\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
W tr®jk©cie prostok©tnym ABC przeciwprostok©tna AB ma d¬ugo¯ª 10 cm,
a przyprostok©tna AC - &1 cm. Ile stopni ma k©t CAB ?
\end text
\begin params
3x3
5¸(3)
¸(3)/2
30
5
1/2
60
5¸(2)
1/¸(2)
45
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
W tr®jk©cie prostok©tnym ABC przeciwprostok©tna AB ma d¬ugo¯ª 10 cm,
a przyprostok©tna AC - &1 cm. Ile stopni ma k©t CAB ?
\end text
\points 5
\begin answer
\evaluate
\length 5
\correct &3
\wrong 2
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.2,0.1,0.8
\draw 0.1,0.2,0.9,0.8
\draw 0.1,0.8,0.9,0.8
\text 0.1,0.1,B
\text 0.9,0.81,A
\text 0.1,0.81,C
\text 0.5,0.82,b
\text 0.50,0.40,c
\text 0.75,0.71,¶
\window 1,11,79,23
\lose 2
\next
Dane s©: ÏABÏ = c = 10 i ÏACÏ = b = &1, a szukany jest k©t ¶.
\next
\lose 2
Poniewa±  cos ¶ = b/c = &1/10 = &2, zatem ¶ = ?
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
W tr®jk©cie ABC dane s©: d¬ugo¯ª boku AC r®wna &1 cm oraz miary k©t®w
BAC i CBA r®wne odpowiednio 30² i 45². Oblicz d¬ugo¯ª boku BC.
\end text
\begin params
4x3
4¸(2)
4
2¸(2)
2¸(2)
2
¸(2)
2¸(3)
s(6)
¸(3)
¸(8)
2
¸(2)
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
W tr®jk©cie ABC dane s©: d¬ugo¯ª boku AC r®wna &1 cm oraz miary k©t®w
BAC i CBA r®wne odpowiednio 30² i 45². Oblicz d¬ugo¯ª boku BC.
\end text
\points 10
\begin answer
\evaluate
\length 5
\correct &2
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.7,1.1,0.7
\draw 0.7,0.3,1.1,0.7
\draw 0.1,0.7,0.7,0.3
\dotted
\draw 0.7,0.3,0.7,0.7
\text 1.1,0.7,B
\text 0.1,0.71,A
\text 0.7,0.225,C
\text 0.7,0.71,C'
\text 0.25,0.475,b
\text 0.71,0.475,h
\text 0.97,0.475,a
\text 0.2,0.61,¶
\text 1.0,0.61,·
\window 1,11,79,23
\lose 3
\next
\lose 1
Dane s©: b = &1, ¶ = 30²,  · = 45². 
Szukamy a.
\next
\lose 3
Zauwa±, ±e tr®jk©t AC'C jest "po¬ow©" tr®jk©ta r®wnobocznego o boku b
(bo ¶ = 30²). St©d mo±esz wyliczyª h
\next
\lose 2
     h = Àb = &3
Rozwa±my teraz tr®jk©t C'BC. Poniewa± · = 45², wi«c jest to r®wnoramienny 
tr®jk©t prostok©tny o przyprostok©tnej h. 
St©d (tw. Pitagorasa) 
     a = h¸(2) = ? 
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
W tr®jk©cie prostok©tnym ABC najd¬u±szy bok ma d¬ugo¯ª 17 cm, a najkr®tszy
8 cm. ¦redni bok tr®jk©ta A'B'C' podobnego do tr®jk©ta ABC ma d¬ugo¯ª &1 cm.
Jaka jest d¬ugo¯ª najd¬u±szego boku tr®jk©ta A'B'C'?
\end text
\begin params
4x2
45
51
60
68
30
34
15
17
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
W tr®jk©cie prostok©tnym ABC najd¬u±szy bok ma d¬ugo¯ª 17 cm, a najkr®tszy
8 cm. ¦redni bok tr®jk©ta A'B'C' podobnego do tr®jk©ta ABC ma d¬ugo¯ª &1 cm.
Jaka jest d¬ugo¯ª najd¬u±szego boku tr®jk©ta A'B'C'?
\end text
\points 15
\begin answer
\length 5
\correct &2
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.3,0.1,0.9
\draw 0.1,0.9,0.9,0.9
\draw 0.1,0.3,0.9,0.9
\draw 0.63,0.1,0.63,0.3
\draw 0.63,0.3,0.9,0.3
\draw 0.9,0.3,0.63,0.1
\text 0.9,0.91,A'
\text 0.1,0.2,B'
\text 0.1,0.91,C'
\text 0.9,0.32,A
\text 0.63,0.0,B
\text 0.63,0.32,C
\text 0.02,0.55,a'
\text 0.5,0.91,b'
\text 0.55,0.55,c'
\text 0.58,0.15,a
\text 0.77,0.32,b
\text 0.82,0.15,c
\window 1,11,79,23
\lose 4
\next
\lose 1
\cont
Najd¬u±szym bokiem w tr®jk©cie prostok©tnym jest przeciwprostok©tna.
Przyjmijmy, ±e najkr®tszym bokiem w tr®jk©cie ABC jest bok BC.
\next
\lose 2
Z podobie­stwa tr®jk©t®w ABC i A'B'C' wynika, ±e:
ÏA'B'Ï/ÏABÏ = ÏA'C'Ï/ÏACÏ = ÏB'C'Ï/ÏBCÏ.
\next
\lose 1
W szczeg®lno¯ci: ¯rednim bokiem tr®jk©ta A'B'C' jest A'C'.
\next
\lose 1
Tak wi«c, dane s©: ÏABÏ = c = 17, ÏBCÏ = a = 8 i ÏA'C'Ï = b' = &1.
Szukamy ÏA'B'Ï = c'.
\next
\lose 2
Z podobie­stwa tr®jk©t®w mamy: c'/c =  b'/b. St©d c' = cÔb'/b.
\next
\lose 2
Korzystaj©c z twierdzenia Pitagorasa w tr®jk©cie ABC dostajemy:
b = ¸(c¾-a¾) = ¸(17¾-8¾) = 15.
\next
\lose 1
Ostatecznie: c' = cÔb'/b = 17Ô&1/15 = ?
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
W tr®jk©cie prostok©tnym jeden z k©t®w ostrych ma miar« ¶. Oblicz obw®d
tego tr®jk©ta wiedz©c, ±e jego pole jest r®wne &1 cm¾, a tg ¶ = &2.
\end text
\begin params
3x5
30
2.4
12
5
30
15
1.2
6
5
11+s(61)
16
2
8
4
12+s(80)
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
W tr®jk©cie prostok©tnym jeden z k©t®w ostrych ma miar« ¶. Oblicz obw®d
tego tr®jk©ta wiedz©c, ±e jego pole jest r®wne &1 cm¾, a tg ¶ = &2.
\end text
\points 18
\begin answer
\evaluate
\length 10
\correct &5
\wrong 4
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.2,0.1,0.8
\draw 0.1,0.2,0.9,0.8
\draw 0.1,0.8,0.9,0.8
\text 0.1,0.14,B
\text 0.9,0.81,A
\text 0.1,0.81,C
\text 0.5,0.82,b
\text 0.50,0.40,c
\text 0.75,0.71,¶
\text 0.03,0.5,a
\text 0.5,0.81,b
\window 1,11,79,23
\lose 4
\next
Niech k©t CAB ma miar« ¶. Mamy dane: pole tr®jk©ta P = &1 oraz tg¶ = &2. 
Szukamy obwodu tr®jk©ta o = a + b + c.
\next
\lose 2
Ze wzoru na pole tr®jk©ta dostajemy P = ÀaÔb. Ponadto tg ¶ = a/b.
\next
\lose 4
Rozwi©zujemy uk¬ad r®wna­ ze wzgl«du na a i b i mamy: a = &2Ôb, podstawiamy 
i mamy &1 = ÀÔ&2Ôb¾. Wyliczamy b, potem a.
\next
\lose 3
     b = &4, a = &3
\next
\lose 2
Z twierdzenia Pitagorasa wynika, ±e c = ¸(a¾ + b¾).
\next
\lose 2
Ostatecznie o = a + b + ¸(a¾ + b¾) = &3 + &4 + ¸(&3¾ + &4¾) = ?
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
W tr®jk©cie ABC, w kt®rym ÏABÏ = &1 cm, wysoko¯ª ÏCC'Ï = &2 cm
przeprowadzono odcinek EF r®wnoleg¬y do boku AB i odleg¬y od niego
o &3 cm. Oblicz pole tr®jk©ta EFC.
\end text
\begin params
4x6
12
15
5
10
8
40
24
12
4
8
16
64
10
12
6
6
5
15
30
12
8
4
10
20
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
W tr®jk©cie ABC, w kt®rym ÏABÏ = &1 cm, wysoko¯ª ÏCC'Ï = &2 cm
przeprowadzono odcinek EF r®wnoleg¬y do boku AB i odleg¬y od niego
o &3 cm. Oblicz pole tr®jk©ta EFC.
\end text
\points 12
\begin answer
\evaluate
\length 5
\correct &6
\wrong 3
\end answer
\begin help
\window 1,3,21,11
\scale 0,0,1,1
\draw 0.33,0,0,1
\draw 0,1,1,1
\draw 1,1,0.33,0
\draw 0.11,0.66,0.77,0.66
\dotted
\draw 0.33,0,0.33,1
\text 0,1.01,A
\text 1,1.01,B
\text 0.33,-0.1,C
\text 0.33,1.01,C'
\text 0.06,0.56,E
\text 0.77,0.56,F
\text 0.34,0.56,M
\window 1,12,79,23
\lose 4
\next
\lose 0
Niech ÏABÏ = &1, ÏCC'Ï = &2, ÏMC'Ï = &3. Szukamy pola P tr®jk©ta EFC.
\next
\lose 1
Oczywi¯cie P = ÀÔÏEFÏÔÏCMÏ oraz ÏCMÏ = ÏCC'Ï - ÏMC'Ï = &2 - &3 = &4.
\next
\lose 2
Zauwa±my, ±e tr®jk©ty ABC i EFC s© podobne.
\next
\lose 2
W szczeg®lno¯ci, ÏEFÏ/ÏABÏ = ÏCMÏ/ÏCC'Ï,
\next
\lose 2
a st©d: ÏEFÏ = ÏABÏÔÏCMÏ/ÏCC'Ï = &1Ô&4/&2 = &5. 
Ostatecznie: P = ÀÔÏEFÏÔÏCMÏ = ?
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
Obw®d tr®jk©ta r®wnoramiennego wynosi &1 cm. Podstawa ma d¬ugo¯ª &2 cm.
Oblicz pole tr®jk©ta.
\end text
\begin params
3x6
40
10
15
5
10¸(2)
50s(2)
24
6
9
3
6¸(2)
18s(2)
36
16
10
8
6
48
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
Obw®d tr®jk©ta r®wnoramiennego wynosi &1 cm. Podstawa ma d¬ugo¯ª &2 cm.
Oblicz pole tr®jk©ta.
\end text
\points 12
\begin answer
\evaluate
\length 8
\correct &6
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.5,0.1,0.1,0.9
\draw 0.5,0.1,0.9,0.9
\draw 0.1,0.9,0.9,0.9
\dotted
\draw 0.5,0.1,0.5,0.9
\text 0.1,0.91,A
\text 0.9,0.91,B
\text 0.5,0.02,C
\text 0.51,0.8,C'
\text 0.2,0.50,a
\text 0.775,0.50,a
\text 0.52,0.50,h
\text 0.5,0.91,c
\window 1,11,79,23
\lose 3
\next
\lose 0
Dane s©: obw®d o = &1 tr®jk©ta ABC i ÏABÏ = c = &2, 
     a szukamy pola P tego tr®jk©ta.
\next
\lose 3
Oczywi¯cie, P = ÀcÔh. Zauwa±my, ±e a = À(o - c) = &3.
\next
\lose 4
Korzystaj©c z twierdzenia Pitagorasa w tr®jk©cie CAC' mamy: 
     h = ¸(a¾ - (Àc)¾) = ¸(&3¾ - &4¾) = &5. 
\next
\lose 1
Ostatecznie, P = ÀcÔh = ÀÔ&2Ô&5 = ?
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
Obw®d tr®jk©ta r®wnoramiennego wynosi &1 cm. D¬ugo¯ª ramienia jest o 30% 
wi«ksza od d¬ugo¯ci podstawy. Oblicz pole tr®jk©ta.
\end text
\begin params
3x3
36
10
60
18
5
15
54
15
135
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
Obw®d tr®jk©ta r®wnoramiennego wynosi &1 cm. D¬ugo¯ª ramienia jest o 30%
wi«ksza od d¬ugo¯ci podstawy. Oblicz pole tr®jk©ta.
\end text
\points 12
\begin answer
\evaluate
\length 5
\correct &3
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.5,0.1,0.1,0.9
\draw 0.5,0.1,0.9,0.9
\draw 0.1,0.9,0.9,0.9
\dotted
\draw 0.5,0.1,0.5,0.9
\text 0.1,0.91,A
\text 0.9,0.91,B
\text 0.5,0.02,C
\text 0.51,0.8,C'
\text 0.2,0.50,a
\text 0.775,0.50,a
\text 0.53,0.50,h
\text 0.5,0.91,c
\window 1,11,79,23
\lose 3
\next
\lose 2
Mamy dany obw®d o = &1 tr®jk©ta ABC oraz wiemy, ±e a = 1.3Ôc. 
Szukamy pola P tego tr®jk©ta.
\next
\lose 3
Oczywi¯cie, P = ÀcÔh. 
Zauwa±my, ±e o = c + 2a = c + 2Ô1.3Ôc = 3.6Ôc, a st©d c = &2.
\next
\lose 3
Z twierdzenia Pitagorasa w tr®jk©cie CAC' wynika, ±e 
     h = ¸(a¾ -(Àc)¾) = ¸((1.3Ôc)¾ - (Àc)¾) = 1.2Ôc. 
Ostatecznie, P = ÀcÔh = 0.6Ôc¾ = 0.6Ô&2¾ = ?
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
Dwa tr®jk©ty prostok©tne s© podobne w skali 1:3. Suma d¬ugo¯ci 
przyprostok©tnych pierwszego tr®jk©ta wynosi &1 cm, za¯ r®±nica d¬ugo¯ci 
przyprostok©tnych drugiego r®wna si« &2 cm. 
Oblicz pole mniejszego z tych tr®jk©t®w.
\end text
\begin params
4x3
8
6
15/2
10
12
21/2
12
6
35/2
10
6
12
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
Dwa tr®jk©ty prostok©tne s© podobne w skali 1:3. 
Suma d¬ugo¯ci przyprostok©tnych pierwszego tr®jk©ta wynosi &1 cm, 
za¯ r®±nica d¬ugo¯ci przyprostok©tnych drugiego r®wna si« &2 cm. 
Oblicz pole mniejszego z tych tr®jk©t®w.
\end text
\points 11
\begin answer
\evaluate
\length 5
\correct &3
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.3,0.1,0.9
\draw 0.1,0.9,0.9,0.9
\draw 0.1,0.3,0.9,0.9
\draw 0.63,0.1,0.63,0.3
\draw 0.63,0.3,0.9,0.3
\draw 0.9,0.3,0.63,0.1
\text 0.9,0.91,A'
\text 0.1,0.2,B'
\text 0.1,0.91,C'
\text 0.9,0.32,A
\text 0.63,0.0,B
\text 0.63,0.32,C
\text 0.02,0.55,a'
\text 0.5,0.91,b'
\text 0.55,0.55,c'
\text 0.58,0.15,a
\text 0.77,0.32,b
\text 0.82,0.15,c
\window 1,11,79,23
\lose 3
\next
\lose 2
Przyjmijmy, ±e d¬u±sz© przyprostok©tn© tr®jk©ta A'B'C' jest A'C'. 
Z warunk®w zadania dostajemy: a + b = &1, b' - a' = &2. 
Mamy obliczyª pole P = ÀaÔb tr®jk©ta ABC. 
\next
\lose 3
Z podobie­stwa tr®jk©t®w wnioskujemy, ±e a' = 3a i b' = 3b. 
Tak wi«c: a + b = &1 i 3b - 3a = &2.
\next
\lose 2
Rozwi©zuj©c ten uk¬ad r®wna­ wyliczamy a i b, 
a nast«pnie obliczamy P = ÀaÔb = ?
\end help
\end question
\end exercise
\type tro
\title Tr®jk©t
\begin exercise
\begin text
\cont
Oblicz d¬ugo¯ª podstawy tr®jk©ta r®wnoramiennego kt®rego rami« ma 
d¬ugo¯ª &1 cm i tworzy z podstaw© k©t o mierze 30².
\end text
\begin params
4x2
4
4s(3)
¸(3)
3
¸(2)
s(6)
¸(27)
9
\end params
\begin question
\title Tr®jk©t
\begin text
\cont
Oblicz d¬ugo¯ª podstawy tr®jk©ta r®wnoramiennego kt®rego rami« ma 
d¬ugo¯ª &1 cm i tworzy z podstaw© k©t o mierze 30².
\end text
\points 8
\begin answer
\evaluate
\length 8
\correct &2
\wrong 2
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\text 0.15,0.8,¶
\draw 0.5,0.5,0.1,0.9
\draw 0.5,0.5,0.9,0.9
\draw 0.9,0.9,0.1,0.9
\dotted
\draw 0.5,0.5,0.5,0.9
\text 0.1,0.91,A
\text 0.9,0.91,B
\text 0.5,0.4,C
\text 0.51,0.8,C'
\text 0.2,0.67,a
\text 0.775,0.67,a
\text 0.5,0.91,c
\solid
\draw 1.5,0.5,1.95,0.5
\draw 1.95,0.5,1.95,0.1
\draw 1.95,0.1,1.5,0.5
\dotted
\draw 1.5,0.5,1.95,0.9
\draw 1.95,0.9,1.95,0.5
\text 1.45,0.5,A
\text 1.97,0.5,C'
\text 1.95,0.9,M
\text 1.95,0.0,C
\text 1.6,0.4,¶
\text 1.7,0.5,Àc
\text 1.65,0.2,a
\window 1,11,79,23
\lose 3
\next
\lose 0
Mamy dane: a = &1 i k©t ¶ = 30². Chcemy obliczyª ÏABÏ = c.
\next
\lose 3
Rozwa±my tr®jk©t CAC'. Poniewa± ¶ = 30², 
     zatem tr®jk©t ten jest "po¬ow©" tr®jk©ta r®wnobocznego MCA o boku a. 
St©d, korzystaj©c ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, mamy 
     Àc = Àa¸(3),
\next
\lose 1
czyli 
     c = aÔ¸(3) = ?
\end help
\end question
\end exercise