GEOMETRIA PLANA AFA 1994 a 2019.pdf

Resoluções elaboradas pelo Prof. Renato Madeira

EXERCÍCIOS DE GEOMETRIA PLANA DA AFA 1994 a 2019 ENUNCIADOS 1) (AFA 1994) Num triângulo ABC, os ângulos B  e C  medem, respectivamente, 45  e 60,  o lado AC mede 2 cm. Então, a medida do lado BC (em cm) é: ˆ

a) 1 +

3 3

b)

1 2

+

3

c) 1 + 3

ˆ

d) 2 + 2

2) (AFA 1995) Na figura, todos os círculos têm raio r. Qual a área da parte hachurada?

a) r 2 ( 2 3 − )

b) r 2 ( 3 3 − )

c) r 2 ( 4 3 − )

d) r 2 ( 5 3 − )

3) (AFA 1995) Num pentágono convexo, os ângulos internos estão em progressão aritmética. Qual o 3º termo, em graus dessa progressão? a) 54 b) 108 c) 162 d) 216 4) (AFA 1995) Dados dois triângulos semelhantes, um deles com 4, 7 e 9 cm de lado, e o outro com 66 cm de perímetro, pode-se afirmar que o menor lado do triângulo maior mede, em cm. a) 9,8 b) 11,6 c) 12,4 d) 13,2 5) (AFA 1995) Na figura abaixo, a razão

a) 5

b) 6

x



c) 2 2

 é:

d) 10

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Resoluções elaboradas pelo Prof. Renato Madeira

6) (AFA 1995) No retângulo ABCD, BC  e PC  medem, respectivamente, 5 cm e 3 cm. Qual a área, em cm 2 ,  do triângulo ABP?

a)

32

b) 16

3

c) 19

d)

62 3

7) (AFA 1995) A razão entre as áreas de um quadrado de lado   e de um círculo de raio r, que possuem o mesmo perímetro, é:     a) b) c) d) 8

6

4

2

8) (AFA 1995) Considere uma circunferência inscrita num quadrado de lado a. A área da região hachurada é:

a)

a2 64

( 4 − )

b)

a2 32

( 4 − )

c)

a2 16

( 4 − )

d)

a2 8

( 4 − )

9) (AFA 1996) Seja ABC um triângulo retângulo com catetos AB e BC. Divide-se AB em 10 partes congruentes e, pelos pontos de divisão, traçam-se retas paralelas a BC, cortando o lado AC e determinando 9 segmentos paralelos a BC. Se BC = 18,  então a

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Resoluções elaboradas pelo Prof. Renato Madeira

10) (AFA 1996) A base maior de um trapézio mede 26 cm, a menor 14 cm e a altura 6 cm. As alturas dos triângulos formados pelos prolongamentos dos lados não paralelos, em cm, são: a) 8 e 9 b) 7 e 13 c) 91 e 14 d) 15 e 18 11) (AFA 1996) Qual a área do triângulo retângulo isósceles que inscreve uma circunferência de raio r = 2 ? a) ( 3 + 2 2 ) b) 2 ( 3 + 2 2 ) c) 3( 2 + 2 ) d) 4 (1 + 2 ) 12) (AFA 1996) Qual a diferença entre a área de um triângulo equilátero de lado a e a área da circunferência nele inscrita? 2 2 a ( 2 3 − ) a ( 3 3 − ) a) b) 12

 b)

a

2

(4

12

3 − )

d)

12

a

2

(5

3 − )

12

13) (AFA 1996) O valor do menor ângulo formado pelos ponteiros de um relógio às 2h e 15min é: a) 15 b) 30 c) 1730' d) 2230' 14) (AFA 1996) Na figura abaixo, OA = 5, AB = 3, AOB = BOC = COD =  e ABO = BCO = CDO = 90.  Se x = cos2 ,  então a área do triângulo CDO é: ˆ

ˆ

a) 3x 2

ˆ

ˆ

ˆ

ˆ

b) 4x 2

c) 6x 2

d) 8x 2

15) (AFA 1996) Os lados de um triângulo ABC medem AB = 20 cm, BC = 15 cm e cm  e traçam-se paralelas DE ao lado AC = 10 cm.  Sobre o lado BC marca-se BD = 3 cm AB (E sobre AC) e DF ao lado AC (F sobre AB). O perímetro do paralelogramo AEDF, em cm, é: a) 24 b) 28 c) 32 d) 36

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Resoluções elaboradas pelo Prof. Renato Madeira

a) 9

c) 6 + 3

b) 5 3

d) 2 ( 2 + 3 )

17) (AFA 1997) Considerando-se a figura abaixo, NÃO se pode afirmar que

a) Se o triângulo ABC é isósceles, então os triângulos ABD, ACE e BCD são sempre, dois a dois, congruentes.  b) Os triângulos ABD e AEC são são congruentes congruentes se os lados AB e BC forem congruentes congruentes e F o incentro do triângulo ABC. c) Os triângulos ABD e AEC são congruentes se os lados AB e BC forem congruentes e F o ortocentro do triângulo ABC. d) Os triângulos BEF e CDF são congruentes se os lados AB e BC forem congruentes e F o baricentro do triângulo ABC. 18) (AFA 1998) Inscreve-se um quadrilátero convexo ABCD  em uma circunferência tal que ABC = x .  Então, ACB + BDC, DC,  em graus, é o a) suplementar de x.  b) suplementar de de 2x. c) complementar de x. d) complementar de 2x. ˆ

ˆ

ˆ

19) (AFA 1998) Dois vértices de um triângulo equilátero pertencem a dois lados de um quadrado cuja área é 1 m2 .   Se o terceiro vértice do triângulo coincide com um dos vértices do quadrado, então, a área do triângulo, tri ângulo, em m 2 ,  é a) 2 3 − 1 b) 2 3 + 1 c) −3 + 2 3 d) 3 + 2 3 20) (AFA 1998) Seja ABCD um quadrado, ABE um triângulo equilátero e E um ponto interior ao quadrado. O ângulo AED  mede, em graus, ˆ

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Resoluções elaboradas pelo Prof. Renato Madeira

21) (AFA 1998) Seja o triângulo equilátero DEF, inscrito no triângulo isósceles ABC, com AB = AC  e DE paralelo a BC (D sobre AB e E sobre AC). Tomando-se ADE = , CEF =   e DFB =   pode-se afirmar que a)  +  = 2 b)  + = 2 c) 2 +  = 3 d)  + 2 = 3 ˆ

ˆ

ˆ

22) (AFA 1998) Um círculo com área 100 cm2  possui uma corda de 16 cm. Qual a área, em cm2 ,   do maior círculo tangente a essa corda e a esse círculo em pontos distintos? a) 36 b) 49 c) 64 d) 81 23) (AFA 1998) O pentágono ABCDE está inscrito em uma circunferência de centro O. Se o ângulo AOB BCD  e AED,  em graus, é AOB  mede 40,  então a soma dos ângulos BCD a) 144 b) 180 c) 200 d) 214 ˆ

ˆ

ˆ

24) (AFA 1999) Na figura abaixo o perímetro do triângulo equilátero ABC é 72 cm , M é o ponto médio de AB e CE = 16 cm . Então, a medida do segmento CN, em cm, é um sétimo de

a) 51

b) 50

c) 49

d) 48

25) (AFA 1999) Na figura abaixo, o lado do quadrado é 1 cm. Então, a área da região sombreada, em cm 2 ,  é

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Resoluções elaboradas pelo Prof. Renato Madeira min a 4h 35min,   o ponteiro das horas de um relógio 26) (AFA 1999) De 2h 45 min  percorre, em radianos, radianos, 11  5 7 a) b) c) d)

36

3

18

24

27) (AFA 1999) A área do quadrado menor, da figura abaixo, vale

a) 2

b) 2

c) 5

d) 8

28) (AFA 1999) Considere um triângulo equilátero, um quadrado e um hexágono regular, todos com o mesmo perímetro. Sejam AT , A Q  e A H  as áreas do triângulo, do quadrado e do hexágono, respectivamente. Então, pode-se afirmar que a) A T  A Q  A H . b) AT = A Q = A H . c) A T  AQ  e A Q  A H . d) A T  A Q  e A Q = A H . 29) (AFA 2000) O valor de x 2 ,  na figura abaixo, é

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Resoluções elaboradas pelo Prof. Renato Madeira

30) (AFA 2000) Seja P um ponto interior a um triângulo equilátero de lado k. Qual o valor de k, sabendo-se que a soma das distâncias de P a cada um dos lados do triângulo é 2? a)

2 3

b) 3

3

c)

4 3

d) 2 3

3

31) (AFA 2000) Na figura, O e M são centros das semicircunferências. O perímetro do triângulo DBC, quando AO = r = 2  AM,  é

a) c)

r (3 2 + 5 ) 2 r ( 2 + 3 10 ) 2

b) d)

r( 2 +3 5) 2 r ( 3 2 + 10 ) 2

32) (AFA 2001) Conforme a figura abaixo, s e t são, respectivamente, retas secante e tangente à circunferência de centro O. Se T é um ponto da circunferência comum às retas tangente e secante, então o ângulo ,  formado por t e s, é

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Resoluções elaboradas pelo Prof. Renato Madeira

a) 115

b) 125

c) 135

d) 145

34) (AFA 2001) Na figura, O é o centro da circunferência de raio r, AD = DE = EB = r   e   é o menor ângulo formado pelos ponteiros de um relógio às 9h 25min.  O valor do ângulo  = CBE  é ˆ

a) 120 120

b) 119,45

c) 126,25

d) 135,50

35) (AFA 2001) A figura abaixo representa um quadrado de 8 cm de lado. A área, em cm 2 ,  da figura sombreada é

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Resoluções elaboradas pelo Prof. Renato Madeira

Quantos metros de comprimento deverá ter o muro que o proprietário do terreno II construirá para fechar o lado que faz frente com a rua B? a) 28 b) 29 c) 32 d) 35

37) (AFA 2002) Na figura abaixo, os pontos A, B e C pertencem à circunferência de centro O e raio r. Se  = 140  e  = 50 ,  então, a área do triângulo BOC é

a)

r 3 2

b)

r2 2 3

c)

r 2 9

d)

r

2

3 4

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Resoluções elaboradas pelo Prof. Renato Madeira

a) 2 + 3

b) 9 3

c) 11 − 3

d) 19 3

39) (AFA 2003) As duas polias da figura giram simultaneamente em torno de seus respectivos centros O e O’, por estarem ligadas por uma correia ine xtensível.

Quantos graus deve girar a menor polia para que a maior dê uma volta completa? a) 1080 b) 120 c) 720 d) 2160 40) (AFA 2003) ABC é um triângulo retângulo em A e CX  é bissetriz do ângulo BCA, onde X é ponto do lado AB. AB.   A medida de CX   é 4 cm e a de BC  é 24 cm. Sendo assim, a medida do lado AC,  em centímetros, é igual a a) 3 b) 4 c) 5 d) 6 ˆ

41) (AFA 2003) Na figura, o triângulo AEC é equilátero e ABCD é um quadrado de lado 2 cm. A distância BE, em cm, vale.

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Resoluções elaboradas pelo Prof. Renato Madeira

a) 2 3

b) 6 − 1

c) 3 + 2

d) 6 − 2

42) (AFA 2003) Na figura, RST é um triângulo retângulo em S. Os arcos RnSpT , RmS RmS  e SqT  são semicircunferências cujos diâmetros são, respectivamente, RT , SR  e RST na ST . A soma das áreas das figuras hachuradas está para a área do triângulo RST razão

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Resoluções elaboradas pelo Prof. Renato Madeira

a) r + 3

b) 2r + r 3

c) r 3

d) r + r 3

45) (AFA 2005) Considere o triângulo ABC, de lados AB = 15, AC = 10, BC = 12 e seu baricentro G. Traçam-se GE e GF paralelos a AB e AC, respectivamente, conforme a figura abaixo. O perímetro do triângulo GEF é um número que, escrito na forma de fração irredutível, tem a soma do numerador com o denominador igual a

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Resoluções elaboradas pelo Prof. Renato Madeira

48) (AFA 2008) Um triângulo ABC é não isósceles. Sejam M,  N  e P, respectivamente, c m  e os pontos médios dos lados AB, BC  e AC  desse triângulo, de forma que AN = 3 cm 2 BP = 6 cm.  Se a área do triângulo ABC mede 3 15 cm ,  então o comprimento da outra

CM ,  em cm, é igual a mediana, CM, a) 3 6 b) 6 15

c) 3

d) 2

49) (AFA 2009) Considere num mesmo plano os pontos da figura abaixo, de tal forma que: (I) AW  CW  EW  GW  IW  LW  NW  PW (II) BW  DW  FW  HW  JW  MW  OW  QW BWC  CW CWD   PWQ  QW QWA (III) AWB  BW (IV) PC  AE  CG  EI  GL  IN  NA  LP  a

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Resoluções elaboradas pelo Prof. Renato Madeira

Se a área sombreada é igual à área não sombreada na figura, é correto afirmar que o raio de  2 , em cm, é um número do intervalo  11 11 23   23 5   5 13  a)  2,  b)  ,  c)  ,   d)  ,   5  10 2   2 5   5 10  51) (AFA 2011) Na figura abaixo têm-se quatro círculos, congruentes de centros O1 , O 2 , O3  e O 4  e de raio igual a 10 cm . Os pontos M, N, P e Q são pontos de tangência entre os círculos e A, B, C, D, E, F, G e H são pontos de tangência entre os círculos e a correia que os contorna.

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Resoluções elaboradas pelo Prof. Renato Madeira

Se os raios das circunferências de centros C1 e C2  medem, respectivamente, respectivamente, 2r   e 3r , então a área da região sombreada vale, em unidades de área, a)

55 8

 r 2

b)

29 4

 r 2

c)

61 8

 r 2 d) 8 r 2

53) (AFA 2013) Um triângulo é tal que as medidas de seus ângulos internos constituem uma progressão aritmética e as medidas de seus lados constituem uma progressão geométrica. Dessa maneira, esse triângulo NÃO é: a) acutângulo. b) equilátero. c) obtusângulo. d) isósceles.

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Resoluções elaboradas pelo Prof. Renato Madeira

a)

17 − 6 3 9

2

r 

b)

11 + 9 3 12

r 2

c)

1 5 − 4 3 9

r 2

d)

13 + 6 3 12

r 2

55) (AFA 2015) Seja o quadrado ABCD  e o ponto E  pertencente ao segmento AB . Sabendo-se que a área do triângulo ADE , a área do trapézio BCDE   e a área do quadrado ABCD  formam juntas, nessa ordem, uma Progressão Aritmética (P.A.) e a soma das áreas desses polígonos é igual a 800 cm 2 , tem-se que a medida do segmento EB

a) é fração própria.  b) é decimal exato. exato. c) é decimal não exato e periódico. d) pertence ao conjunto A = *+ −

+.

56) (AFA 2017) Considere, no triângulo ABC abaixo, os pontos P  AB, Q  BC, R  AC  e os segmentos PQ e QR   paralelos, respectivamente, a AC e AB. AB.  Sabendo cm   e que a área do triângulo ABC é 8 cm2 ,  então a área do que BQ = 3 cm, QC = 1 cm  paralelogramo hachurado, hachurado, em cm 2 ,  é igual a

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Resoluções elaboradas pelo Prof. Renato Madeira

A medida de AC,  uma das diagonais do pentágono regular, em cm, é igual a a) 1 + 5

b) −1 + 5

c) 2 +

5 2

d) 2 5 − 1

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Resoluções elaboradas pelo Prof. Renato Madeira

RESPOSTAS 1) c (Relações métricas nos triângulos) 2) a (Áreas de regiões circulares) 3) b (Polígonos angular e progressão aritmética) 4) d (Semelhança de triângulos) 5) d (Triângulos retângulos) 6) a (Relações métricas nos triângulos retângulos) 7) c (Áreas) 8) b (Áreas de regiões circulares) 9) a (Semelhança de triângulos) 10) b (Semelhança de triângulos) 11) b (Triângulos retângulos e áreas) 12) b (Áreas) 13) d (Ângulos no relógio) 14) c (Triângulos retângulos e áreas) 15) d (Semelhança de triângulos) 16) d (Semelhança de triângulos)

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Resoluções elaboradas pelo Prof. Renato Madeira

48) a (Pontos notáveis e relações métricas nos triângulos) 49) d (Áreas) 50) c (Áreas) 51) c (Comprimento da circunferência) 52) c (Áreas) 53) c (Relações métricas nos triângulos) tr iângulos) 54) d (Áreas) 55) c (Áreas) 56) b (Áreas) 57) a (Relações métricas nos polígonos regulares)

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Resoluções elaboradas pelo Prof. Renato Madeira

RESOLUÇÕES 1) (AFA 1994) Num triângulo ABC, os ângulos B  e C  medem, respectivamente, 45  e 60,  o lado AC mede 2 cm. Então, a medida do lado BC (em cm) é: ˆ

a) 1 +

3 3

RESOLUÇÃO: c

b)

1 2

+

3

c) 1 + 3

ˆ

d) 2 + 2

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Resoluções elaboradas pelo Prof. Renato Madeira

a) r 2 ( 2 3 − )

b) r 2 ( 3 3 − )

c) r 2 ( 4 3 − )

d) r 2 ( 5 3 − )

RESOLUÇÃO: a

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A soma dos ângulos internos do pentágono convexo é S5 = 180  (5 − 2 ) = 540,   então esse é o valor da soma dos termos da PA. Assim, temos:

( a − 2r ) + ( a − r ) + a + ( a + r ) + ( a + 2r ) = 540  5a = 54 540  a = 108 Portanto, o 3º termo da PA é a = 108.

4) (AFA 1995) Dados dois triângulos semelhantes, um deles com 4, 7 e 9 cm de lado, e o outro com 66 cm de perímetro, pode-se afirmar que o menor lado do triângulo maior

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Resoluções elaboradas pelo Prof. Renato Madeira

Se os raios das circunferências de centros C1 e C2  medem, respectivamente, respectivamente, 2r   e 3r , então a área da região sombreada vale, em unidades de área, a)

55 8

 r 2

b)

29 4

 r 2

c)

61 8

 r 2 d) 8 r 2

RESOLUÇÃO: c Seja R 3  o raio da circunferência de centro C3 . PG : 2r, 3r, R 3

 ( 3r ) = ( 2r )  R 3  R 3  = 2

9r  2

A área da região sombreada é igual à metade da área da circunferência de centro C3 menos metade da área da circunferência de centro C2   mais metade da área da circunferência de centro C1 . Logo, 2

1 2 2  9r  1  81 9    r 2 = S =      −    ( 3r ) +    ( 2r 2 r ) =  − + 2   2 2 2 2 8 2  1

61 8

  r2 .

 

53) (AFA 2013) Um triângulo é tal que as medidas de seus ângulos internos constituem uma progressão aritmética e as medidas de seus lados constituem uma progressão geométrica. Dessa maneira, esse triângulo NÃO é: a) acutângulo. b) equilátero. c) obtusângulo. d) isósceles. RESOLUÇÃO: c x, x + r )  , com r  0 , cujos elementos são os ângulos internos do Sejam a PA ( x − r, x,  y y, yq  yq   , com q  1 , cujos elementos são os lados do triângulo. triângulo, e a PG  , y,  q  

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Resoluções elaboradas pelo Prof. Renato Madeira 2

1  y  y y = ( yq ) +   − 2  ( yq )    cos 60  1 = q2 + 1 q4 − 2q2 + 1= 0 q= 1. − q 2 q  q  Portanto, os lados do triângulo são todos iguais, ou seja, o triângulo é equilátero, isósceles e acutângulo, porém não é obtusângulo. 2

2

54) (AFA 2014) Na figura abaixo, os três círculos têm centro sobre a reta AB  e os dois de maior raio têm centro sobre a circunferência de menor raio.

A expressão que fornece o valor da área sombreada é 17 − 6 3 2 11 + 9 3 2 1 5 − 4 3 2 13 + 6 3 2 r  b) r  c) r  d) r  a) 9

RESOLUÇÃO: d

12

9

12

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Resoluções elaboradas pelo Prof. Renato Madeira

Sejam O , O1 e O 2   os centros da circunferência menor de raio

r  2

  e das duas

circunferências maiores de raio r , respectivamente. respectivamente. r 

Como O1O = O2O = , a corda CD  é o lado do triângulo equilátero inscrito nas 2

circunferências maiores e determina segmentos circulares de d e 120 . Dessa forma, a área sombreada é igual à área de duas circunferências de raio r  menos a área de dois segmentos circulares de 120   e raio r   e menos a área de uma circunferência de raio

r  2

.

2   r2 r  r  r 2 1  S = 2   r − 2  − sen 120  −     =  2 − −   r2 + 2  2   3 4  3  2

3 2

r

2

=

13 + 6 3 12

r

2

55) (AFA 2015) Seja o quadrado ABCD  e o ponto E  pertencente ao segmento AB . Sabendo-se que a área do triângulo ADE , a área do trapézio BCDE   e a área do quadrado ABCD  formam juntas, nessa ordem, uma Progressão Aritmética (P.A.) e a soma das áreas desses polígonos é igual a 800 cm 2 , tem-se que a medida do segmento EB

a) é fração própria.  b) é decimal exato. exato. c) é decimal não exato e periódico. d) pertence ao conjunto A = *+ −

+.

RESOLUÇÃO: c

SADE = S − r  800  PA : SBCDE = S  SADE + SBCDE + SABCD = 3  S = 800  S =

 

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Resoluções elaboradas pelo Prof. Renato Madeira SABCD

=

SBCDE

=

2

=

800 3

+

400 3

( 20 + EB )  20

Assim, EB =

2 20 3

=

= 400  = 20 cm 800 3

 EB =

20 3

cm

cm  é um decimal não exato e periódico.

56) (AFA 2017) Considere, no triângulo ABC abaixo, os pontos P  AB, Q  BC, R  AC  e os segmentos PQ e QR   paralelos, respectivamente, a AC e AB. AB.  Sabendo cm   e que a área do triângulo ABC é 8 c m2 ,  então a área do que BQ = 3 cm, QC = 1 cm  paralelogramo hachurado, hachurado, em cm 2 ,  é igual a

a) 2

b) 3

c) 4

d) 5

RESOLUÇÃO: b

Como os segmentos PQ e QR    paralelos, respectivamente, a AC e AB,   então os triângulos ABC, PBQ e RQC são semelhantes. Sabendo que, no caso de figuras semelhantes, a razão entre as áreas é igual ao quadrado da razão de semelhança. Assim, podemos escrever: SPBQ BQ

2

=

SRQC QC

2

=

SABC BC BC

2



SPBQ 2

3

=

SRQC 2

1

=

8 4

2

 SPBQ =

9 2

 SRQC =

1 2

A área do paralelogramo APQR é igual à área do triângulo ABC menos a área dos triângulos PBQ e RQC. S

=S

S

S

=8

9

1

= 3 cm 2

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Resoluções elaboradas pelo Prof. Renato Madeira

57) (AFA 2018) A figura a seguir é um pentágono regular de lado 2 cm.

Os triângulos DBC e BCP são semelhantes. A medida de AC,  uma das diagonais do pentágono regular, em cm, é igual a a) 1 + 5

b) −1 + 5

c) 2 +

5 2

d) 2 5 − 1

RESOLUÇÃO: a

O pentágono regular divide a circunferência em 5 arcos iguais de

360 5

= 72.  Sabendo

que a medida de um ângulo inscrito é metade da medida do arco por ele determinado,  podemos identificar identificar os ângulos da figura.  No triângulo ABP, temos ABP = APB = 72,   então o triângulo é isósceles e ˆ

ˆ

AP = AB = 2.

Todas as diagonais do pentágono regular são iguais, pois são cordas que determinam arcos de 2  72 = 144. Sejam AC = BD = x  e, como ABC BPC,  temos: x 2

=

2 x−2

 x 2 − 2x − 4 = 0  x = 1 

5

Como x  0,  então x = AC = 1 + 5 cm. Alternativamente, poderia ter sido usado o teorema de Ptolomeu diretamente no