,\left( 7,\,5 \right),\,\left( 7,\,6 \right),\,\left( 7,\,8 \right) \right\}$, card $B\times D=12$
$C\times A=\left\{ \left( 1,\,3 \right),\,\left( 1,\,5 \right),\,\left( 1,\,7 \right),\,\left( 7,\,3 \right),\,\left( 7,\,5 \right),\,\left( 7,\,7 \right) \right\}$, card $C\times A=6$
$C\times B=\left\{ \left( 1,\,1 \right),\,\left( 1,\,2 \right),\,\left( 1,\,3 \right),\,\left( 1,\,7 \right),\,\left( 7,\,1 \right),\,\left( 7,\,2 \right),\,\left( 7,\,3 \right),\,\left( 7,\,7 \right) \right\}$, card $C\times B=8$
$C\times C=\left\{ \left( 1,\,1 \right),\,\left( 1,\,7 \right),\,\left( 7,\,1 \right),\,\left( 7,\,7 \right) \right\}$, card $C\times C=4$
$C\times D=\left\{ \left( 1,\,5 \right),\,\left( 1,\,6 \right),\,\left( 1,\,8 \right),\,\left( 7,\,5 \right),\,\left( 7,\,6 \right),\,\left( 7,\,8 \right) \right\}$, card $C\times D=6$
$D\times A=\left\{ \left( 5,\,3 \right),\,\left( 5,\,5 \right),\,\left( 5,\,7 \right),\,\left( 6,\,3 \right),\,\left( 6,\,5 \right),\,\left( 6,\,7 \right),\,\left( 8,\,3 \right),\,\left( 8,\,5 \right),\,\left( 8,\,7 \right) \right\}$, card $D\times A=9$
$D\times B=\left\{ \left( 5,\,1 \right),\,\left( 5,\,2 \right),\,\left( 5,\,3 \right),\,\left( 5,\,7 \right),\,\left( 6,\,1 \right),\,\left( 6,\,2 \right),\,\left( 6,\,3 \right),\,\left( 6,\,7 \right),\,\left( 8,\,1 \right),\,\left( 8,\,2 \right),\,\left( 8,\,3 \right),\,\left( 8,7 \right) \right\}$, card $D\times B=12$
$D\times C=\left\{ \left( 5,\,1 \right),\,\left( 5,\,7 \right),\,\left( 6,\,1 \right),\,\left( 6,\,7 \right),\,\left( 8,\,1 \right),\,\left( 8,\,7 \right) \right\}$, card $D\times C=6$
$D\times A=\left\{ \left( 5,\,5 \right),\,\left( 5,\,6 \right),\,\left( 5,\,8 \right),\,\left( 6,\,5 \right),\,\left( 6,\,6 \right),\,\left( 6,\,8 \right),\,\left( 8,\,5 \right),\,\left( 8,\,6 \right),\,\left( 8,\,8 \right) \right\}$, card $D\times D=9$
3. $P_{ABC}=P_{BCD}=12$ u.

4. $A_{ABC}=12,5$ u$^2$.

5. a) 32; b) 7,25.

6.

7. a) $A\times B=\{ \left( 1,\,1 \right),\,\left( 1,\,4 \right),\,\left( 1,\,9 \right),\,\left( 1,\,16 \right),\,\left( 2,\,1 \right),\,\left( 2,\,4 \right),\,\left( 2,\,9 \right),\,\left( 2,\,16 \right),\,\left( 3,\,1 \right),\,\left( 3,\,4 \right),\, \left( 3,\,9 \right),\,\left( 3,\,16 \right),\,\left( 4,\,1 \right),\,\left( 4,\,4 \right),\,\left( 4,\,9 \right),\,\left( 4,\,16 \right) \}$; b) 4,2.

8.

9. a) $A(-4,\,4)$, $B(-3,\,1)$, $C(-5,\,-2)$, $D(4,\,0)$, $E(1,\,3)$.

b) $AB=\sqrt{10}$, $BC=\sqrt{13}$, $CA=\sqrt{37}$, $DE=\sqrt{18}$. c) Podul ar trebui să fie cât mi scurt, deci să unească $BE$.
10. Graficul 3.

Unitatea 6 - Patrulaterul

Patrulaterul convex. Suma măsurilor unghiurilor unui patrulater convex

2. a) convexe: Figura 1, Figura 2, Figura 4, concave: Figura. 3; b) Figura 1, laturi: $AB$, $BC$, $CD$, $DA$, diagonale: $AC$, $BD$, unghiuri: $\sphericalangle DAB$, $\sphericalangle ABC$, $\sphericalangle BCD$, $\sphericalangle CDA$; Figura 2, laturi: $EF$, $FG$, $GH$, $HE$, diagonale: $EG$, $FH$, unghiuri: $\sphericalangle HEF$, $\sphericalangle EFG$, $\sphericalangle FGH$, $\sphericalangle GHE$; Figura 4, laturi: $MN$, $NO$, $OP$, $PM$, diagonale: $MO$, $NP$, unghiuri: $\sphericalangle PMN$, $\sphericalangle MNO$, $\sphericalangle NOP$, $\sphericalangle OPM$.

3. a) $\sphericalangle D=40^{\circ}$; b) $\sphericalangle A=\sphericalangle 103^{\circ}$; c) $\sphericalangle C=157^{\circ}$; d) $B=192^{\circ}$.

4. a) $60^{\circ}$, $120^{\circ}$, $140^{\circ}$, $40^{\circ}$; b) $90^{\circ}$, $40^{\circ}$, $100^{\circ}$, $130^{\circ}$; c) $90^{\circ}$, $30^{\circ}$, $60^{\circ}$, $20^{\circ}$; d) $72^{\circ}$, $144^{\circ}$, $48^{\circ}$, $96^{\circ}$.

6. Dacă aeroportul este în zonele 5 sau 7, patrulaterul format este concav. Dacă aeroportul este în zona 6, patrulaterul format este convex.

JOC - Numere și patrulatere

Paralelogramul: proprietăți

1. Da, 2.congruente şi paralele, 3.Cele două triunghiuri sunt congruente (echerele sunt identice), deci unghiurile alterne interne sun congruente, ceea ce implică că laturile sunt paralele.

2. a) F; b) A; c) F; d) F; e) F; f) F; g) A; h) A.

3. a) $\sphericalangle B=113^{\circ}$, $\sphericalangle C=67^{\circ}$, $\sphericalangle D=113^{\circ}$; b) $\sphericalangle A=78^{\circ}$, $\sphericalangle C=78^{\circ}$, $\sphericalangle D=102^{\circ}$; c) $\sphericalangle A=110^{\circ}$, $\sphericalangle B=70^{\circ}$, $\sphericalangle D=70^{\circ}$; d) $\sphericalangle A=148^{\circ}$, $\sphericalangle C=148^{\circ}$, $\sphericalangle D=32^{\circ}$.

4. a)

b)

c)

d)

6. $AE=\cfrac{AB}{3}=\cfrac{CD}{3}$ (din faptul că $ABCD$ paralelogram) $=CD-CF=FD$ (1)

$AE\parallel AB\parallel CD\parallel FD$ (din faptul că $ABCD$ paralelogram) (2) Din construcţie, patrulaterul AEFD este convex (3). $(1)+(2)+(3)\Rightarrow AEFD $ paralelogram. Analog, $BEFC$ paralelogram.

7. Comparăm triunghiurile $\triangle ABD$ şi $\triangle CDB$: $\left. \begin{array}{l}BD\equiv DB \,(\text{latură comună})\\ \sphericalangle ABD\equiv\sphericalangle CDB\\\sphericalangle A\equiv\sphericalangle C\end{array}\right\} \xRightarrow[]{ULU} \triangle ABD\equiv \triangle CDB\Rightarrow$ $\left. \begin{array}{l}AB\equiv CD\\AD\equiv CB\end{array}\right\} \Rightarrow ABCD$ paralelogram.

9. Comparăm $\triangle AD'A'$ şi $\triangle CB'C'$: $\left.\begin{array}{l}AA'=2AB=2CD=CC'\\AD'=AD=BC=CB'\\\sphericalangle D'AA'=180^{\circ}-\sphericalangle BAD=180^{\circ}-\sphericalangle BCD=\sphericalangle B'CC'\end{array}\right\}\xRightarrow[]{LUL}$ $\triangle AD' A'\equiv \triangle CB' C\Rightarrow$ $ D' A'\equiv B' C'$. Analog $A' B'\equiv C' D'$. Din cele două congruenţe rezultă că $A' B' C' D'$ paralelogram.

10. a) $\left. \begin{array}{l}AA’=BB’,\,AB=A’B’\Rightarrow ABB’A’\text{ paralelogram}\Rightarrow AA’\parallel BB'\\ BB’= CC’,\,BC=B’C’\Rightarrow BCC’B’\text{ paralelogram}\Rightarrow BB'\parallel CC'\end{array}\right\}\Rightarrow AA'\parallel CC'$ $\xRightarrow[]{AA'=CC'} ACC’A’$ paralelogram.

b) $CC’=DD’$, $CD=C’D’\Rightarrow CDD’C’$ paralelogram $\Rightarrow DD’\parallel CC'$ $\xRightarrow[]{AA'\parallel CC'}$ $AA'\parallel DD'$ $\xRightarrow[]{AA'=DD'}ADD’A’$ paralelogram.

c)Analog cu b). $ABB’A’$ paralelogram $\Rightarrow\sphericalangle ABB'=180^{\circ}-\sphericalangle A'AB=180^{\circ}-\sphericalangle C'CB=\sphericalangle B' BC$ ($BCC’B’$ paralelogram).
$\sphericalangle ABC=360^{\circ}-\sphericalangle ABB'-\sphericalangle B'BC=2\cdot\sphericalangle A'AB$.

Triunghiul $ABC$ este isoscel, deci $\sphericalangle BCA=\cfrac{1}{2} (180^{\circ}-\sphericalangle ABC)=90^{\circ}-\sphericalangle A' AB=90^{\circ}-\sphericalangle BCC'$ $\Rightarrow CC'\perp AC$. Analog, $CC'\perp CE$, deci $A$, $C$, $E$ coliniare.
$\sphericalangle ABC=2\cdot\sphericalangle A'AB=\sphericalangle BCD+\sphericalangle C'CD=\sphericalangle BCD$ $\Rightarrow AB\parallel CD\xRightarrow[]{AB=CD} ACDB$ paralelogram.

$\left.\begin{array}{l}\text{Analog e), se demonstrează că }BCED\text{ paralelogram, deci }BC\parallel ED\\DD’=EE’,\,DE=D’E’\Rightarrow DED’E’ \text{ paralelogram}\Rightarrow DE\parallel D'E'\end{array}\right\}$ $\Rightarrow BC\parallel D'E'\xRightarrow[]{BC=DE} BCE'D'$ paralelogram.

11. b) pentru mai multe locuri de parcare alegem Figura 12. Pentru mai mult spaţiu, alegem Figura 14. Figura 13 optimizează cele două cerinţe.

Aplicații în geometria triunghiului: linie mijlocie în triunghi, centrul de greutate al unui triunghi

4. $P=15$ cm.

5. $MP$ linie mijlocie $\Rightarrow \left\{\begin{array}{l}MP=\cfrac{1}{2} DF=DN\\MP\parallel DF\parallel DN\end{array}\right\}$ $\Rightarrow DMPN$ paralelogram.

6. a) $P=13$ cm; b) $P=28$ cm; c) $P=9$ cm; d) $P=16,5$ cm.

7. a) $GA=10$ cm, $AA’=15$ cm; b) $GB’=3$ cm, $BB’=9$ cm; c) $GC=6$ cm, $GC’=3$ cm.

8. a) $F$ este centrul de greutate al $\triangle BCD$; b) $DN=CN$.

9. Fie $ABCD$ paralelogram, $M$, $N$, $P$, $Q$ mijloacele laturilor $AB$, $BC$, $CD$, respectiv $DA$.

$\left. \begin{array}{l}MN\text{ linie mijlocie în }\triangle ABC\Rightarrow MN\parallel AC\text{ şi }MN=\cfrac{1}{2} AC \\ QP\text{ linie mijlocie în }\triangle ADC\Rightarrow QP\parallel AC şi QP=\cfrac{1}{2} AC\end{array}\right\}$ $\Rightarrow MNPQ$ paralelogram.

10. $MN\parallel BC\parallel CD\parallel NP\Rightarrow M,\,N,\,P$ coliniare. $M$ mijlocul lui $BC$, $MN\parallel BC\Rightarrow MN$ linie mijlocie $\Rightarrow$ $N$ mijlocul lui $AC$ $\xRightarrow[]{NP\parallel CD}P$ mijlocul lui $AD\Rightarrow NP$ linie mijlocie $\Rightarrow 2MP=2MN+2NP=BC+CD$.

11. a) $MD=\cfrac{1}{2}AD=\cfrac{1}{2}BC=BQ=BP$. $MD\parallel AD\parallel BC\parallel BP$, deci $MPBD$ paralelogram. b) $AM=\cfrac{1}{2}AD=\cfrac{1}{2}BC=BQ=BP$. $AM\parallel AD\parallel BC\parallel BP$, deci $APBM$ paralelogram.

12. a) $\left.\begin{array}{l}PN\text{ linie mijlocie}\Rightarrow PN\parallel BM,\text{ deci }\sphericalangle APN\equiv \sphericalangle PBM .\\MN\text{ linie mijlocie }\Rightarrow MN\parallel PB,\text{ deci }\sphericalangle PBM\equiv \sphericalangle NMC\\PN\text{ linie mijlocie}\Rightarrow PN\parallel CM,\text{ deci }\sphericalangle NMC\equiv \sphericalangle PNM\end{array}\right\}\Rightarrow$\\$ \sphericalangle APN\equiv \sphericalangle PBM\equiv \sphericalangle NMC\equiv \sphericalangle MNP$. Analog $\sphericalangle ANP\equiv \sphericalangle PMB\equiv \sphericalangle NCM\equiv \sphericalangle NPM$.

$PN$ linie mijlocie $\Rightarrow PN=\cfrac{1}{2} BC=BM=MC$ ($M$ mijlocul lui $BC$).

Din cazul de congruenţă LUU, $\triangle APN\equiv \triangle PBM\equiv \triangle NMC\equiv \triangle MNP$.

b) $MN$ linie mijlocie, $\Rightarrow \left\{\begin{array}{l}MN=\cfrac{1}{2} AB=AP\rule[-4mm]{0mm}{2mm}\\MN\parallel AB\parallel AP\end{array}\right\}\Rightarrow APMN$ paralelogram.

c) Analog punctului anterior, $PNCM$ paralelogram, deci diagonalele sale se intersectează în mijloc, i.e. $S$ mijlocul lui MN. d) Analog punctului anterior, $Q$ mijlocul lui $NP$ şi $R$ mijlocul lui $MP$, deci $MQ$, $PS$, $NR$ mediane, i.e. $G$ centrul de greutate al triunghiului $MNP$. e) La punctul anteriror am demonstrat că triunghiul median al unui triunghi are acelaşi centru de greutate cu acesta. $QRS$ este triunghiul median al triunghiului $MNP$, deci are acelaşi centru de greutate cu acesta, i.e. $G$.

Paralelograme particulare: dreptunghi; proprietăți

2. $AB+BC+CA=DC+CB+BD$, deci cele două creaturi parcurg aceeaşi distanţă.

3. $AO=DO\Rightarrow AC=DB\Rightarrow ABCD$ dreptunghi (un paralelogram cu diagonalele congruente este dreptunghi).

4. Fie $\{O\}=AM\cap BC$, adică $O$ mijlocul lui $BC\Rightarrow AO$ mediană deci $BO=OC$. $M$ simetricul lui $A$ faţă de $O\Rightarrow MO=AO$, deci $O$ mijlocul $MA$ şi $BC$, deci $ABMC$ paralelogram. Acesta are un unghi drept ($\sphericalangle BAC$), deci $ABMC$ dreptunghi.

5. a) $\sphericalangle ADO=30^{\circ}$. b) $\sphericalangle ADO=\sphericalangle DAO\Rightarrow$ triunghiul $ADO$ este isoscel, deci $AO=OD\Rightarrow 2AO=2OD\Rightarrow AC=BD\Rightarrow ABCD$ dreptunghi.

6. a) $\sphericalangle ABD=30^{\circ}$.

7. a) $\sphericalangle DAE=150^{\circ}$. Comparăm $\triangle DAE$ şi $\triangle CBE$: $\left.\begin{array}{l} DA=BC\,(ABCD\text{ dreptunghi})\\EA=EB\, (ABE\text{ triungh echilateral})\\\sphericalangle DAE=\sphericalangle DAB+\sphericalangle BAD=60^{\circ}+90^{\circ}=\sphericalangle EBA+\sphericalangle ABC=\sphericalangle EBC\end{array}\right\} \xRightarrow[]{LUL}$ $\triangle DAE \equiv \triangle CBE\Rightarrow DE=CE (1)\Rightarrow\triangle DEC$ isoscel. c) Comparăm $\triangle DBE$ şi $\triangle CAE$: $\left.\begin{array}{l}DB=CA\, (\text{ABCD dreptunghi})\\(1)\\BE=AE\, (\triangle BAE\text{ triunghi echilateral})\end{array}\right\}\xRightarrow[]{LLL} \triangle DBE \equiv \triangle CAE$.

9. $AC$ diametru $\Rightarrow \sphericalangle ABC=\sphericalangle CDA=\cfrac{180^{\circ}}{2}=90^{\circ}$.

$BD$ diametru $\Rightarrow\sphericalangle BCD=\sphericalangle DAB=\cfrac{180^{\circ}}{2}=90^{\circ}$, deci $ABCD$ dreptunghi.

10. $ABCD$ dreptunghi $\xLeftrightarrow[]{AC\cap BD=\{O\}}AO=OC=BO=OD\Leftrightarrow\cfrac{1}{2}AO=\cfrac{1}{2}OC=\cfrac{1}{2}BO=\cfrac{1}{2}OD$ $\Leftrightarrow MO=OP=ON=OQ$ $\xLeftrightarrow[]{MP\cap NQ=\{O\}} MNPQ$ dreptunghi.

11. a) $FI\parallel FA\parallel EB\parallel DC\parallel DR$, $FI=\cfrac{1}{2} FA=\cfrac{1}{2} EB=\cfrac{1}{2} DC=RD$, deci $FIRD$ paralelogram, deci $IR=FD=FE+ED$. b) În prima situaţie $IR=IE+ER>FE+ED$ (ipotenuzele sunt mai mari decât catetele), dei distanţa dintre Ioana şi Radu este mai mare decât distanţa dintre cele două vagoane.

Paralelograme particulare: romb; proprietăți

2. a) $\sphericalangle DAB=40^{\circ}$, $\sphericalangle ABC=140^{\circ}$, $\sphericalangle BCD=40^{\circ}$, $\sphericalangle CDA=140^{\circ}$; b) $\sphericalangle DAB=110^{\circ}$, $\sphericalangle ABC=70^{\circ}$, $\sphericalangle BCD=110^{\circ}$, $\sphericalangle CDA=70^{\circ}$; c) $\sphericalangle DAB=80^{\circ}$, $\sphericalangle ABC=100^{\circ}$, $\sphericalangle BCD=80^{\circ}$, $\sphericalangle CDA=100^{\circ}$; d) $\sphericalangle DAB=80^{\circ}$, $\sphericalangle ABC=100^{\circ}$, $\sphericalangle BCD=80^{\circ}$, $\sphericalangle CDA=100^{\circ}$.

3. a) $\sphericalangle DAB=130^{\circ}$, $\sphericalangle ABC=50^{\circ}$, $\sphericalangle BCD=130^{\circ}$, $\sphericalangle CDA=50^{\circ}$; b) $\sphericalangle MNP=80^{\circ}$, $\sphericalangle NPQ=100^{\circ}$, $\sphericalangle PQM=80^{\circ}$, $\sphericalangle QMN=100^{\circ}$; c) $\sphericalangle PRS=70^{\circ}$, $\sphericalangle RST=110^{\circ}$, $\sphericalangle STP=70^{\circ}$, $\sphericalangle TPR=110^{\circ}$.

4. Fie $O$ punctul de intersecţie al diagonalelor. $\sphericalangle AOB=180^{\circ}-\sphericalangle CAB-\sphericalangle DBA=180^{\circ}-70^{\circ}-20^{\circ}=90^{\circ}$, deci $AC\perp BD$, deci $ABCD$ romb.

5. Fie $O$ punctul de intersecţie al diagonalelor. $\sphericalangle AOB=180^{\circ}-\sphericalangle CAB-\sphericalangle DBA=180^{\circ}-70^{\circ}-20^{\circ}=90^{\circ}$, deci $AC\perp BD$, deci $ABCD$ romb.

6. $\sphericalangle DAB=60^{\circ}$, $\sphericalangle ABC=120^{\circ}$, $\sphericalangle BCD=60^{\circ}$, $\sphericalangle CDA=120^{\circ}$.

7. $\sphericalangle DAB=120^{\circ}$, $\sphericalangle ABC=60^{\circ}$, $\sphericalangle BCD=120^{\circ}$, $\sphericalangle CDA=60^{\circ}$.

8. Deoarece $ABCD$ romb, $BT$ mediatoarea lui $AC$, deci $TC=AT$. Dar $AT=AC$, deci $\triangle ATC$ echilateral.

9. Comparăm $\triangle AOB$, $\triangle COB$, $\triangle AOD$, $\triangle COD$ dreptunghice (deoarece $ABCD$ romb): $\left.\begin{array}{l}AO=CO=AO=CO\\OB=OB=OD=OD\end{array}\right\} \xRightarrow[]{CC} \triangle AOB\equiv \triangle COB\equiv \triangle AOD\equiv \triangle COD$.

10. $MP$ linie mijlocie, deci $MP=\cfrac{1}{2} YZ=YN=\cfrac{1}{2} YZ=\cfrac{1}{2} XY=MY=\cfrac{1}{2} XY=NP$ ($NP$ linie mijlocie), deci $MPNY$ romb. Analog $XMNP$, $ZPMN$ romburi.

11. $BD=DB'$ şi $AD\perp BB'$, deci $AD$ mediatoare $\Rightarrow\triangle BAB'$ isoscel $\Rightarrow AB=AB'$.

$AD=DA'$ şi $AD\perp BB'$, deci $BD$ mediatoare $\Rightarrow \left\{\begin{array}{l}\triangle AB'A'\text{ isoscel}\Rightarrow AB'=B'A'\\ \triangle ABA' \text{ isoscel}\Rightarrow AB=BA'\end{array}\right\} \xRightarrow[]{AB=AB'} ABA'B'$ romb. Analog $AC'A'C$ romb.

12. $MN$ linie mijlocie, deci $MN=\cfrac{1}{2} AC=PQ= \cfrac{ 1}{2} AC=\cfrac{1}{2} BD =NP= \cfrac{ 1}{2} BD=MQ$ ($MQ$ linie mijlocie), deci $MNPQ$ romb.

13. $\left.\begin{array}{l}MN\text{ linie mijlocie, deci }MN=\cfrac{1}{2} AC=PQ\, (PQ\text{ linie mijlocie})\rule[-5mm]{0mm}{2mm}\\NP\text{ linie mijlocie, deci }NP=\cfrac{ 1}{2} BD=MQ\, (MQ\text{ linie mijlocie})\end{array}\right\}\Rightarrow MNPQ$ paralelogram.

$\left.\begin{array}{l}MB=\cfrac{1}{2} AB=1/2 CD=CP,\text{ deci }MBCP\text{ paralelogram}\Rightarrow MP=BC\rule[-5mm]{0mm}{2mm}\\QD=\cfrac{1}{2} AD=\cfrac{1}{2} BC=CN,\text{ deci }DQNC\text{ paralelogram}\Rightarrow QN=CD\rule[-4mm]{-0mm}{2mm}\end{array}\right\}\xRightarrow[]{BC=CD}$ $ MP=QN\xRightarrow[]{MNPQ\text{ paralelogram}} MNPQ$ romb.

14. $MN=\cfrac{1}{2} AB=\cfrac{1}{2}BC=NP=\cfrac{1}{2}BC=\cfrac{1}{2} CD=PQ=\cfrac{1}{2} CD=\cfrac{1}{2} DA=MQ\Rightarrow MNPQ$ romb.

15. a) $ABCD$ romb şi $\sphericalangle ADC=\sphericalangle ABC=120^{\circ}$, $ABCD$ romb, deci $BD$ bisectoarea $\sphericalangle ABC$, $\sphericalangle ADC \Rightarrow \sphericalangle ABD=60^{\circ}=\sphericalangle BDC$.

$\left.\begin{array}{l}BE\text{ bisectoare, deci }\sphericalangle EBD=30^{\circ}\\ DF\text{ bisectoare, deci }\sphericalangle BDF=30^{\circ}\end{array}\right\}\Rightarrow DF\parallel BE$. $BF\parallel BC\parallel DA\parallel DE$, deci $BFDE$ paralelogram.

b) $ABD$ echilateral ($AB=AD$ şi $\sphericalangle ABD=60^{\circ}$) şi $BCD$ echilateral ($BC=CD$ şi $\sphericalangle BDC=60^{\circ}$). Ele sunt congruente deoarece au latura comună $BD$. $BE$, $AO$ mediane, deci $G$ centru de greutate al $\triangle ABD$. $DF$, $CO$ mediane, deci $H$ centru de greutate al $\triangle CBD$. $DG=BG=\cfrac{2}{3}$ din mediana triunghiului $ABD$ (triunghiul echilateral, deci medianele sunt congruente) $=\cfrac{2}{3}$ din mediana $\triangle BCD=BH=HD$, deci $BGDH$ romb.

16. a) Comparăm triunghiurile $AGE$, $EGC$ şi $CGA$: $\left.\begin{array}{l}AG=EG=CG\text{ (laturi ale romburilor)}\\GE=GC=GA\text{ (laturi ale romburilor)}\\\sphericalangle AGE=120^{\circ}=180^{\circ}-60^{\circ}=\sphericalangle AGC=120^{\circ}=360^{\circ}-120^{\circ}-120^{\circ}=\sphericalangle EGC\end{array}\right\}\xRightarrow[]{LUL}$ $\triangle AGE\equiv \triangle EGC\equiv \triangle CGA\Rightarrow AE\equiv EC\equiv CA\Rightarrow\triangle ACE$ echilateral.

b) $GA=GC=GE$ (laturi ale romburilor), deci $G$ centrul cercului circumscris $\triangle ACE$. Dar triunghiul $ACE$ e echilateral, deci $G$ e şi centrul de greutate al $\triangle ACE$.

c) Analog cu punctul b) $G$ e şi centrul de greutate al $\triangle BDF$.

Paralelograme particulare: pătrat; proprietăți

1. a) sunt congruente; b) sunt drepte; c)

3. a) A; b) F; c) A; d) A; e) F.

4. Diagonala $AC$ este bisectoarea $\sphericalangle BAD$, deci $ABCD$ este romb şi dreptunghi, deci pătrat.

6. $90^{\circ}$.

7. $135^{\circ}$.

8. a) $ADE$ isoscel; b) $75^{\circ}$; c) $150^{\circ}$.

9. Fie $\{O\}=DN\cap MC$. Comparăm triunghiurile dreptunghice $\triangle MBC$ şi $\triangle NCD$: $\left.\begin{array}{l}CD=BC\text{ (laturi ale pătratului)}\\CN=\cfrac{1}{2} CB=\cfrac{1}{2} AB=BM\end{array}\right\} \xRightarrow[]{CC} \triangle MBC\equiv \triangle NCD$ $\Rightarrow \sphericalangle DNC=\sphericalangle CMB=$ $=90^{\circ}-\sphericalangle MCB\Rightarrow\sphericalangle CON=90^{\circ}\Rightarrow DN\perp CM$.

10. Am demonstrat la Problema 10 de la dreptunghi că $ABCD$ dreptunghi. Deoarece diagonalele sunt perpendiculare rezultă că $ABCD$ romb, deci $ABCD$ pătrat.

11. $BA\parallel CD$. $\sphericalangle DAC=\sphericalangle ACB\Rightarrow DA\parallel BC\Rightarrow ABCD$ paralelogram. $AD\perp AB$, deci $\sphericalangle DAB=90^{\circ}$, $AC$ bisectoare, deci $\sphericalangle DAC=45^{\circ}$. $AD\perp AB\parallel CD$, deci $AD\perp CD\Rightarrow\sphericalangle ADC=90^{\circ}\xRightarrow[]{\sphericalangle DAC=45^{\circ}}\sphericalangle DCA=45^{\circ}$, aşadar $DAC$ isoscel. $DA=DC\xRightarrow[]{ABCD\text{ paralelogram}} ABCD$ romb $\xRightarrow[]{\sphericalangle DAB=90^{\circ}} ABCD$ pătrat.

12. a) $\left.\begin{array}{l}MN\perp AB\perp AC\Rightarrow MN\parallel AC\parallel AP\\MP\perp AC\perp AB\Rightarrow MP\parallel AB\parallel AN\end{array}\right\}\Rightarrow MPAN$ paralelogram $\xRightarrow[]{\sphericalangle BAC=90^{\circ}} MPAN$ dreptunghi. b) Comparăm $\triangle ABM$ şi $\triangle ACM$: $\left.\begin{array}{l}AB=AC\\BM=MC\\AM\text{ latură comună}\end{array}\right\} \xRightarrow[]{LLL} \triangle ABM\equiv \triangle ACM$ $\Rightarrow \sphericalangle BAM\equiv\sphericalangle CAM\Rightarrow AM\text{ bisectoare}\xRightarrow[]{APMN\text{ dreptunghi}} APMN$ pătrat. c) $\left.\begin{array}{l}MP\parallel AN\xRightarrow[]{M\text{ mijlocul lui }BC} MP\text{ linie mijlocie}\Rightarrow AP=\cfrac{1}{2} AC\rule[-5mm]{0mm}{2mm}\\ MN\parallel AP\xRightarrow[]{M\text{ mijlocul lui }BC} MN\text{ linie mijlocie}\Rightarrow AN=\cfrac{1}{2} AB\end{array}\right\}\xRightarrow[]{AP=AN} AC=AB$.

13. Comparăm triunghiurile dreptunghice $\triangle DAB$ şi $\triangle DCN$: $\left.\begin{array}{l}AM=AB-BM=BC-BN=CN\\DA=DC\end{array}\right\} \xRightarrow[]{CC} \triangle DAB\equiv \triangle DCN\Rightarrow DM=DN\Rightarrow DMN$ isoscel. $\left.\begin{array}{l}\triangle BMN\text{ dreptunghic isoscel}\Rightarrow\sphericalangle BNM=\sphericalangle BMN=45^{\circ}\\\triangle BCA\text{ dreptunghic isoscel}\Rightarrow\sphericalangle BCA=\sphericalangle BAC=45^{\circ}\end{array}\right\}\Rightarrow MN\parallel AC$ ($\sphericalangle BNM$, $\sphericalangle BCA$ unghiuri corespondente).

14. $\triangle BAE$ dreptunghic isoscel $\Rightarrow\sphericalangle BEA=\sphericalangle EBA=45^{\circ}$. $CE$ bisectoarea lui $\sphericalangle BED\Rightarrow\sphericalangle BEF=45^{\circ}$. $BD$ bisectoarea lui $\sphericalangle CBE\Rightarrow\sphericalangle DBE=45^{\circ}$. $\sphericalangle AEF=\sphericalangle AEB+\sphericalangle BEF=45^{\circ}+45^{\circ}=90^{\circ}=45^{\circ}+45^{\circ}=\sphericalangle FBE+\sphericalangle EBA=\sphericalangle FBA$. Cum $\sphericalangle BAE=90^{\circ}$, rezultă că $ABFE$ dreptunghi $\xRightarrow[]{AB=AE} ABFE$ pătrat.

15. a) $\sphericalangle CAD=30^{\circ}$, $\sphericalangle ACB=15^{\circ}$.

b) Comparăm triunghiurile $ABC$ şi $AED$: $\left.\begin{array}{l}AB=AE\text{ (laturi ale triunghiului echilateral)}\\BC=ED\text{ (laturi ale pătratului)}\\\sphericalangle ABC=\sphericalangle ABE+\sphericalangle EBC=60^{\circ}+90^{\circ}=150^{\circ}=\\=60^{\circ}+90^{\circ}=\sphericalangle AEB+\sphericalangle BED=\sphericalangle AED\end{array}\right\}\xRightarrow[]{LUL} \triangle ABC\equiv \triangle AED\Rightarrow AC=AD\Rightarrow \triangle ACD$ isoscel.

c) $\sphericalangle ACE=\sphericalangle BCE-\sphericalangle ACB=45^{\circ}-15^{\circ}=30^{\circ}=\sphericalangle CAF\Rightarrow\triangle FAC$ isoscel $\Rightarrow AF=FC$.

16. a) $\sphericalangle ABE=180^{\circ}-\sphericalangle BAF=180^{\circ}-45^{\circ}=135^{\circ}=180^{\circ}-45^{\circ}=180^{\circ}-\sphericalangle BAD=\sphericalangle ABC$.

$\sphericalangle EBC=360^{\circ}-\sphericalangle ABE-\sphericalangle ABC=360^{\circ}-135^{\circ}-135^{\circ}=90^{\circ}\xRightarrow[]{BCGE\text{ romb}} BCGE$ pătrat.

b) $BCGE$ romb $\Rightarrow BG$ bisectoarea $\sphericalangle CBE\Rightarrow \sphericalangle CBG=45^{\circ}=180^{\circ}-\sphericalangle ABC\Rightarrow A$, $B$, $G$ coliniare.

$\left.\begin{array}{l}DC=AB=EF\\DC\parallel AB\parallel EF\end{array}\right\}\Rightarrow DCEF$ paralelogram.

$\triangle DAF$ dreptungic isoscel $\Rightarrow \sphericalangle ADF=45^{\circ}\xRightarrow[]{\sphericalangle ADC=135^{\circ}} FDC=90^{\circ}\xRightarrow[]{DCEF\text{ paralelogram}} DCEF$ dreptunghi.

Trapezul, clasificare, proprietăţi. Linia mijlocie în trapez. Trapezul isoscel, proprietăţi

1. a) Nu. Dacă laturile paralele ar fi congruente, patrulaterul ar fi paralelogram, deci laturile opuse ar fi paralele, aşadar nu ar fi paralelogram. b) Da.

c) Da, poate avea maxim două unghiuri drepte. Dacă ar avea 3, ar fi dreptunghi, deci paralelogram.

2. a) $\left.\begin{array}{l}MO\text{ linie mijlocie, deci }MO\parallel AB\\NO\text{ linie mijlocie, deci } NO\parallel CD\end{array}\right\}\xRightarrow[]{AB\parallel CD} MO\parallel NO\Rightarrow M$, $O$, $N$ coliniare.

b) $\left.\begin{array}{l}MO\text{ linie mijlocie, deci }MO=\cfrac{1}{2} AB\rule[-5mm]{0mm}{2mm}\\NO\text{ linie mijlocie, deci }NO=\cfrac{1}{2} CD\end{array}\right\}\xRightarrow[]{MN=MO+ON} MN=\cfrac{AB+CD}{2}$.

4. a) baza mare: $CD$, baza mică: $AB$, $\sphericalangle A=90^{\circ}$, $\sphericalangle B=120^{\circ}$; b) baza mare: $GH$, baza mică: $EF$, $\sphericalangle E=135^{\circ}$, $\sphericalangle G=78^{\circ}$;

c) baza mare: $IJ$, baza mică: $KL$, $\sphericalangle J=76^{\circ}$, $\sphericalangle K=\sphericalangle L=104^{\circ}$; d) baza mare: $OP$, baza mică: $MN$, $\sphericalangle O=45^{\circ}$, $\sphericalangle M=105^{\circ}$.

5. I. b)

II. b) sau d)

III. c)

IV. c)

V. b)

7. a) $\sphericalangle D=135^{\circ}$, $\sphericalangle C=115^{\circ}$; b) $\sphericalangle A=70^{\circ}$, $\sphericalangle C=120^{\circ}$; c) $\sphericalangle B=118^{\circ}$, $\sphericalangle D=80^{\circ}$.

8. a) $\sphericalangle B=105^{\circ}$, $\sphericalangle C=75^{\circ}$, $\sphericalangle D=75^{\circ}$; b) $\sphericalangle A=40^{\circ}$, $\sphericalangle C=140^{\circ}$, $\sphericalangle D=140^{\circ}$; c) $\sphericalangle A=110^{\circ}$, $\sphericalangle B=110^{\circ}$, $\sphericalangle D=70^{\circ}$; d) $\sphericalangle A=112^{\circ}$, $\sphericalangle B=112^{\circ}$, $\sphericalangle C=68^{\circ}$.

10. $\sphericalangle DEF=\sphericalangle EFC=\sphericalangle FCD=\sphericalangle CDE=90^{\circ}\Rightarrow FEDC$ dreptunghi $\Rightarrow ED=CF$. Comparăm triunghiurile dreptunghice $AED$, $BFC$: $\left.\begin{array}{l}ED=FC\\AD=BC\end{array}\right\}\xRightarrow[]{CI} \triangle AED\equiv\triangle BFC\Rightarrow AE=BF$.

11. $MN\parallel BC\Rightarrow MNCB$ trapez. $\triangle ABC$ isoscel $\Rightarrow \sphericalangle B=\sphericalangle C\Rightarrow MNCB$ trapez isoscel $\Rightarrow MC=BN$ (diagonalele sunt congruente).

12. $ABCD$ trapez isoscel $\Rightarrow \sphericalangle A=\sphericalangle B\Rightarrow\triangle VAB$ isoscel și $\sphericalangle DCV=\sphericalangle CDV$ (unghiuri corespondente) $\Rightarrow \triangle VDC$ isoscel.

13. Comparăm $\triangle MPN$ și $\triangle NQM$: $\left.\begin{array}{l}MP=QN\, (MNPQ\text{ trapez isoscel})\\MN=NM\text{ (latură comună)}\\\sphericalangle PNM=\sphericalangle QMN\, (MNPQ\text{ trapez isoscel})\end{array}\right\}\xRightarrow[]{LUL} \triangle MPN\equiv \triangle NQM\Rightarrow \sphericalangle PMN=\sphericalangle QNM\Rightarrow \triangle TMN$ isoscel.

Comparăm $\triangle MPQ$ și $\triangle NQP$: $\left.\begin{array}{l}MQ=PN\, (MNPQ\text{ trapez isoscel})\\PQ=QP\text{ (latură comună)}\\\sphericalangle MQP=\sphericalangle QPN\, (MNPQ\text{ trapez isoscel})\end{array}\right\}\xRightarrow[]{LUL}$\\ $\triangle MPQ\equiv \triangle NQP\Rightarrow \sphericalangle QPM=\sphericalangle PQN\Rightarrow \triangle TPQ$ isoscel.

14. $\left.\begin{array}{l}MP\parallel AN\xRightarrow[]{M\text{ mijlocul lui }BC} MP\text{ linie mijlocie}\Rightarrow P\text{ mijlocul lui }AC\, (1)\\MN\parallel AP\xRightarrow[]{M\text{ mijlocul lui }BC} MN\text{ linie mijlocie}\Rightarrow N\text{ mijlocul lui }AB\, (2) \end{array}\right\}$ $\Rightarrow NP$ linie mijlocie $\Rightarrow NP\parallel BC\Rightarrow NBCP$ trapez.

$\left.\begin{array}{l}(1)\Rightarrow PC=\cfrac{1}{2}AC\rule[-5mm]{0mm}{2mm}\\(2)\Rightarrow NB=\cfrac{1}{2}AB\end{array}\right\}\xRightarrow[]{NBCP\text{ trapez isoscel}}AC=AB$.

15. a) $\sphericalangle D=90^{\circ}$, $\sphericalangle A=90^{\circ}$, $\sphericalangle CEA=90^{\circ}$ $\Rightarrow AECD$ dreptunghi $\sphericalangle CEB=90^{\circ}$, $\sphericalangle CBE=45^{\circ}\Rightarrow \sphericalangle ECB=45\Rightarrow \triangle CEB$ isoscel $\Rightarrow CE=EB$ (1) $\sphericalangle ACB=90^{\circ}\xRightarrow[]{\sphericalangle CBE=45^{\circ}}\sphericalangle ACE=45^{\circ}$. Dar $\sphericalangle CEA=90^{\circ}$ $\Rightarrow \triangle CEA$ isoscel $\Rightarrow CE=EA\xRightarrow[]{AECD\text{ dreptunghi}} AECD$ pătrat (2) $(2)\Rightarrow CE=CD\xRightarrow[]{CE=EB} CD=EB$. Dar $CD\parallel AB\parallel BE\Rightarrow EBCD$ paralelogram.

16. $AD\parallel MQ\parallel NP\parallel BC\Rightarrow AD\parallel BC$. $AB\parallel MN\parallel QP\parallel CD\Rightarrow AB\parallel CD$, deci $ABCD$ paralelogram. $\left.\begin{array}{l}MQ\perp QP\\AD\parallel MQ\\CD\parallel QP\end{array}\right\}\Rightarrow MQ\perp QP\Rightarrow MNPQ$ dreptunghi. b) $WX$, $XY$, $YZ$, $ZW$ linii mijlocii în trapeze isoscele, deci $MNYX$, $NXYP$, $PYZQ$, $ZQMW$ trapeze isoscele și problema devine analoagă cu cea anterioară.

Perimetre și arii: paralelogram, paralelograme particulare, triunghi, trapez

1. a) Numărăm câte laturi ale pătrăţelelor sunt pe $AD$ şi câte laturi sunt pe $AB$ şi înmulţim cele două numere: $8\cdot14=112u^2$; b) 28 pătrate haşurate.

2. a) $\triangle ABD$, $\triangle CDB$ dreptunghice: $\left.\begin{array}{l}AB=CD\\AD=CB\end{array}\right\}\xRightarrow[]{CC} \triangle ABD\equiv\triangle CDB$. b) $\triangle ABD\equiv \triangle CDB\Rightarrow A_{\triangle ABD}=A_{\triangle CDB}$.

c) $A_{ABCD}=A_{\triangle ABD}+A_{\triangle CDB}=2A_{\triangle ABD}\Rightarrow AB\cdot AD=2A_{\triangle ABD}\Rightarrow A_{\triangle ABD}=\cfrac{AB\cdot AD}{2}$.

3. a) $\triangle ABD$, $\triangle CDB$: $\left.\begin{array}{l}AB=CD\\AD=CB\\BD=DB\end{array}\right\}\xRightarrow[]{LLL} \triangle ABD\equiv \triangle CDB$

b) $\triangle ABD\equiv \triangle CDB\Rightarrow A_{\triangle ABD}=A_{\triangle CDB}$. c) $A_{ABCD}=A_{\triangle ABD}+A_{\triangle CDB}=2A_{\triangle ABD}\Rightarrow 2 (AB\cdot DE):2=A_{ABCD}\Rightarrow A_{ABCD}=AB\cdot DE$.

4. a) 40 cm$^2$; b) 21 cm$^2$; c) 54 cm$^2$.

5. a) $P=28$ cm; b) $P=34$ cm; c) $P=78$ cm.

6. a) $A_{ABC}=AB\cdot AC:2\Rightarrow AC=4$ cm $\xRightarrow[]{\text{teorema lui Pitagora}} BC=5$ cm. b) $P_{ABC}=12$ cm; c) $h_A=2,4$ cm.

7. a) 60 cm$^2$; b) 42 cm$^2$; c) 60 dm$^2$.

8. a) $A=48$ cm$^2$; b) $A=60$ m$^2$; c) $A=192$ dam$^2$.

9. a) $A=36$ cm$^2$, $P=24$ cm; b) $A=25$ dm$^2$, $P=20$ dm; c) $A=81$ m$^2$, $P=36$ m.

10. a) 25 cm$^2$; b) 9 dm$^2$; c) $6,25$ m$^2$.

11. a) 24 cm; b) 20 dm; c) 40 m.

12. a) 84 cm$^2$; b) 57 cm$^2$; c) 33 cm$^2$.

13. 72 cm$^2$.

14. Fie $h_A$ înălţimea din $A$ a $\triangle ABC\Rightarrow \cfrac{1}{2}A_{\triangle ABC}=\cfrac{1}{2}\cdot\cfrac{1}{2} BC\cdot h_A=\cfrac{1}{2} BM\cdot h_A=A_{\triangle ABM}$.

Similar, $\cfrac{1}{2}A_{\triangle ABC}=\cfrac{1}{2}\cdot\cfrac{1}{2} BC\cdot h_A=\cfrac{1}{2} CM\cdot h_A=A_{\triangle ACM}$. Deci, $A_{\triangle ABM}=A_{\triangle ACM}=\cfrac{1}{2} A_{\triangle ABC}$.

15. a) Construim $AE$ și $BF$ înălţimile din $A$ respectiv $B$ ale trapezului, $E,\,F\in CD\Rightarrow \sphericalangle EAB=\sphericalangle ABF=\sphericalangle BFD=90^{\circ}\Rightarrow ABFE$ dreptunghi $\Rightarrow AE=BF\Rightarrow \cfrac{1}{2} AE\cdot CD=\cfrac{1}{2} BF\cdot CD\Rightarrow A_{\triangle ACD}=A_{\triangle BCD}\Rightarrow A_{\triangle AOD}+A_{\triangle OCD}=A_{\triangle BOC}+A_{\triangle OCD}\Rightarrow A_{\triangle AOD}=A_{\triangle BOC}$.

b) $A_{\triangle AOD}=A_{\triangle BOC}\Rightarrow OP\cdot AD=OR\cdot BC\xRightarrow[]{ABCD\text{ trapez isoscel}}OP=OR\Rightarrow\triangle OPR$ isoscel.

16. Construim $CE$ și $DF$ înălţimile din $C$ respectiv $D$ ale trapezului, $E,\,F\in AB$, $MG$ înălţimea din $M$ a $\triangle MAB\Rightarrow CEGM$, $MGFD$ dreptunghiuri $\Rightarrow CE=MG=DF$.

17. Fie $BM\perp AC$, $M\in AC$, $DN\perp AC$, $N\in AC$. $A_{\triangle BOC}=\cfrac{1}{2} BM\cdot OC=\cfrac{1}{2} BM\cdot OA=$

$=A_{\triangle BOA}=\cfrac{1}{2} BM\cdot OA=\cfrac{1}{4} BM\cdot AC=\cfrac{1}{2} A_{\triangle ABC}=\cfrac{1}{2} A_{\triangle ACD} =\cfrac{1}{4} DN\cdot AC=\cfrac{1}{2} DN\cdot OA=$

$=A_{\triangle DOA}=\cfrac{1}{2} DN\cdot OA=\cfrac{1}{2} DN\cdot OC=A_{\triangle COD}$.

18. Fie $BM\perp AC$, $M\in AC$, $AN\perp BD$, $N\in BD$.

$A_{\triangle AOB}=A_{\triangle BOC}\Rightarrow BM\cdot OA=BM\cdot OC\Rightarrow OA=OC$.

$A_{\triangle AOB}=A_{\triangle AOD}\Rightarrow AN\cdot OB=AN\cdot OD\Rightarrow OA=OD$.

Deci, $O$ mijlocul $AC$ și $BD$, deci $ABCD$ paralelogram.

19. $\triangle ADE$ dreptunghic $\Rightarrow A_{\triangle ADE}=\cfrac{AD\cdot DE}{2}=\cfrac{AD\cdot AB}{10}=\cfrac{A_{ABCD}}{10}=\cfrac{A_{\triangle ADE}+A_{ABCE}}{10}\Rightarrow A_{ABCE}=9\cdot A_{\triangle ADE}$.

Recapitulare

1. a) $\sphericalangle Q=140^{\circ}$; b) $\sphericalangle P=93^{\circ}$.

2. a) $\sphericalangle A=50^{\circ}$, $\sphericalangle B=130^{\circ}$, $\sphericalangle C=50^{\circ}$, $\sphericalangle D=130^{\circ}$.

3. a) $\sphericalangle A=120^{\circ}$, $\sphericalangle B=60^{\circ}$, $\sphericalangle C=120^{\circ}$, $\sphericalangle D=60^{\circ}$.

4. $DM=\cfrac{1}{2} DA=\cfrac{1}{2} BC=BN$.

5. a) romb; b) 12 cm.

7. $TACD$ paralelogram.

8. Figura 44: $\large\mathcal{A} =252$ m$^2$, Figura 45: $\large\mathcal{A}=297$ m$^2$, Figura 46: $\large\mathcal{A}=112,5$ m$^2$.

10. Fie $ABCD$ patrulater convex şi $\sphericalangle A=\sphericalangle B=\sphericalangle C=90^{\circ}$. Deoarece $\sphericalangle A+\sphericalangle B=180^{\circ}$, rezultă că $AD\parallel BC$ (1) (unghiuri interne de aceeaşi parte a secantei). Deoarece $\sphericalangle B+\sphericalangle C=180^{\circ}$, rezultă că $AB\parallel CD$ (2) (unghiuri interne de aceeaşi parte a secantei). Din (1) şi (2) decurge că $ABCD$ paralelogram. Deoarece are un unghi drepte, rezultă că $ABCD$ dreptunghi.

11. $CD\parallel AB\parallel BE$ şi $CE\parallel DB$, deci $BECD$ paralelogram. În plus, $CD=AB=BD$, deci $BECD$ romb $\Rightarrow DE\perp BC$.

14. Comparăm $\triangle AMN$, $\triangle MBP$, $\triangle NPC$, $\triangle PNM$: $\left.\begin{array}{l}AM=BM=NP=PN\\MN=BP=PC=NM\\AN=MP=NC=PM\end{array}\right\}\xRightarrow[]{LLL} \triangle AMN\equiv \triangle MBP\equiv \triangle NPC\equiv \triangle PNM\Rightarrow $ $\large\mathcal{A}_{\triangle AMN}=\large\mathcal{A}_{\triangle MBP}=\large\mathcal{A}_{\triangle NPC}=\large\mathcal{A}_{\triangle PNM}$. Dar $\large\mathcal{A}_{\triangle AMN}+\large\mathcal{A}_{\triangle MBP}+\large\mathcal{A}_{\triangle NPC}+\large\mathcal{A}_{\triangle PNM}=\large\mathcal{A}_{\triangle ABC}\Rightarrow \large\mathcal{A}_{\triangle AMN}=\large\mathcal{A}_{\triangle MBP}=\large\mathcal{A}_{\triangle NPC}=\large\mathcal{A}_{\triangle PNM}=\cfrac{1}{4} \large\mathcal{A}_{\triangle ABC}$.

Exersezi și progresezi

1. a) $ABCD$ dreptunghi; b) $ABCD$ romb; c) $ABCD$ romb.

2. a) linie mijlocie; b) $90^{\circ}$.

4. a) 28 cm; b) 44 cm.

5. $4,48$ kg, prin rotunjire $5$ kg.

6. a) 25 m$^2$; b) 50 dm$^2$; c) $22,5$ cm$^2$.

8. a) Dacă $A$, $B$, $C$ coliniare atunci $\sphericalangle CAF=\sphericalangle AFD=\sphericalangle FDC=\sphericalangle DCA=90^{\circ}\Rightarrow ACDF$ dreptunghi. b) i) $\sphericalangle EBA=\sphericalangle EBC\Rightarrow 180^{\circ}-\sphericalangle EBA=180^{\circ}-\sphericalangle EBC\Rightarrow \sphericalangle BEF=\sphericalangle BED$. (1) Comparăm $\triangle BEF$, $\triangle BED$: $\left.\begin{array}{l}(1)\\FE=ED\\BE=BE\text{ (latură comună)}\end{array}\right\} \xRightarrow[]{LUL} \triangle BEF\equiv \triangle BED\Rightarrow BF=BD\Rightarrow \triangle BDF$ isoscel. ii) $\triangle BEF\equiv \triangle BED \Rightarrow \sphericalangle EBF=\sphericalangle EBD$. Dar $\sphericalangle EBA=\sphericalangle EBC\Rightarrow \sphericalangle FBA=\sphericalangle DBC$. (2) Comparăm $\triangle BEF$, $\triangle BED$ dreptunghice:$\left.\begin{array}{l}BF=BD\\(2)\end{array}\right\} \xRightarrow[]{IU} \triangle ABF\equiv \triangle BCD\Rightarrow$ $AF=CD$. iii) Comparăm $\triangle AEF$, $\triangle CED$ dreptunghice: $\left.\begin{array}{l}EF=ED\\AF=CD\end{array}\right\}\xRightarrow[]{CC} \triangle AEF\equiv \triangle CED\Rightarrow$ $AE=CE\Rightarrow \triangle AEC$ isoscel. iv) $\triangle ABF\equiv \triangle BCD\Rightarrow AB=BC\Rightarrow \triangle ABC$ isoscel. $\sphericalangle EBA=\sphericalangle EBC\Rightarrow EB$ bisectoare $\Rightarrow EB\perp AC$. $\triangle BDF$ isoscel, $\sphericalangle EBF=\sphericalangle EBD\Rightarrow EB$ bisectoare $\Rightarrow EB\perp DF\xRightarrow[]{EB\perp AC} DF\perp AC$.

9. $AM=AB-MB=CD-MB=CD-BP=CD-DR=CD-DS=CS$. $AM\parallel AB\parallel CD\parallel CS$, deci $AMCS$ paralelogram. $RQ=DR=BP=PN$. $RQ\parallel DS \parallel DC\parallel AB\parallel MP\parallel PN$, deci $RQPN$ paralelogram. $SQ=DR=BP=MN$. $SQ\parallel DR \parallel DA\parallel BC\parallel BP\parallel MN$, deci $MNSQ$ paralelogram. Comparăm triunghiurile $DSN$ şi $BMQ$: $\left.\begin{array}{l}SN=QM\, (MQSN\text{ paralelogram})\\DS=DR=BP=MB\\\sphericalangle DSN=\sphericalangle DSQ+\sphericalangle NSQ=90^{\circ}+\sphericalangle NSQ=\\=\sphericalangle BMN+\sphericalangle NSQ=\sphericalangle BMN+\sphericalangle NMQ\end{array}\right\} \xRightarrow[]{LUL} \triangle DSN \equiv\triangle BMQ\Rightarrow DN=BQ$. $DQ=DR\sqrt{2}\,(DSQR\text{ pătrat})= BP\sqrt{2}=BN$ ($BMPN$ pătrat), deci $DQBN$ paralelogram.

Unitatea 7- Cercul

Unghi înscris în cerc. Coarde şi arce în cerc; proprietăţi

1. 1.c, 2.a, 3.b, 4.e, 5.f.

2. $40^{\circ}$.

3. a) unghi exterior; b) $60^{\circ}$; c) $60^{\circ}$.

4.

5. a) $45^{\circ}$; b) 3 cm; c) $45^{\circ}$; d) $40^{\circ}$; e) 6 cm.

6. a) $n=4$, arcele au $90^{\circ}$.

b) $n=5$, arcele au $72^{\circ}$

c) $n=6$, arcele au $60^{\circ}$

d) $n=8$, arcele au $45^{\circ}$

7. $145^{\circ}$

8. $\widehat{AB}=\widehat{BC}=\widehat{CD}=\widehat{DE}=\widehat{EF}=36^{\circ}$, $\widehat{FG}=\widehat{GH}=\widehat{HA}=60^{\circ}$

9. a) Matei câștigă deoarece cel mai scurt drum dintre două puncte este segmentul de dreaptă care le unește, deci Matei parcurge un drum mai scurt. b) Sunt egale, deoarece $\widehat{ABC}=\widehat{BCD}\Rightarrow \widehat{AB}+\widehat{BC}=\widehat{BC}+\widehat{CD}\Rightarrow \widehat{AB}=\widehat{CD}$. La arce egale corespund coarde egale, deci $AB=CD$.

10. a) $\sphericalangle BOC=60^{\circ}$, $\sphericalangle BAC=30^{\circ}$; b) $\sphericalangle BAC=96^{\circ}$; c) $\sphericalangle DFE=28^{\circ}$.

11. unghiurile marcate au $72^{\circ}$

12. $90^{\circ}$

13. $45^{\circ}$

14. a) $35^{\circ}$; c) $40^{\circ}$

15. Două coarde egal depărtate de centru sunt egale, deci $AB=CD$. De asemenea, $d(O,AB)=d(O,CD)$. $A_{\triangle OAB}=\cfrac{1}{2}\cdot d(O,AB)\cdot AB=\cfrac{1}{2}\cdot d(O,CD)\cdot CD={{A}_{\triangle OCD}}$.

Tangente dintr-un punct exterior la un cerc

1. a) exterioară; b) mai mare; c) tangentă; d) egală; e) secantă; f) mai mică.

2. Diametru-negru, secantă-verde, tangentă-albastru, coardă-galben, rază-roșu.
3.

4. Cele două tangente sunt paralele deoarece sunt ambele perpendiculare pe diaemtrul AB. Proprietatea nu se modifică la mărirea razei.

6. 24 cm$^2$

7. $A{{T}_{1}}$ tangentă $\Rightarrow A{{T}_{1}}\perp O{{T}_{1}}$, deci $\sphericalangle AT_1 O=90^{\circ}$. Analog $\sphericalangle AT_2 O=90^{\circ}$. Comparăm triunghiurile dreptunghice $AT_1 O$ și $AT_2O$: $\left.\begin{array}{l} AO\text{ latură comună}\\ T_1O=T_2O\text{ (raze în cerc)}\end{array}\right\}\xRightarrow[]{CI}\triangle AT_1O\equiv \triangle AT_2O\Rightarrow \left\{\begin{array}{l} \sphericalangle {{T}_{1}}AO\equiv \sphericalangle {{T}_{2}}AO\Rightarrow AO\text{ bisectoare} \\ \quad A{{T}_{1}}\equiv A{{T}_{2}}\Rightarrow \triangle A{{T}_{1}}{{T}_{2}}\text{ isoscel} \end{array}\right\} \Rightarrow AO$ mediatoare $\Rightarrow AO\perp T_1T_2$

8. Folosind problema anterioară, $AO\perp {{T}_{1}}{{T}_{2}}$, deci $AO$ mediatoare în triunghiul isoscel $O{{T}_{1}}{{T}_{2}}\left( O{{T}_{1}}=O{{T}_{2}}\text{ ca raze în cerc} \right)$, deci bisectoare. Rezultă că $\sphericalangle BO{{T}_{1}}\equiv \sphericalangle BO{{T}_{2}}\Rightarrow \widehat{B{{T}_{1}}}\equiv \widehat{B{{T}_{2}}}\Rightarrow \left[ B{{T}_{1}} \right]\equiv \left[ B{{T}_{2}} \right]$. $BC$ diametru $\Rightarrow\sphericalangle BT{{C}_{1}}=\sphericalangle B{{T}_{2}}C=90^{\circ} $. Comparăm triunghiurile dreptunghice $B{{T}_{1}}C$ și $BT_2C$: $\left. \begin{array}{l} BC\text{ latură comună}\\ \left[ B{{T}_{1}} \right]\equiv \left[ B{{T}_{2}} \right] \\ \end{array} \right\}\xRightarrow[]{CI}\triangle B{{T}_{1}}C\equiv \triangle B{{T}_{2}}C\Rightarrow \left[ C{{T}_{1}} \right]\equiv \left[ C{{T}_{2}} \right]$.

9. O secantă, o tangentă, o dreaptă exterioară.

10. Pot trasa două tangente printr-un punct exterior unei drepte.

11. a) $90^{\circ}$; b) $45^{\circ}$.

12. $150^{\circ}$.

Poligoane regulate înscrise în cerc - construcție, măsuri de unghiuri

4. a) $108^{\circ}$; b) $120^{\circ}$; c) $135^{\circ}$.

5. a) $36^{\circ}$; b) $60^{\circ}$.

6. Pentagon regulat. Laturile pentagonului sunt coarde care corespund la unghiuri congruente ale stelei care sunt înscrise în cerc. Folosind problema 11b, se obține concluzia.

8. Octogon regulat.

9. a) $n=12$; b) $n=18$; c) $n=24$.

10. a) $n=10$; b) $n=12$; c) $n=20$.

11. a) Nu, un romb care nu e pătrat nu este poligon regulat, deoarece are toate laturile egale fără a avea toate unghiurile egale. b) Da. Fie ${{A}_{1}}{{A}_{2}}\ldots {{A}_{n}}$ un poligon cu toate laturile egale înscris în cerc. Introducem următoarea notație: $n+1=1$. Comparăm triunghiurile $ O{{A}_{i}}{{A}_{i+1}}$, $O{{A}_{j}}{{A}_{j+1}}$: $\left. \begin{array}{l} O{{A}_{i}}=O{{A}_{j}}=O{{A}_{i+1}}=O{{A}_{j+1}}\left( \text{raze în cerc} \right) \\ {{A}_{i}}{{A}_{i+1}}={{A}_{j}}{{A}_{j+1}} \\ \end{array} \right\} \xRightarrow[]{LLL},\triangle O{{A}_{i}}{{A}_{i+1}}\equiv \triangle O{{A}_{j}}{{A}_{j+1}}$ isoscele pentru orice $i$ de la 1 la $n\Rightarrow \sphericalangle O{{A}_{i}}{{A}_{i+1}}= \sphericalangle O{{A}_{j}}{{A}_{j+1}}$ și $\sphericalangle O{{A}_{i+1}}{{A}_{i}}=\sphericalangle O{{A}_{j+1}}{{A}_{j}}$ pentru orice $i$ de la 1 la $n\Rightarrow \sphericalangle O{{A}_{i}}{{A}_{i-1}}= \sphericalangle O{{A}_{j}}{{A}_{j-1}}$. Dar $\sphericalangle {{A}_{i+1}}{{A}_{i}}{{A}_{i-1}}=\sphericalangle O{{A}_{i}}{{A}_{i-1}}+\sphericalangle O{{A}_{i}}{{A}_{i+1}}=\sphericalangle O{{A}_{j}}{{A}_{j-1}}+\sphericalangle O{{A}_{j}}{{A}_{j+1}}=\sphericalangle {{A}_{j+1}}{{A}_{j}}{{A}_{j-1}}$, deci toate unghiurile poligonului sunt regulate, deci poligonul este regulat.

12. Rapoartele vor fi egale cu numărul de aur, deoarece figurile desenate sunt asemenea.

Lungimea cercului și aria discului

2. a) $L=6\pi$ cm, $A=9\pi$ cm$^2$; b) $L=14\pi$ cm, $A=49\pi$ cm$^2$; c) $L=2,4\pi$ m, $A=1,44\pi$ m$^2$; d) $L=4,2\pi$ dm, $A=4,41\pi$ dm$^2$.

3. a) $L=4\pi$ cm, $A=4\pi$ cm$^2$; b) $L=10\pi$ m, $A=25\pi$ cm$^2$; c) $L=9\pi$ cm, $A=20,25\pi$ cm$^2$.

8. Distanța parcursă de Maria $=50\cdot2\cdot2\pi=200\pi$ m. Distanța parcursă de Vlad $= 30\cdot2\cdot3,5\pi=$ $=210\pi$ m, deci Vlad parcurge mai mult cu $10\pi\text{ m}\sim31,4$ m.

9. Suprafață iarbă $\geq$ Suprafață grădină - 8 $\cdot$ Suprafața disc 1m $=(100\pi -8\pi )$ m$^2=92$ m$^2$. Suprafața ierbii este mai mare dacă cel puțin 2 dintre cei 8 copăcei sunt la o distanță de 1 m, deci cercurile din jurul lor sunt secante.

10. $a) A=(19^2\pi -182\pi )$ cm$^2=37\pi$ cm$^2$; b) $A=(122-36\pi)$ cm$^2=36(4-\pi)$ cm; d) $A=(72-9\pi)$ cm$^2$; e) $A=(144-36\pi )$ cm$^2=36(4-\pi )$ cm$^2$.

11. a) 2500; b) 6000; c) 17000.

12. a) $\sim1273$; b) $\sim1591$; c) $\sim1446$.

13. a) diametru $=2 \cdot 315 \cdot \cfrac{80}{100} + 22 \cdot 25,4 = 1062,8$ mm
circumferința $\sim 3338,88$ mm, rotații $\sim 17970,07$

b) diametru $=2 \cdot 215 \cdot \cfrac{65}{100} + 16 \cdot 25,4 = 685,9$ mm
circumferința $\sim 2154,81$ mm, rotații $\sim 27844,57$

a) diametru $=2 \cdot 195 \cdot \cfrac{65}{100} + 15 \cdot 25,4 = 634,5$ mm
circumferința $\sim 1993,34$ mm, rotații $\sim 30100,22$

Recapitulare

1. $\widehat{ABC}=\widehat{DCB}\Leftrightarrow \widehat{AB}+\widehat{BC}=\widehat{DC}+\widehat{CB}\Leftrightarrow \widehat{AB}=\widehat{DC}\Leftrightarrow AB=DC$.

2. $\triangle OCF$ dreptunghic în $C$, $\triangle ECF$ dreptunghic în $E$ $\Rightarrow \sphericalangle COF=90^{\circ} -\sphericalangle CFO=90^{\circ} -\sphericalangle CFE=\sphericalangle ECF$.

3. $ \widehat{AB}=180^{\circ} \Rightarrow \widehat{BC}=70^{\circ} \Rightarrow \widehat{BC}=\widehat{BD}\Rightarrow BC=BD$. $\widehat{AC}=\widehat{AD}\Rightarrow \sphericalangle ABC=\sphericalangle ABD$. $\triangle BCD\text{ isoscel și }BA\text{ bisectoare }\Rightarrow BA\text{ mediatoare }\Rightarrow \text{BA}\perp \text{CD}$.

4. a) $20^{\circ}$; b) $90^{\circ}$; c) $80^{\circ}$.

5. $180^{\circ}$.

6. Fie $\left\{ M \right\}=CD\cap {{O}_{1}}{{O}_{2}}$. Deoarece $CD$ tangentă $\Rightarrow\left\{ \begin{array}{l} CD\perp C{{O}_{1}}\Rightarrow \sphericalangle MC{{O}_{1}}=90^{\circ} \\ CD\perp D{{O}_{2}}\Rightarrow \sphericalangle MD{{O}_{2}}=90^{\circ} \\ \end{array} \right.$

Comparăm triunghiurile dreptunghice $C{{O}_{1}}M$ și $DO_2M$: $\left. \begin{array}{l} C{{O}_{1}}=C{{O}_{2}}\text{ }\left( \text{raze egale} \right) \\ \sphericalangle CM{{O}_{1}}=\sphericalangle DM{{O}_{2}}\text{ (opuse la vârf)} \\ \end{array} \right\} \xRightarrow[]{CU}\triangle C{{O}_{1}}M\equiv \triangle D{{O}_{2}}M\Rightarrow {{O}_{1}}M={{O}_{2}}M\xRightarrow[]{M\in {{O}_{1}}{{O}_{2}}}M$ mijlocul ${{O}_{1}}{{O}_{2}}$, deci $C$, $D$ și mijlocul lui ${{O}_{1}}{{O}_{2}}$ sunt coliniare.

7. a) triunghi echilateral; b) pătrat; c) hexagon regulat; d) decagon regulat.

8. a) $L=16\pi $ cm, $A=64\pi$ cm$^2$; b) $L=6\pi \sqrt{5}\text{ cm, }A=45\pi \text{ c}{{\text{m}}^{2}}$; c) $L=4\pi \sqrt{7}\text{ cm, }A=28\pi \text{ c}{{\text{m}}^{2}}$; d) $L=22\pi \text{ cm, }A=121\pi \text{ c}{{\text{m}}^{2}}$.

9. a)

b)

c)

d)

10. a) $A_{\text{dreptunghi}}=3200$ m$^2$, $A_{\text{semicercuri}}=400\pi$ m$^2$, deci $A_{\text{teren}}=400(8+\pi)$ m$^2$. b) $P_{\text{teren}}=160+40\pi<160+40\cdot 3,15=286$ m, deci suficienți cei 286 m de sârmă.

Exersezi și progresezi

1. centrul cercului $=O$, raze $=OB$, $OD$, $OF$, tangente $=AB$, $AC$, unghiuri la centru $=\sphericalangle BOD$, $\sphericalangle DOF$, $\sphericalangle BOF$, unghi înscris în cerc $=\sphericalangle BDE$.

2. a) dodecagon; b) $150^{\circ}$; c) $30^{\circ}$; d) i) A, ii) A, iii) F, iv) A.

3. a) $160^{\circ}$; b) $144^{\circ}$; c) $140^{\circ}$; d) $120^{\circ}$.

4. a) $L=12\pi$ m, $A=36\pi$ m$^2$; b) $L=20\pi$ cm, $A=100\pi$ cm$^2$; c) $L=7\pi$ dm, $A=12,25\pi$ dm$^2$; d) $L=5\pi$ m, $A=6,25\pi$ m$^2$.

5. 5 cm.

6. $A_{\text{roșie}}=A_{ABCD}-4A_{\text{sector de cerc}}= A_{ABCD}-A_{\text{cerc}}=(162-64\pi)\text{ cm}^2=64(4-\pi)\text{ cm}^2$.

7. Fie latura pătratelor $=4l$, Raza cerc albastru $=2l$, Raza cercuri verzi $=l$, A albastră $=4l2\pi=$ A verde.

8. $\cfrac{\pi}{4} cm.$

10. a) $\pi$ cm; b) $\pi$ cm; c) $\pi$ cm. Toate rezultatele sunt egale.

Unitatea 8 - Teorema lui Thales

Segmente proporționale. Teorema paralelelor echidistante

1. a) $BC=2$ cm, $BD=4$ cm; b) $\cfrac{1}{2}$; c) $\cfrac{1}{2}$.

2. a) 1; b) $\cfrac{1}{2}$; c) $\cfrac{1}{3}$; d) 2; e) 2; f) 3.

3. a) $\cfrac{AB}{CD}=1=\cfrac{MN}{PQ}\Rightarrow \cfrac{AB}{MN}=\cfrac{CD}{PQ}$; b) $\cfrac{AC}{BD}=1=\cfrac{MP}{NQ}\Rightarrow \cfrac{AC}{MP}=\cfrac{BD}{NQ}$;

c) $\cfrac{AB}{BD}=\cfrac{2}{3}=\cfrac{1}{3}+\cfrac{1}{3}=\cfrac{MN}{NP}+\cfrac{PQ}{NP}=\cfrac{MN+PQ}{NP}\Rightarrow \cfrac{AB}{MN+PQ}=\cfrac{BD}{NP}$;

d) $\cfrac{BD}{AD}=\cfrac{3}{5}=\cfrac{NP}{MQ}\Rightarrow \cfrac{BD}{NP}=\cfrac{AD}{MQ}$; f) $\cfrac{AC}{BD}=1=\cfrac{NQ}{MP}\Rightarrow AC\cdot MP=BD\cdot NQ$.

4. a) $AB=BC=CD=3$ cm $\Rightarrow$ paralelele sunt echidistante; b) $AB=BC=CD\Rightarrow$ paralelele sunt echidistante; c) $AB=BC=CD \Rightarrow$ paralelele sunt echidistante.

5. a) Traseele sunt egale; b) Traseele sunt egale.

6. $MN$ linie mijlocie $\Rightarrow MN\parallel AB\parallel CD$. $AM=MD\Rightarrow MN$, $AB$, $CD$ paralele echidistante.

7. a) Cele 8 porţiuni de pe Broadway sunt egale între ele, de asemenea cele de pe bulevardele 6th Ave și 7th Ave. b) Străzile 34thW, 38thW, 42thW paralele echidistante, deci $B_1B_5=B_5B_9$. Dar $NB_5=B_5Y$, deci $NB_1YB_9$ paralelogram. Străzile 35thW, 38thW, 41hW paralele echidistante, deci $B_2B_5=B_5B_8$. Dar $NB_5=B_5Y$, deci $NB_2YB_8$ paralelogram. Străzile 36thW, 38thW, 40thW paralele echidistante, deci $B_3B_5=B_5B_7$. Dar $NB_5=B_5Y$, deci $NB_3YB_7$ paralelogram. Străzile 37thW, 38thW, 39thW paralele echidistante, deci $B_4B_5=B_5B_6$. Dar $NB_5=B_5Y$, deci $NB_4YB_6$ paralelogram. c) Fie punctul $Q$ la intersecţia 7th Ave – W 34th, iar punctul $R$ la intersecția 6th Ave – W42nd. $\sphericalangle SQT=\sphericalangle QTR=\sphericalangle TRS=\sphericalangle RSQ=90^{\circ}$, deci $TRSQ$ dreptunghi, deci $TQ=RS$. Străzile 34thW, 38thW,42thW paralele echidistante, deci $TN=NQ=\cfrac{TQ}{2}=\cfrac{RS}{2}=SY=YR$. Comparăm $\triangle B_9TN$ și $\triangle B_1YS$ dreptunghice: $\left.\begin{array}{l}NB_1 YB_9\text{ paralelogram, deci }NB_9=B_1 Y\\TN=YS\end{array}\right\}\xRightarrow[]{CI} \triangle B_9 TN\equiv \triangle B_1 YS\Rightarrow TB_9=B_1 S$ $A_{NTB_9 B_5}=\cfrac{(TB_9+NB_5 )\cdot TN}{2}=\cfrac{(SB_1+YB_5 )\cdot YS}{2}=A_{YSB_1 B_5}$.

Teorema lui Thales. Împărțirea unui segment în părți proporționale cu numere date

1. 10 m. 2. a) $AM=32$ mm, $BN=20$ mm, $BP=40$ mm.

b) $AN=40$mm, $AM=8$ mm, $BN=50$ mm.

c) $CP=9,6$ mm, $CM=14,4$ mm, $BN=36$ mm.

d) $CP=6$ mm, $BP=15$ mm, $BN=45$ mm.

e) $AN=28$ mm, $CP=14$ mm, $CN=10$ mm.

f) $AM=8$ mm, $AN=40$ mm, $CM=4$ mm.

3. a)

b)

4. a) $AM=6$ mm, $BM=18$ mm, $AN=8$ mm, $CN=24$ mm; b) $AM=12$ mm, $BM=36$ mm, $AN=16$ mm, $CN=48$mm.

5. a) $\triangle OAB$, $CD\parallel AB$, $C\in OA$, $D\in OB$ $\xRightarrow[]{\text{Thales}}\cfrac{OC}{CA}=\cfrac{OD}{DB}\Leftrightarrow \cfrac{OC}{CA-OC}=\cfrac{OD}{DB-OD}\Leftrightarrow \cfrac{OC}{OA}=\cfrac{OD}{OB}\Leftrightarrow OA\cdot OD=OC\cdot OC$.

b) $2x\cdot 6=(x-4)\cdot 16\Rightarrow 12x=16x-64\Rightarrow 4x=64\Rightarrow x=16$.

6. $\triangle VAB$, $DC\parallel AB$, $D\in VA$, $C\in VB\xRightarrow[]{\text{Thales}} \cfrac{VD}{DA}=\cfrac{VC}{CB}\Leftrightarrow \cfrac{VD}{DA+VD}=\cfrac{VC}{CB+VC}\Leftrightarrow \cfrac{VD}{VA}=\cfrac{VC}{VB}$.
9. a)

b)

10. a) Fie $x=\sphericalangle BAC$. $\sphericalangle BAE=180^{\circ}-x$. $BE\parallel AD\Rightarrow \sphericalangle DAB=\sphericalangle EBA$ (unghiuri alterne interne) $\Rightarrow \cfrac{x}{2}=\sphericalangle EBA$.

$\sphericalangle AEB=180^{\circ}-\sphericalangle BAE-\sphericalangle EBA=180^{\circ}-(180^{\circ}-x)-\cfrac{x}{2}=x-\cfrac{x}{2}=\cfrac{x}{2}\Rightarrow \sphericalangle AEB=\sphericalangle EBA\Rightarrow \triangle ABE$ isoscel. b) $\triangle EBC$, $DA\parallel EB$, $D\in CB$, $A\in CE\xRightarrow[]{\text{Thales}} \cfrac{AE}{AC}=\cfrac{BD}{DC}\xRightarrow[AE=AB]{\triangle ABE\text{ isoscel}} \cfrac{AB}{AC}=\cfrac{BD}{DC}$.

12. a) $\triangle ABC$, $RQ\parallel BC$, $R\in AB$, $Q\in AC\xRightarrow[]{Thales} \cfrac{AR}{BR}=\cfrac{AQ}{QC} \xRightarrow[]{\cfrac{AQ}{QC}=\cfrac{BP}{AP}} \cfrac{AR}{BR}=\cfrac{BP}{AP}\Leftrightarrow\cfrac{AP+PR}{BR}=\cfrac{BR+RP}{AP}\Rightarrow AP^2+PR\cdot AP=BR^2+BR\cdot RP\Rightarrow AP^2-BR^2+PR\cdot AP-BR\cdot RP = 0\Rightarrow (AP-BR)(AP+BR+RP)=0\Rightarrow(AP-BR)\cdot AB=0\Rightarrow AP=BR$.

b) $M$ mijlocul lui $AB\Rightarrow AM=MB\Rightarrow AP+PM=MR+RB\xRightarrow[]{AP=RB} PM=MR$.

$MN$ linie mijlocie în $\triangle ABC\Rightarrow MN\parallel BC\xRightarrow[]{RQ\parallel BC} MN\parallel RQ\Rightarrow MS\parallel RQ\xRightarrow[]{M\text{ mijlocul lui }PR} S$ mijlocul lui $PQ$.

13. Fie $D'\in (EF)$, $ED'=D' F\Rightarrow G\in AD'$, $\cfrac{DG}{GD'}=2$.

$\triangle D' DF$, $GQ\parallel EF$, $G\in DD' \xRightarrow[]{\text{Thales}} \cfrac{FQ}{QD}=\cfrac{D'G}{GD}=\cfrac{1}{2}$.

Celelalte rapoarte se calculează analog.

Reciproca teoremei lui Thales

1. a) Nu. b) Cei doi nu au realizat că au ajuns la o contadicţie, deoarece nu există două puncte în interiorul segmentului $AC$ care împart segmentul în același raport.

2. a) $NA=MN-MA=25$ mm, $BP=NP-BN=6$ mm, $CM=MP-CM=20$ mm

$\cfrac{NA}{AM}=\cfrac{NB}{BP}\xRightarrow[]{\text{reciproca Thales}} AB\parallel MP$, $\cfrac{PC}{CM}=\cfrac{PB}{BN}\xRightarrow[]{\text{reciproca Thales}} BC\parallel MN$

b) $NA=MN-MA=25$ mm, $BP=NP-BN=6$ mm, $CM=MP-CM=4$ mm $\cfrac{NA}{AM}=\cfrac{PC}{CM}\xRightarrow[]{\text{reciproca Thales}} AC\parallel NP$, $\cfrac{MA}{AN}=\cfrac{PB}{BN}\xRightarrow[]{\text{reciproca Thales}} AB\parallel MP$

c) $NA=MN-MA=25$ mm, $BP=NP-BN=30$ mm, $CM=MP-CM=4$ mm $\cfrac{NA}{AM}=\cfrac{PC}{CM}\xRightarrow[]{\text{reciproca Thales}} AC\parallel NP$, $\cfrac{PC}{CM}=\cfrac{PB}{BN}\xRightarrow[]{\text{reciproca Thales}} BC\parallel MN$

3. a) $AM=\cfrac{AD}{2}\Rightarrow M$ mijlocul lui $AD$, $CN=\cfrac{BC}{2}\Rightarrow N$ mijlocul lui $BC$, deci $MN$ linie mijlocie în trapezul $ABCD\Rightarrow MN\parallel AB$.

b) $PM=DP-DM=DP-(DA-AM)=6=\cfrac{AM}{2}\Rightarrow P$ mijlocul lui $MA$, $BN=BC-CN=8$ cm $\Rightarrow BQ=\cfrac{BN}{2}\Rightarrow Q$ mijlocul lui $BN$, deci $PQ$ linie mijlocie în trapezul $MNBA\Rightarrow PQ\parallel AB$.

c) $MN$ linie mijlocie în trapezul $ABCD\Rightarrow MN=\cfrac{AB + CD}{2}=22,5$ cm.

d) $PQ$ linie mijlocie în trapezul $MNPQ\Rightarrow PQ=\cfrac{MN + AB}{2}=23,75$ cm.

e) i) A; ii) A; iii) A; iv) A.

f) $RS$ linie mijlocie în trapezul $MNCD\Rightarrow PQ=\cfrac{MN+CD}{2}=\cfrac{\cfrac{AB+CD}{2}+CD}{2}=\cfrac{AB+3CD}{4}=\cfrac{AB\cdot1+CD\cdot3}{1+3}$.

4. $AP=CA-CP=CN+NA-CP$ Fie $O$ mijlocul lui $AC$, deci $OA=OC$ și $ON=OP$. Există două situaţii care se tratează similar:

1. ordinea punctelor pe $(AC)$ este $A$, $P$, $O$, $N$, $C$.

$NA-CP=NO+OA-CO-OP=0$.

2. ordinea punctelor pe $(AC)$ este $A$, $N$, $O$, $P$, $C$.

$NA-CP=OA-ON-CO+OP=0$.

Deci, în ambele cazuri, $AP=CN=AM\xRightarrow[]{AB=AC} \cfrac{AM}{AB}=\cfrac{AP}{AC}\Rightarrow \cfrac{AM}{MB}=\cfrac{AP}{PC} \xRightarrow[]{\text{reciproca Thales}} MP\parallel BC$.

5. $MB=AB-AM=12$ cm, $AN=AC-CN=4$ cm, $PC=BC-BP=20$ cm, $CQ=BC-BQ=5$ cm.

$\cfrac{AM}{MB}=\cfrac{AN}{NC}\xRightarrow[]{\text{reciproca Thales}} MN\parallel BC$. (1)

$\cfrac{MB}{MA}=\cfrac{BQ}{QC}\xRightarrow[]{\text{reciproca Thales}}MQ\parallel BC$. (2)

$\cfrac{CN}{NA}=\cfrac{CP}{PB}\xRightarrow[]{\text{reciproca Thales}} NP \parallel B$. (3)

$MN\parallel BC\parallel PQ\Rightarrow MN\parallel PQ\xRightarrow[]{\text{Thales}}\cfrac{ON}{NP}=\cfrac{OM}{MQ}\Rightarrow \cfrac{ON}{OP}=\cfrac{OM}{OQ}\Rightarrow ON\cdot OQ=OM\cdot OP$.

$PQ=BC-CQ-BP=BQ-BP=15$ cm

$(1)\Rightarrow MN\parallel BP$, $(3)\Rightarrow NP\parallel BM\Rightarrow MNPB$ paralelogram $\Rightarrow BP=MN$

$(1)\Rightarrow MN\parallel CQ$, $(2)\Rightarrow MQ\parallel CN\Rightarrow MNCQ$ paralelogram $\Rightarrow CQ=MN$, deci $PQ=BC-2\cdot MN$.

Dacă $AM=7,5$ cm și $AN=10$ cm, atunci $PQ=0$ cm.

În acest caz, $MN$ este linie mijlocie. Cum $MN=BP=CQ\Rightarrow P=O=Q=$ mijlocul lui $BC$.

Recapitulare

1. a) $U\in(ST)\Rightarrow US=18$ cm, $UT=12$cm, $U\notin(ST)\Rightarrow US=90$ cm, $UT=60$ cm.

2. a) $\cfrac{1}{2}$; b) $1$; c) $3$; d) $1$; e) $2$; f) $\cfrac{1}{3}$.

3. $AE=DE-DA=25$ cm, $PF=DF-DP=15$ cm, $\cfrac{DA}{AE}=\cfrac{DP}{PF}\xRightarrow[]{\text{reciproca Thales}}AP\parallel EF$

$MD=DE-ME=25$ cm, $BF=EF-EB=35$ cm, $\cfrac{ME}{MD}=\cfrac{BE}{BF}\xRightarrow[]{\text{reciproca Thales}}MB\parallel DF$

$NE=FE-FN=35$ cm, $DC=DF-CF=15$ cm, $\cfrac{FN}{NE}=\cfrac{FC}{CD}\xRightarrow[]{\text{reciproca Thales}}CN\parallel DE$

4. a)

b)

d)

e)

5. a) În $\triangle MAN$, $BC\parallel AN$, $B\in (AM)$, $C\in(NM)\xRightarrow[]{\textit{Thales}} \cfrac{MB}{BA}=\cfrac{MC}{CN}$. (1)

b) În $\triangle MAN$, $CD\parallel MA$, $D\in(AN)$, $C\in(NM)\xRightarrow[]{\textit{Thales}} \cfrac{ND}{DA}=\cfrac{NC}{CM}$. (2)

c) $(1)$, $(2)\Rightarrow \cfrac{MB}{BA}=\cfrac{MC}{CN}=\cfrac{DA}{ND}\Rightarrow \cfrac{MB}{BA}\cdot\cfrac{ND}{DA}=1$.

d) $(1)\Rightarrow \cfrac{MB}{BA+MB}=\cfrac{MC}{CM+NC}\Rightarrow \cfrac{MB}{AM}=\cfrac{MC}{MN}$. (3)

e) $(2)\Rightarrow \cfrac{ND}{ND+DA}=\cfrac{NC}{CM+NC}\Rightarrow \cfrac{ND}{AN}=\cfrac{NC}{MN}$. (4)

f) $(3)$, $(4)\Rightarrow \cfrac{MB}{AM}+\cfrac{ND}{AN}=\cfrac{MC}{MN}+\cfrac{NC}{MN}=\cfrac{MN}{MN}=1$.

7. $\cfrac{BN}{BP}=2,5=\cfrac{5}{2}\Rightarrow \cfrac{BP}{BN}=\cfrac{2}{5}=\cfrac{AM}{AN}\xRightarrow[]{\text{reciproca Thales}} AB\parallel MP$; b) 2,5; c) $\cfrac{5}{9}$.

8. a) $NP\parallel AB$, $PA\parallel NB\Rightarrow PABN$ paralelogram $\Rightarrow PA=BN\xRightarrow[]{AD=BC} PD=CN$. $\left.\begin{array}{l}PQ\parallel AC\xRightarrow[]{\text{Thales}} \cfrac{QD}{QC}= \cfrac{DP}{PA}\\MN\parallel AC\xRightarrow[]{\text{Thales}} \cfrac{BN}{NC}=\cfrac{BM}{MA} \end{array}\right\}\Rightarrow \cfrac{QD}{QC}=\cfrac{AM}{MB}\Rightarrow \cfrac{QD}{DC}=\cfrac{AM}{AB}\xRightarrow[]{AB=CD}$ $ QD=AM\xRightarrow[]{QD\parallel AM} QDAM$ paralelogram $\Rightarrow QM\parallel AD$.

b) $\left.\begin{array}{l}\text{Fie }\{S'\}=BD\cap NP\, (S'\in(BD))\xRightarrow[]{NP\parallel AB} \cfrac{DP}{PA}=\cfrac{DS'}{S'B} \\ \text{Fie }{S'' }=BD\cap MQ\,(S''\in(BD))\xRightarrow[]{MQ\parallel AD\parallel BC} \cfrac{DQ}{QC}=\cfrac{DS''}{S''B}\end{array}\right\}\Rightarrow \cfrac{DS'}{S'B}=\cfrac{DS''}{S''B}\Rightarrow S'=S''=S$, deci $S\in(BD)$.

9. $\left.\begin{array}{l}QN\perp CD\perp AD\Rightarrow QN\parallel AD\xRightarrow[]{\text{Thales}} \cfrac{QA}{QC}=\cfrac{ND}{NC}\Rightarrow \cfrac{QA}{AC}=\cfrac{DN}{DC}\\MP\perp AB\perp BC\Rightarrow MP\parallel BC\xRightarrow[]{\text{Thales}} \cfrac{CP}{PA}=\cfrac{BM}{MA} \Rightarrow \cfrac{CP}{CA}=\cfrac{BM}{BA} \end{array}\right\}\Rightarrow$ $\xRightarrow[]{DN=BM,\,CD=BA} AQ=CP$. Fie $O$ mijlocul lui $AC\Rightarrow AO=OC\Rightarrow AQ+QO=CP+PO\Rightarrow QO=OP\Rightarrow$ punctul $P$ este simetricul punctului $Q$ față de mijlocul segmentului $AC$.

Exersezi și progresezi

1. Fie $\{V\}=AD\cap BC$. $\left.\begin{array}{l}CD\parallel AB\xRightarrow[]{\text{Thales}} \cfrac{VD}{DA}=\cfrac{VC}{CB} \Rightarrow \cfrac{VA}{DA}=\cfrac{VB}{CB}(1) \\ MN\parallel AB\xRightarrow[]{\text{Thales}} \cfrac{VM}{MA}=\cfrac{VN}{NB}\Rightarrow \cfrac{VA}{MA}=\cfrac{VB}{NB} \end{array}\right\}\Rightarrow \cfrac{MA}{DA}=\cfrac{NB}{CB}$ (2)

$\left.\begin{array}{r}\text{punctele }M\text{ și }Q\text{ de pe latura }AD\text{ sunt simetrice față de mijlocul laturii }AD\\ \text{punctele }N\text{ și }P\text{ de pe latura }BC\text{ sunt simetrice față de mijlocul laturii }BC\\ MA=QD,\,NB=CP\end{array}\right\}\xRightarrow[]{(2)}$ $\cfrac{QA}{DA}=\cfrac{PB}{CB}\xRightarrow[]{(1)} \cfrac{VA}{QA}=\cfrac{VB}{PB}\Rightarrow \cfrac{VQ}{QA}=\cfrac{VP}{PB}\xRightarrow[]{\text{reciproca Thales}}PQ\parallel AB$.

b) Fie $X$ mijlocul lui $AD$, $Y$ mijlocul lui $BC$. $QX=XM$, $NY=YP$ (punctele $M$ și $Q$ de pe latura $AD$ sunt simetrice față de mijlocul laturii $AD$, iar punctele $N$ și $P$ de pe latura $BC$ sunt simetrice față de mijlocul laturii $BC$), deci cele două linii mijlocii coincid.

2. a) $\cfrac{AM}{MB}=\cfrac{CP}{DP}\Rightarrow \cfrac{AM}{AB}=\cfrac{CP}{CD}\xRightarrow[]{AB=CD} AM=CP$

$\cfrac{BN}{CN}=\cfrac{DQ}{AQ}\Rightarrow \cfrac{BN}{BC}=\cfrac{DQ}{DA}\xRightarrow[]{BC=DA} BN=DQ$

b) $\left.\begin{array}{l}AM=CP\\BN=DQ\Rightarrow AQ=CN\\\sphericalangle A=\sphericalangle C\end{array}\right\}\xRightarrow[]{LUL} \triangle MAQ\equiv \triangle PNC\Rightarrow MQ=PN$.

Analog, $MN=PQ\Rightarrow MNPQ$ paralelogram.

3. $\left.\begin{array}{l}MN\parallel CA\xRightarrow[]{\text{Thales}} \cfrac{AM}{MB}=\cfrac{CN}{NB}\rule[-5mm]{0mm}{2mm}\\ NP\parallel AB\xRightarrow[]{\text{Thales}} \cfrac{BN}{NC}=\cfrac{AP}{PC}\rule[-5mm]{0mm}{2mm}\\ PM\parallel BC\xRightarrow[]{\text{Thales}} \cfrac{CP}{PA}=\cfrac{BM}{MA}\end{array}\right\}\Rightarrow \cfrac{AM}{MB}=\cfrac{MB}{AM}\Rightarrow \cfrac{AM}{AB}=\cfrac{MB}{AB} \Rightarrow AM=BM\Rightarrow$

$M$ mijlocul lui $AB\Rightarrow MN$, $MP$ linii mijlocii $\Rightarrow P$ mijlocul lui $AC$, $N$ mijlocul lui $BC$.

4. $M$ mijlocul lui $AB$.

5. a) $MQ\perp BC\perp AB\Rightarrow MQ\parallel BC\xRightarrow[]{\text{Thales}} \cfrac{AM}{MB}=\cfrac{AQ}{QC}$ (1)

$NQ\perp CD\perp AD\Rightarrow NQ\parallel AD\xRightarrow[]{\text{Thales}} \cfrac{CN}{ND}=\cfrac{CQ}{QA}\xRightarrow[]{(1)} \cfrac{DN}{CN}=k$

$QS\perp BC\perp AB\Rightarrow QS\parallel AB\xRightarrow[]{\text{Thales}} \cfrac{CS}{SB}=\cfrac{CQ}{QA} \xRightarrow[]{(1)} \cfrac{BS}{CS}=k$

$QT\perp AD\perp CD\Rightarrow QT\parallel CD\xRightarrow[]{\text{Thales}} \cfrac{DT}{TA}=\cfrac{CQ}{QA}\xRightarrow[]{(1)} \cfrac{AT}{DT}=k$.

b) $A_{AMQT}=\left(\cfrac{k}{1+k}\right)^2\cdot L\cdot l$, $A_{MBSQ}=\cfrac{k}{(1+k)^2}\cdot L\cdot l=A_{DNQT}=\cfrac{k}{(1+k)^2}\cdot L\cdot l$, $A_{SCNQ}=\cfrac{1}{(1+k)^2}\cdot L\cdot l$.

6. Consideră $P$ simetricul lui $A$ față de $E$, $F$ simetricul lui $A$ față de $D$. $AD=DC$, $BD=DF\Rightarrow ABFC$ paralelogram.

$\cfrac{AE}{ED}=\cfrac{AE}{DA-EA}=\cfrac{2AE}{2(DA-EA)}=\cfrac{AP}{2DA-2EA}=\cfrac{AP}{FA-PA}=\cfrac{AP}{PF}$.

$MN\parallel BC\xRightarrow[]{E\in MN,\,D\in BC} ME\parallel BD\xRightarrow[]{\text{Thales}} \cfrac{AM}{MB}=\cfrac{AE}{ED}=\cfrac{AP}{PF} \xRightarrow[]{\text{reciproca Thales}} MP\parallel BF\Rightarrow MP\parallel AN$

$MN\parallel BC\xRightarrow[]{E\in MN,\,D\in BC} NE\parallel CD\xRightarrow[]{\text{Thales}} \cfrac{AN}{NC}=\cfrac{AE}{ED}=\cfrac{AP}{PF} \xRightarrow[]{\text{reciproca Thales}} NP\parallel CF\Rightarrow NP\parallel AM$

Deci $AMNP$ paralelogram, rezultă $ME=EN$.

Unitatea 9 - Asemănarea triunghiurilor

Triunghiuri asemenea; teorema fundamentală a asemănării

1. a) 4 cm; b) $90^{\circ}$; c) șapte ori.

2. d) $\sphericalangle ADE\equiv \sphericalangle ABC$ (unghiuri corespondente pentru $DE\parallel BC$ cu secanta $BD$);

e) $\sphericalangle AED\equiv\sphericalangle ACB$ (unghiuri corespondente pentru $DE\parallel BC$ cu secanta $CD$);

f) $DE\parallel BC\xRightarrow[]{\text{Thales}} \cfrac{AD}{DB}=\cfrac{AE}{EC}\Rightarrow \cfrac{AD}{AD+DB}=\cfrac{AE}{AE+EC}\Rightarrow \cfrac{AD}{AB}=\cfrac{AE}{AC}$.

3. a) $MN$; b) $BC$; c) $60^{\circ}$; d) $50^{\circ}$; e) $\cfrac{AB}{MN}=\cfrac{BC}{NP}=\cfrac{CA}{MP}$.

5. $AB=9$ cm, $BC=15$ cm, $CA=21$ cm.

6. $\sphericalangle A=\sphericalangle M=50^{\circ}$, $\sphericalangle B=\sphericalangle N=60^{\circ}$, $\sphericalangle C=\sphericalangle P=70^{\circ}$.

7. a) 36 cm, b) 6 cm.

8. a) Da, deoarece relaţia de asemănare este tranzitivă.

b) $\cfrac{AB}{DE}=3$, $\cfrac{DF}{AC}=\cfrac{1}{3}$, $\cfrac{EF}{NP}=2$, $\cfrac{MN}{DE}=\cfrac{1}{2}$, $\cfrac{AB}{MN}=6$, $\cfrac{NP}{BC}=\cfrac{1}{6}$.

9. a) A; b) F, s-a amestecat teoerma lui Thales cu teorema fundamentală a asemănării: $\cfrac{AE}{AC}=\cfrac{DE}{BC}$; c) A; d) A; e) F; f) F rapoartele sunt inversate.

10. a) $17,5$ cm; b) $5,625$ cm; c) 8 cm.

11. a) $BC=12$ cm, $AC=12$ cm.

Criterii de asemănare a triunghiurilor

2. $\sphericalangle C=\sphericalangle X$, $\sphericalangle B=\sphericalangle Y\xRightarrow[]{UU} \triangle ABC\sim \triangle ZYX$.

$\cfrac{MN}{TR}=\cfrac{NP}{RS}=\cfrac{PM}{ST}\xRightarrow[]{LLL} \triangle MNP\sim \triangle TRS$

$\cfrac{GI}{DF}=\cfrac{GH}{DE}$ și $\sphericalangle G=\sphericalangle D$, deci $\triangle IGH\sim\triangle FDE$.

4. a) $\left.\begin{array}{l}\cfrac{AO}{BO}=\cfrac{OC}{OD}\\\sphericalangle AOC=\sphericalangle BOD\end{array}\right\}\xRightarrow[]{LUL} \triangle AOC\sim\triangle BOD\Rightarrow\left\{\begin{array}{l}\sphericalangle OAC=\sphericalangle OBD\\\sphericalangle AOC=\sphericalangle BOD\\\sphericalangle OCA=\sphericalangle ODB\\AC\parallel BD\end{array}\right.$

b) $\left.\begin{array}{l}\cfrac{AO}{DO}=\cfrac{OC}{OB}\\\sphericalangle AOC=\sphericalangle DOB\end{array}\right\}\xRightarrow[]{LUL} \triangle AOC\sim\triangle DOB\Rightarrow\left\{\begin{array}{l}\sphericalangle OAC=\sphericalangle OBD\\\sphericalangle AOC=\sphericalangle BOD\\\sphericalangle OCA=\sphericalangle ODB\end{array}\right.$

5. b) $AB=10$cm, $AD=7,2$ cm.

6. a) Da, triunghiul este asemenea cu cel din realitate, deoarece laturile sunt proporţionale. b) Laturile ţarcului sunt: 30 m, 25 m, 27,5 m.

7. a) $\left.\begin{array}{l}\sphericalangle BAC=\sphericalangle BDA\,\,(=90^{\circ})\\\sphericalangle ABC=\sphericalangle ABD\text{ (unghi comun)}\end{array}\right\}\xRightarrow[]{UU} \triangle ABC\sim \triangle DBA$

b) $\triangle ABC\sim \triangle DBA\Rightarrow \cfrac{AB}{DB}=\cfrac{BC}{BA}\Rightarrow AB^2=BD\cdot BC$.

8. a) $DE=DF\Rightarrow\sphericalangle DEF=\sphericalangle DFE=\cfrac{180^{\circ}-\sphericalangle EDF}{2}=72^{\circ}$. $EN$ bisectoare $\Rightarrow\sphericalangle NEF=\cfrac{72^{\circ}}{2}=36^{\circ}$.

$\left.\begin{array}{l}\sphericalangle EDF=\sphericalangle NEF\\\sphericalangle DEF=\sphericalangle NFE\end{array}\right\} \xRightarrow[]{UU} \triangle DEF\sim\triangle ENF$

10. Folosim indicaţia din problemă. $P-C-M$ coliniare $PM\parallel AB\xRightarrow[]{TFA} \triangle CNP\sim \triangle BNA\Rightarrow \cfrac{CN}{NB}=\cfrac{CP}{AB}=\cfrac{CM}{CD}=\cfrac{1}{2}\Rightarrow \cfrac{CN}{CB}=\cfrac{1}{3}\Rightarrow N$ este situat la 10 m de $C$.

Raportul ariilor a două triunghiuri asemenea; aproximarea în situații practice a distanţelor folosind asemănarea

8. a) $\cfrac{A_{\triangle ABC}}{A_{\triangle DEF}}=\cfrac{1}{4}$; b) $\cfrac{A_{\triangle ABC}}{A_{\triangle DEF}} =\cfrac{4}{9}$; c) $\cfrac{A_{\triangle ABC}}{A_{\triangle DEF}} =\cfrac{16}{25}$.

9. 9.

10. a) $\cfrac{1}{2}$; b) $\cfrac{2}{3}$; c) 4.

12. 15.

13. 16,875 cm$^2$.

Exersezi şi progresezi

1. a) congruente; b) proporţionale; c) $\cfrac{4}{9}$.

2. a) $\triangle ABC\sim \triangle FED$, $\sphericalangle A=\sphericalangle F$, $\sphericalangle B=\sphericalangle E$, $\sphericalangle C=\sphericalangle D$.

4. a) $NC=6$ cm, $MN=\cfrac{14}{3}$ cm; b) $MC=14$ cm, $NB=6$ cm.

Unitatea 10 - Relații metrice în triunghiul dreptunghic

Proiecţii ortogonale pe o dreaptă. Teorema înălţimii. Teorema catetei

4. c) În $\triangle ABC$ și $\triangle DAC$: $\left. \begin{array}{l} \sphericalangle BAC\equiv \sphericalangle ADC\,\,\left( =90^{\circ} \right) \\ \sphericalangle ACB\equiv \sphericalangle ACD\text{ (unghi comun)} \\ \end{array} \right\}\Rightarrow \triangle ABC\sim \triangle DAC$.

d) Din $\triangle ABC\sim \triangle DAC\Rightarrow \cfrac{AB}{DA}=\cfrac{BC}{AC}=\cfrac{AC}{DC}$ De aici $\cfrac{BC}{AC}=\cfrac{AC}{DC}\Rightarrow A{{C}^{2}}=BC\cdot CD$.

5. f) Din $\triangle DBA\sim \triangle DAC\Rightarrow \cfrac{DB}{DA}=\cfrac{BA}{AC}=\cfrac{DA}{DC}$. De aici $\cfrac{DB}{DA}=\cfrac{DA}{DC}\Rightarrow D{{A}^{2}}=DB\cdot DC$.

6. a) $p{{r}_{BC}}A=D$, $p{{r}_{CA}}A=A$; b) $p{{r}_{AD}}C=D$; c) $p{{r}_{BC}}AB=AD$, $p{{r}_{AD}}AC=AD$.

7.

$p{{r}_{MP}}N={{N}{'}}$, $p{{r}_{NP}}MP={{M}{'}}P$.

8. 4 cm.

9. a) $p{{r}_{MP}}N=A$; b) $p{{r}_{NP}}N=N$; c) $p{{r}_{MP}}NM=AM$, $p{{r}_{NP}}NM=AP$.

10.

11. a) $R=5$ cm; b) $BD=32$ cm.

Teorema lui Pitagora. Reciproca teoremei lui Pitagora

2. a) $BC=15$ cm; b) $AC=10$ cm; c) $AB=a$.

3. a) $\triangle ABC$ dreptunghic în $A$; b) $\triangle ABC$ dreptunghic în $C$; c) $\triangle ABC$ nu e dreptunghic; d) $\triangle ABC$ dreptunghic în $B$.

4. a) $\cfrac{5\sqrt{3}}{2}$ cm; b) $\cfrac{7\sqrt{3}}{2}$ cm.

5. a) $5\sqrt{2}$ cm; b) $7\sqrt{2}$ cm.

6. 60 cm$^2$

7. 18 cm$^2$

9. a) dreptunghic; b) 15 cm$^2$.

10. a) isoscel; b) 90 cm$^2$.

11. $16\sqrt{7}$ cm$^2$.

Noţiuni de trigonometrie în triunghiul dreptunghic: sinusul, cosinusul, tangenta şi cotangenta unui unghi ascuţit

1. $\triangle OAM$, $\triangle OBN$: $\left. \begin{array}{l} \sphericalangle OAM\equiv \sphericalangle OBN\,\,(=90^{\circ}) \\ \sphericalangle AOM\equiv \sphericalangle BON\text{ (unghi comun)}\\ \end{array} \right\}\Rightarrow \triangle OAM\sim \triangle OBN$

4. 1-c, 2-g, 3-a, 4-h, 5-b, 6-d, 7-e, 8-f.

5. a) 1; 1, b) 1; 1, c) 1; 1.

6. ${{\sin }^{2}}x+{{\cos }^{2}}x=\cfrac{A{{B}^{2}}+A{{C}^{2}}}{B{{C}^{2}}} = \cfrac{B{{C}^{2}}}{B{{C}^{2}}}=1$ (teorema lui Pitagora)

$\text{tg }x\cdot \text{ctg }x=\cfrac{AC}{AB}\cdot \cfrac{AB}{AC}=1$. $\cfrac{\sin x}{\cos x}=\cfrac{\cfrac{AC}{BC}}{\cfrac{AB}{BC}}=\cfrac{AC}{BC}\cdot \cfrac{BC}{AB}=\cfrac{AC}{AB}=\text{tg }x $

$\cfrac{\cos x}{\sin x}=\cfrac{\cfrac{AB}{BC}}{\cfrac{AC}{BC}}=\cfrac{AB}{BC}\cdot \cfrac{BC}{AC}=\cfrac{AB}{AC}=\text{ctg }x$

7. $\cos x=\cfrac{\sqrt{15}}{4}$, $\text{tg }x=\cfrac{1}{\sqrt{15}}$, $\text{ctg }x=\sqrt{15}$

8. $BC=\cfrac{16\sqrt{3}}{3}$, $AC=\cfrac{8\sqrt{3}}{3}$

9. $30^{\circ}$, $60^{\circ}$. 10. $AC=24$ cm, $AB=32$ cm.

11. $\sim 67^{\circ}$, $\sim 23^{\circ} $.

Rezolvarea triunghiului dreptunghic

2.

4. ${{h}_{A}}=8\text{ cm}\Rightarrow 2A_{\triangle ABC}=96\text{ c}{{\text{m}}^{2}}\Rightarrow \sin A=\cfrac{96}{100}$.

5. $AC=8$ cm, $AB=\cfrac{8\sqrt{6}}{3}\text{ cm},$ $BC=4\sqrt{2}\left( 1+\cfrac{\sqrt{3}}{3} \right)\Rightarrow {{P}_{\triangle ABC}}=4\left( 2+\sqrt{2}+\sqrt{6} \right)\text{ cm}.$

6. a) $3\sqrt{3\text{ }}\text{cm}$; b) $5\text{ cm}$; c) $2\sqrt{13}\text{ cm}$.

7. a) $3\sqrt{3}\text{ cm}$; b) $11\text{ cm}$; c) $2\sqrt{37}\text{ cm}$.

Calculul elementelor în triunghiul echilateral, în pătrat și în hexagonul regulat

2.

3.

4.

Aproximarea în situații practice a distanţelor folosind relaţii metrice

2. a) $\sim1,26$ m; b) $1,62$ m.

3. $\sim23^{\circ}$.

4. $\sim 5^{\circ}$.

5. $\sim 4825$ m/min.

Recapitulare

1. a) 8; b) 16; c) 12; d) 24; e) 17; f) 16.

2. a) $\triangle ABC\text{ dreptunghic, }\sphericalangle A=90^{\circ}$; b) $\triangle ABC\text{ nu e dreptunghic;}$ c) $\triangle ABC\text{ nu e dreptunghic}.$

3. a) A; b) A.

4. a) $AC=3$ cm, $AC=3\text{ cm, }AB=6\sqrt{2}\text{ cm, }P=\left( 2+\sqrt{2} \right)\text{ cm}$;

b) $BC=4\text{ cm, }AB=2\sqrt{3}\text{ cm, }P=2\left( 3+\sqrt{3} \right)\text{ cm}$; c) $AB=3\text{ cm, }AC=3\sqrt{15}\text{ cm, }P=3\left( 5+\sqrt{15} \right)\text{ cm}$; d) $AC=7\text{ cm, }BC=\sqrt{149}\text{ cm, }P=17+\sqrt{149}\text{ cm}$.

6. 24 cm.

7. $A=27\sqrt{3}\text{ c}{{\text{m}}^{2}}$, $P=18\sqrt{3}\text{ cm}$.

8. $A=18\sqrt{3}\text{ c}{{\text{m}}^{2}}$, $P=12\sqrt{3}\text{ cm}$.

9. $CD=3$ cm, $BD=9$ cm, $BC=12$ cm, $AC=6$ cm, $AB=6\sqrt{3}$ cm, $P=6\left( 3+\sqrt{3} \right)$ cm $A=18\sqrt{3}\text{ c}{{\text{m}}^{2}}$.

10. $P=2\left( 5+\sqrt{13}+\sqrt{10} \right)\text{ cm}$.

13. $16\sqrt{3}\text{ c}{{\text{m}}^{2}}$.

14. 50 cm$^2$.

15. $A=39\sqrt{3}\text{ c}{{\text{m}}^{2}}$, $P=26+12\sqrt{3}\text{ cm}$.

16. Fie $CE\perp AB,E\in AB$. Din faptul că trapezul e isoscel $BE=\cfrac{AB-CD}{2}=18\text{ cm}$. $\text{Din teorema lui Pitagora pentru dreptunghic }\triangle CEB$, $CE=24\text{ cm}\text{.}$ $AE=AB-BE=32$ cm. $\text{Din teorema lui Pitagora pentru dreptunghic }\triangle CEA$, $CA=40\text{ cm}\text{.}$ Deci $AB^2=AC^2+BC^2$, din reciproca teoremei lui Pitagora, $\triangle ABC$ dreptunghic în $C$, deci $AC\perp CB$.

17. $6\sqrt{2}\text{ cm}$.

18. $\sphericalangle B=90^{\circ}$, $\sphericalangle A=60^{\circ}$, $\sphericalangle C=30^{\circ}$.

19. $AC=6\text{ cm}$, $BC=6\left( 1+\sqrt{3} \right)\text{ cm}$, $P=6\left( 3+\sqrt{3}+\sqrt{6} \right)\text{ cm}$, $A=18\left( 3+\sqrt{3} \right)\text{ c}{{\text{m}}^{2}}$.

20. a) $\left. \begin{array}{l} AB=CD=EF\text{ (laturile hexagonului regulat)}\\ BC=DE=FA\text{ (laturile hexagonului regulat)}\\ \sphericalangle ABC=\sphericalangle CDE=\sphericalangle EFA\text{ (unghiurile hexagonului regulat)}\\ \end{array} \right\}\xRightarrow[]{LUL}$ $\Rightarrow\triangle ABC\equiv \triangle CDE\equiv \triangle EFA\Rightarrow AC=CE=EA\Rightarrow \triangle ACE$ echilateral.

b) Putem privi $ABCDEF$, $ACE$ ca hexagon respectiv triunghi echilateral înscrise în același cerc de rază $R\Rightarrow {{A}_{ABCDEF}}=\cfrac{3{{R}^{2}}\sqrt{3}}{2}={{A}_{\text{flori}}}+{{A}_{\text{gazon}}}$, ${{A}_{\text{flori}}}={{A}_{ACE}}=\cfrac{3{{R}^{2}}\sqrt{3}}{4}=\cfrac{{{A}_{ABCDEF}}}{2}\Rightarrow {{A}_{\text{flori}}}={{A}_{\text{gazon}}}$.

Exersezi și progresezi

1. lungimea ipotenuzei $=26$ cm, lungimea înălțimii $=12$ cm, lungimile catetelor $=4\sqrt{13}\text{ cm}$, $6\sqrt{13}\text{ cm}$.

2. lungimea ipotenuzei $=50$ cm, lungimea înălțimii $=24$ cm, lungimea catetei $=40$cm.

3. $AB=24$ cm, $BC=26$ cm

4. 0.

5. $P=16\sqrt{2}\text{ cm }<23\text{ cm}$.

6. $A=32\sqrt{2}\text{ c}{{\text{m}}^{2}}$.

7. a) $32+8\sqrt{3}\text{ cm}$; b) $\sphericalangle A=60^{\circ}$ ,$\sphericalangle B=30^{\circ}$, $\sphericalangle C=150^{\circ}$, $\sphericalangle D=120^{\circ}$.

8. a) $\triangle ABC\text{ echilateral}\Rightarrow \left\{ \begin{array}{l} AB=MB(1) \\ \sphericalangle ABM=60^{\circ} (2) \\ \end{array} \right.$

$ABCD\text{ pătrat}\Rightarrow \left\{ \begin{array}{l} AB=CB(3) \\ \sphericalangle ABC=90^{\circ} (4) \\ \sphericalangle DCB=90^{\circ} \\ \end{array} \right.$

(2), (4) $\Rightarrow \sphericalangle MBC=30^{\circ} $

(1), (3) $\Rightarrow MB=CB\Rightarrow \sphericalangle BMC=\sphericalangle BCM=\cfrac{180^{\circ} -\sphericalangle MBC}{2}=\cfrac{180^{\circ} -30^{\circ} }{2}=75^{\circ} $

$\sphericalangle DCM=\sphericalangle DCB-\sphericalangle MCB=90^{\circ} -75^{\circ} =15^{\circ} $.

b) $\triangle ABC$ echilateral, $MP$ mediană $\Rightarrow MP\perp AB$.

Analog cu punctul anterior demonstrăm că $\sphericalangle CDM=15^{\circ} \Rightarrow \triangle CDM$ isoscel cu baza $CD$, $MP$ mediană $\Rightarrow NM\perp CD$. Dar $AB\parallel CD\Rightarrow M-N-O$ coliniare. În plus, $NP\perp PB\perp BC\perp CN\Rightarrow PBCN$ dreptunghi.

$MP$ înălțime în triunghiul echilateral $\Rightarrow MN=\cfrac{l\sqrt{3}}{2}$ unde $l$ este lungimea laturii pătratului $\Rightarrow MN=NP-\cfrac{l\sqrt{3}}{2}=BC-\cfrac{l\sqrt{3}}{2}=l-\cfrac{l\sqrt{3}}{2}=l\cfrac{2-\sqrt{3}}{2}$.

c) $\text{tg }15^{\circ} =\cfrac{MN}{CN}=\cfrac{2MN}{CD}=2-\sqrt{3}$

9. a) $\sim 1953,65$ m; b) $\sim5$ min.

Unitatea 11 - Bun venit, vacanță!

Recapitulare finală

1. $2,23$; $3,16$; $4,79$; $6,08$; $9,94$; $31,62$.

2. a) $\sqrt{2}<1,5<\sqrt{3}<\left|-\sqrt{7}\right|$; b) $\sqrt{\cfrac{5}{2}}<\sqrt{3,5}<\sqrt{6}<(-2)^2$; c) $5\sqrt{3}<4\sqrt{5}<3\sqrt{10}<10$.

3. $A\cap \mathbb{N}=\varnothing$, $A\cap \mathbb{Z}=\left\{-\cfrac{7}{3}\sqrt{729}\right\}$, $A\cap \mathbb{Q}=\left\{\sqrt{1,44},\,\cfrac{5}{2}\sqrt{169},\,-\cfrac{7}{3}\sqrt{729},\,-\sqrt{3.24}\right\}$, $A\cap (\mathbb{R}\setminus\mathbb{Q})=\left\{-\sqrt{3},\,\sqrt{288}\right\}$.

4. a)

b)

c)

5. c)

6. al doilea traseu este mai scurt

7. a) $6-2\sqrt{6}+6\sqrt{2}$; b) $7\sqrt{3}-3\sqrt{7}+\sqrt{21}$; c) $\cfrac{2942\sqrt{2}}{195}$; d) $\cfrac{3177\sqrt{3}}{1001}$.

8. a) Într-o lună nufărul își crește suprafaţa de $\sqrt{2}$ ori. În 11 luni nufărul își crește suprafaţa de $2^5\sqrt{2}$ ori.

9. a) $6\sqrt{2}$; b) $\cfrac{19\sqrt{15}}{27}$; c) $8\sqrt{5}(\sqrt{2}+\sqrt{3})$; d) $\cfrac{\sqrt{6}}{3}-\cfrac{\sqrt{15}}{3}$.

10. 6 ori

11. $3\sqrt{7}$, $\sqrt{7}$, $-3\sqrt{7}$, $-\sqrt{7}$

12. $x=\cfrac{2\cdot4\cdot6\cdot...\cdot98}{3\cdot5\cdot7\cdot...\cdot99}$, $y=\cfrac{1\cdot3\cdot5\cdot...\cdot99}{2\cdot4\cdot6\cdot...\cdot100}\Rightarrow x\cdot y=\cfrac{1}{100}\Rightarrow\sqrt{x\cdot y}=10\in\mathbb{Q}$.

13. a) $A\times B=\{(\sqrt{2},2),\,(\sqrt{2},3),\,(\sqrt{2},5),\,(\sqrt{3},2),\,(\sqrt{3},3),\,(\sqrt{3},5),\,(\sqrt{5},2),\,(\sqrt{5},3),\,(\sqrt{5},5)\}$,
$B\times C=\{(2,0),\,(2,1),\,(3,0),\,(3,1),\,(5,0),\,(5,1)\}$
$A\times C=\{(\sqrt{2},0),\,(\sqrt{2},1),\,(\sqrt{3},0),\,(\sqrt{3},1),\,(\sqrt{5},0),\,(\sqrt{5},1)\}$;
b) 3.

14.

15. $OA^2=3^2+(-4)^2=5^2$, $OB^2=5^2+0^2=5^2$, $OC^2=(-2)^2+(\sqrt{21})^2=5^2$,
$OD=(\sqrt{17})^2+(2\sqrt{2})^2=5^2$, $OE^2=(-\sqrt{11})^2+(-\sqrt{14})^2=5^2$, $OF^2=(\sqrt{2})^2+(-\sqrt{23})^2$, $OG^2=(-\sqrt{5})^2+(2\sqrt{5})^2=5^2$, $OH=(-2\sqrt{3})^2+(-\sqrt{13})^2=5^2$,
deci $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$ $\in C(0,5)$.

16. $\left.\begin{array}{l}(AB^2=(3-1)^2+(2-1)^2=5 \\ BC^2=(0-3)^2+(3-2)^2=10\\CA^2=(0-1)^2+(3-1)^2=5 \end{array}\right\}$ $\Rightarrow\left\{\begin{array}{l}BC^2=AB^2+CA^2\\AB=CA\end{array}\right\}$ $\xRightarrow[]{\text{reciproca Pitagora}} \triangle ABC$ dreptunghic isoscel.

17. $\left.\begin{array}{l} MN^2=(0+3)^2+(1+1)^2=13 \\ NP^2=(2-0)^2+(4-1)^2=13 \\ PQ^2=(-1-2)^2+(2-4)^2=13 \\ QN^2=(-3+1)^2+(-1-2)^2=13 \end{array}\right\}$ $\Rightarrow MN=NP=PQ=QN\Rightarrow MNPQ$ romb.

18. $\left.\begin{array}{l}ST^2=(1+4)^2+(-4+2)^2=29\\TU^2=(3-1)^2+(1+4)^2=29\\UV^2=(-2-3)^2+(3-1)^2=29\\VS^2=(-3+1)^2+(-1-2)^2=29\\SU^2=(3+4)^2+(1+2)^2=58\\TV^2=(-2-1)^2+(3+4)^2=58\end{array}\right\}$ $\Rightarrow\left.\begin{array}{l}ST=TU=UV=SU\Rightarrow STUV \text{romb}\\SU=TV\end{array}\right\}$ $\Rightarrow STUV$ pătrat.

19. a) $-\cfrac{1}{9}$; b) $-\cfrac{\sqrt{3}}{9}$; c) $\cfrac{5}{9}-\cfrac{\sqrt{2}}{9}$; d) $\cfrac{\sqrt{15}}{9}-\cfrac{2}{3}$.

20. a) $(x,y)=(1,1)$; b) $(x,y)=\left(\cfrac{15\sqrt{2}-10\sqrt{3}}{6},\cfrac{15\sqrt{2}-10\sqrt{3}}{6}\right)$; c) $(x,y)=\left(\cfrac{\sqrt{30}}{5},\cfrac{\sqrt{30}}{5}\right)$; d) $(x,y)=(\sqrt{15}-\sqrt{10},\sqrt{15}-\sqrt{10})$.

21.

a)

b)

c) Punctul de intersecţie al celor două segmente este $(1,-2)$ care verifică sistemul de ecuaţii.

22. $\sim153$ litri/mil

23. a) $AB=CD=EF=FG=HI=\sqrt{5}$, $BC=DE=GH=\sqrt{13}$;

b)Pe porţiunile de tip $AB:4\sqrt{5}$ $\cfrac{l}{100km}$ $BC:2\sqrt{13}$ $\cfrac{l}{100km}$

24. 16 cm$^2$

25. a) $DC\parallel AB\Rightarrow\sphericalangle DMA=\sphericalangle MAB$ (unghiuri alterne interne; secanta $MA$)$=\sphericalangle DAM$ ($AM$ bisectoare) $\Rightarrow\triangle DMA$ isoscel cu baza $MA$. $DC\parallel AB\Rightarrow\sphericalangle CMB=\sphericalangle MBA$ (unghiuri alterne interne; secanta $MB$)$=\sphericalangle MBC$($BM$ bisectoare)$\Rightarrow\triangle BMC$ isoscel cu baza $MB$.

b) $\sphericalangle AMB=180^{\circ}-\sphericalangle MAB-\sphericalangle MBA=180^{\circ}-\cfrac{\sphericalangle DAB}{2}-\cfrac{\sphericalangle DBA}{2}=180^{\circ}-\cfrac{\sphericalangle DAB+\sphericalangle DBA}{2}=$ $=180^{\circ}-\cfrac{180^{\circ}}{2}$ ($\sphericalangle DAB$, $\sphericalangle DBA$ interne de aceeași parte a secantei)$=90^{\circ}$.

c) $A=48$ cm$^2$, $P=30$ cm.

26. Folosim indicaţia din problemă. Fie $\{G\}=FO\cap BC$. $\left.\begin{array}{l}BO=OA (\triangle AOB\text{ dreptunghic isoscel})\\\sphericalangle OBC=90^{\circ}-\sphericalangle GOB=180^{\circ}-90^{\circ}-\sphericalangle GOB=\sphericalangle AOF\\\sphericalangle OCB=90^{\circ}-\sphericalangle GOC=180^{\circ}-90^{\circ}-\sphericalangle GOC=\sphericalangle DOF=\sphericalangle OFA (\text{alterne interne})\end{array}\right\}\xRightarrow[]{LUU}$ $\triangle AOF\equiv \triangle OBC$ $\Rightarrow AF=OC=OD$ $\xRightarrow[]{AF\parallel OD} AFDO$ paralelogram $\Rightarrow OF$ trece prin mijlocul lui $AD$.

27. $\left.\begin{array}{l}A_2 B_2 \text{ linie mijlocie în }\triangle ACB\Rightarrow A_2 B_2\parallel AB,\,A_2 B_2=AB/2\\A_3 B_3 \text{ linie mijlocie în }\triangle AHB\Rightarrow A_3 B_3\parallel AB,\,A_3 B_3=AB/2\end{array}\right\}$ $\Rightarrow A_2 B_2 A_3 B_3$ paralelogram

$A_3 B_2$ linie mijlocie în $\triangle BHC\Rightarrow A_3 B_2\parallel CC_1\Rightarrow B_2 A_3 A_1=\sphericalangle C_1 HA$ (unghiuri alterne interne)

$\sphericalangle BA_3 B_2=\sphericalangle BA_3 B_1+\sphericalangle A_1 A_3 B_2=\sphericalangle BAA_1$ (unghiuri corespondente)$+\sphericalangle C_1 HA=180^{\circ}-\sphericalangle AC_1 H=90^{\circ}$, deci $A_2 B_2 A_3 B_3$ dreptunghi. Analog $A_2 C_2 A_3 C_3$ dreptunghi.

28. $\left.\begin{array}{l}\sphericalangle A=180^{\circ}-\sphericalangle B-\sphericalangle C=36^{\circ}.\,\sphericalangle DBA=36^{\circ}\Rightarrow BD=DA\\\sphericalangle BDC=\sphericalangle A+\sphericalangle DBA=72^{\circ}=\sphericalangle C\Rightarrow BC=BD\end{array}\right\}$ $\Rightarrow BC=DA$ (1).

$\sphericalangle B=\sphericalangle C=90^{\circ}\Rightarrow AB=AC$ (2).

$BD$ bisectoare $\xRightarrow[]{\text{teorema bisectoarei}} \cfrac{BC}{BA}=\cfrac{DC}{DA}\Rightarrow BC\cdot DA=BA\cdot DC\xRightarrow[]{{(1),\,(2)}} BC^2=AC\cdot DC$.

30. Fie $\{V\}=s_1\cap s_2$. $\left.\begin{array}{l}d_1\parallel d_3\xRightarrow[]{\text{Thales}} \cfrac{VA_1}{A_1 C_1}=\cfrac{VA_2}{A_2 C_2} \\ d_1\parallel d_2 \xRightarrow[]{\text{Thales}} \cfrac{VA_1}{A_1 B_1} = \cfrac{VA_2}{A_2 B_2} \end{array}\right\}$ $\Rightarrow \cfrac{A_1 B_1}{A_1 C_1}=\cfrac{A_2 B_2}{A_2 C_2}$ $\Rightarrow \cfrac{A_1 B_1}{A_1 C_1-A_1 B_1}=\cfrac{A_2 B_2}{A_2 C_2-A_2 B_2}$ $\Rightarrow \cfrac{A_1 B_1}{B_1 C_1}=\cfrac{A_2 B_2}{B_2 C_2}$.

31. Demonstrăm următoarea lemă: Fie $XYZT$ trapez $XY\parallel T$ și $W\in(TX)$, $Q\in(ZY)$, astfel încât $WQ\parallel XY\Rightarrow \cfrac{TW}{WX}=\cfrac{ZQ}{QY}$.

Într-adevăr, fie $\{V\}=XT\cap ZY$. $\left.\begin{array}{l}WQ\parallel XY\xRightarrow[]{\text{Thales}} \cfrac{VT}{TW}=\cfrac{VZ}{ZQ}\\TZ\parallel XY\xRightarrow[]{\text{Thales}} \cfrac{VT}{TX}=\cfrac{VZ}{ZY}\end{array}\right\}$ $\Rightarrow \cfrac{TW}{TX}=\cfrac{ZQ}{ZY}$ $\Rightarrow \cfrac{TW}{WX}=\cfrac{ZQ}{QY}$

Aplicăm lema pentru trapezele: $\left.\begin{array}{l}ABB_1 A_1 \Rightarrow \cfrac{AM}{MB}=\cfrac{A_1 R}{RB_1} \\ACC_1 A_1\Rightarrow \cfrac{CP}{PA}=\cfrac{C_1 R}{RA_1} \\NRB_1 B\Rightarrow \cfrac{CB}{CN}=\cfrac{C_1 B_1}{C_1R} \Rightarrow \cfrac{BN}{NC}=\cfrac{RB_1}{C_1R} \end{array}\right\}$ $\Rightarrow \cfrac{AM}{MB}\cdot \cfrac{BN}{NC}\cdot\cfrac{CP}{PA}=1$.

32. Fie $\{N’\}=MP\cap BC$. Din teorema lui Menelaus, $\cfrac{AM}{MB}\cdot\cfrac{BN'}{CN'}\cdot\cfrac{CP}{PA}=1$, deci $\cfrac{BN}{NC}=\cfrac{BN'}{N' C}$, i.e. $N$ și $N’$ sunt pe dreapta $BC$ în afara segmentului $BC$ și dau același raport, deci $N=N'\Rightarrow M,\,N,\,P$ coliniare.

33. Din teorema lui Menelaus pentru $\triangle ABN$ cu secanta $M$, $O$, $C$: $\cfrac{AM}{MB}\cdot \cfrac{BC}{CN}\cdot \cfrac{NO}{OA}=1$.

Din teorema lui Menelaus pentru $\triangle ACN$ cu secanta $B$, $O$, $P$: $\cfrac{AO}{ON}\cdot \cfrac{NB}{BC}\cdot \cfrac{CP}{PA}=1$.

Înmulţind cele două relaţii, se obţine: $\cfrac{AM}{MB}\cdot \cfrac{BN}{NC}\cdot \cfrac{CP}{PA}=1$.

34. Fie $\{O’\}=NA\cap CM$ și $\{P' \}=BO'\cap AC$. Din teorema lui Ceva, $\cfrac{AM}{MB}\cdot \cfrac{BN}{NC}\cdot \cfrac{CP'}{P'A}=1$, deci $\cfrac{CP'}{P'C}=\cfrac{BP'}{P' C}$, i.e. $P$ și $P’$ sunt pe latura $CA$ și îl împart în același raport, deci $P=P'\Rightarrow B,\,O,\,P'$ coliniare $\Rightarrow AM,\,BN,\,CP$ concurente.

35. dreptunghic.

36. a) $AB\parallel CD\xRightarrow[]{\text{TFA}} \triangle DOC\sim \triangle BOA$ $\Rightarrow \cfrac{DO}{BO}=\cfrac{DC}{BA}=\cfrac{DC:2}{BA:2}=\cfrac{DN}{BM}$ $\Rightarrow \cfrac{DO}{DN}=\cfrac{BO}{BM}$.

b) $\left.\begin{array}{l} \cfrac{DO}{BO}=\cfrac{DN}{BM}\\ \sphericalangle NDO=\sphericalangle OBM\end{array}\right\}$ $\xRightarrow[]{\text{LUL}} \triangle DON\sim \triangle BOM$

c) $\triangle DON\sim \triangle BOM\Rightarrow\sphericalangle DON=\sphericalangle BOM\Rightarrow M,\,O,\,N$ coliniare.

d) $AB\parallel CD\xRightarrow[]{\text{TFA}} \triangle DPC\sim \triangle APB$ $\Rightarrow \cfrac{DP}{AP}=\cfrac{DC}{AB}=\cfrac{DC:2}{AB:2}=\cfrac{DN}{AM}$ $\Rightarrow \cfrac{DP}{DN}=\cfrac{AP}{AM}$.

e) $\left.\begin{array}{l} \cfrac{DP}{AP}=\cfrac{DN}{AM}\\ \sphericalangle NDO=\sphericalangle OBM\end{array}\right\}$ $\xRightarrow[]{\text{LUL}} \triangle DON\sim \triangle BOM$

f) $\triangle DON\sim \triangle BOM\Rightarrow\sphericalangle DON=\sphericalangle BOM\Rightarrow M,\,O,\,N$ coliniare.

37. 18 cm.

38. 27 cm$^2$.