\type tra
\title Trapez
\begin exercise
\begin text
\cont
Dany jest trapez r®wnoramienny ABCD. Oblicz pole tego trapezu wiedz©c,
±e k©t przy podstawie wynosi 30², ÏABÏ = &1 cm, a d¬ugo¯ª podstawy CD
jest 1/3 d¬ugo¯ci podstawy AB.
\end text
\begin params
4x3
12
32s(3)/3
4
3
2s(3)/3
1
9
6s(3)
3
6
8s(3)/3
2
\end params
\begin question
\title Trapez
\begin text
\cont
Dany jest trapez r®wnoramienny ABCD. Oblicz pole tego trapezu wiedz©c,
±e k©t przy podstawie wynosi 30², ÏABÏ = &1 cm, a d¬ugo¯ª podstawy CD
jest 1/3 d¬ugo¯ci podstawy AB.
\end text
\points 15
\begin answer
\evaluate
\length 12
\correct &2
\wrong 4
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,1.0,0.9
\draw 1.0,0.9,0.6,0.6
\draw 0.6,0.6,0.5,0.6
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\text 1.0,0.9,B
\text 0.63,0.5,C
\text 0.47,0.5,D
\text 0.6,0.89,a
\text 0.55,0.5,b
\text 0.5,0.9,D'
\text 0.52,0.7,h
\text 0.2,0.8,¶
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\text 1.95,0.5,D'
\text 1.9,0.1,D
\text 1.9,0.8,M
\text 1.6,0.4,¶
\text 1.92,0.3,h
\window 1,11,79,23
\lose 4
\next
\lose 2
Mamy dane ÏABÏ = a = &1, ÏCDÏ = b = Âa oraz k©t BAD = ¶ = 30². 
Niech DD' oznacza wysoko¯ª trapezu; h = ÏDD'Ï. Chcemy obliczyª pole trapezu 
P = À(a + b)h = À(a + Âa)Ôh = ÃaÔh. Trzeba obliczyª h.
\next
\lose 3
Zauwa±my, ±e trapez jest r®wnoramienny, a wi«c 
ÏAD'Ï = À(a - b) = À(a - Âa) = &3.
\next
\lose 4
Rozwa±my tr®jk©t DAD'. Poniewa± ¶ = 30², zatem jest on "po¬ow©" tr®jk©ta 
r®wnobocznego MDA o boku 2h. St©d      ÏAD'Ï = h¸(3), czyli h = &3/¸(3).
\next
\lose 1
Ostatecznie, 
     P = ÃaÔh = ?
\end help
\end question
\end exercise
\type tra
\title Trapez
\begin exercise
\begin text
\cont
Podstawy trapezu r®wnoramiennego maj© d¬ugo¯ª &1 cm i &2 cm. 
Rami« trapezu tworzy z d¬u±sz© podstaw© k©t &4. 
Oblicz pole tego trapezu.
\end text
\begin params
4x6
10
6
16s(3)/3
30²
2
2/¸(3)
5
4
9/4
45²
À
À
8
2
15s(3)
60²
3
3¸(3)
8
4
4s(3)
30²
2
2/¸(3)
\end params
\begin question
\title Trapez
\begin text
\cont
Podstawy trapezu r®wnoramiennego maj© d¬ugo¯ª &1 cm i &2 cm. 
Rami« trapezu tworzy z d¬u±sz© podstaw© k©t &4. 
Oblicz pole tego trapezu.
\end text
\points 15
\begin answer
\evaluate
\length 12
\correct &3
\wrong 4
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,1.0,0.9
\draw 1.0,0.9,0.6,0.5
\draw 0.6,0.5,0.5,0.5
\draw 0.5,0.5,0.1,0.9
\dotted
\draw 0.5,0.4,0.5,0.9
\text 0.1,0.9,A
\text 1.0,0.9,B
\text 0.63,0.4,C
\text 0.47,0.4,D
\text 0.6,0.89,a
\text 0.55,0.4,b
\text 0.5,0.9,D'
\text 0.52,0.7,h
\text 0.2,0.8,¶
\solid
\draw 1.5,0.5,1.9,0.5
\draw 1.5,0.5,1.9,0.2
\draw 1.9,0.5,1.9,0.2
\text 1.45,0.5,A
\text 1.7,0.5,d
\text 1.95,0.5,D'
\text 1.9,0.1,D
\text 1.6,0.4,¶
\text 1.92,0.3,h
\window 1,11,79,23
\lose 3
\next
\lose 2
Niech h oznacza wysoko¯ª trapezu; h = ÏDD'Ï. 
Mamy dane ÏABÏ = a = &1, ÏCDÏ = b = &2 oraz k©t BAD = ¶ = &4. 
Chcemy obliczyª pole trapezu P = À(a + b)h. Brakuje nam h.
\next
\lose 3
Poniewa± trapez jest r®wnoramienny, zatem 
     ÏAD'Ï = À(a - b) = &5.
\next
\lose 4
Rozwa±my tr®jk©t AD'D i oznaczmy d = ÏAD'Ï. 
Widaª, ±e
\half
     h
     È = tg &4,   czyli  h = tg &4Ôd = &6
     d
\cr
\full
\next
\lose 1
Ostatecznie, 
     P = À(a + b)h = ?
\end help
\end question
\end exercise
\type tra
\title Trapez
\begin exercise
\begin text
\cont
Oblicz pole trapezu, kt®rego boki r®wnoleg¬e maj© d¬ugo¯ci &1 cm 
i &2 cm, a nier®wnoleg¬e &3 cm i &4 cm.
\end text
\begin params
2x9
16
44
17
25
30
28
12
20
15
2
7
¸(10)
5
4.5
5
3
4
3
\end params
\begin question
\title Trapez
\begin text
\cont
Oblicz pole trapezu, kt®rego boki r®wnoleg¬e maj© d¬ugo¯ci &1 cm i &2 cm, 
a nier®wnoleg¬e &3 cm i &4 cm.
\end text
\points 22
\begin answer
\evaluate
\length 5
\correct &5*&9
\wrong 5
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\dotted
\draw 0.9,0.1,0.9,0.9
\text 0.88,0.9,C'
\text 0.82,0.5,h
\text 0.99,0.5,c
\text 0.13,0.5,d
\solid
\draw 0.1,0.9,1.0,0.9
\draw 1.0,0.9,0.9,0.1
\draw 0.9,0.1,0.3,0.1
\draw 0.3,0.1,0.1,0.9
\dotted
\draw 0.3,0.1,0.3,0.9
\text 0.1,0.9,A
\text 1.0,0.9,B
\text 0.9,0.0,C
\text 0.3,0.0,D
\text 0.6,0.89,a
\text 0.6,0.0,b
\text 0.3,0.9,D'
\text 0.32,0.5,h
\window 1,11,79,23
\lose 4
\cont
\next
\lose 2
Mamy dane ÏABÏ = a = &2, ÏCDÏ = b = &1, ÏBCÏ = c = &3, ÏADÏ = d = &4. 
Szukamy pola trapezu P = À(a + b)h = À(&2 + &1)h = &5h. 
\next
\lose 3
Rozwa±my dwie wysoko¯ci CC' i DD' oraz dwa tr®jk©ty DAD' i BCC'. 
Z twierdzenia Pitagorasa wynika, ±e d¾ = ÏAD'Ͼ + h¾ oraz c¾ = ÏC'BϾ + h¾.
\next
\lose 6
Odejmuj©c stronami mamy: 
d¾ - c¾ = ÏAD'Ͼ - ÏC'BϾ = (ÏAD'Ï - ÏC'BÏ)(ÏAD'Ï + ÏC'BÏ). 
Zauwa±my, ±e ÏAD'Ï + ÏC'BÏ = a - b = &6. 
Tak wi«c, ÏAD'Ï - ÏC'BÏ = (d¾ - c¾)/(a - b) = (&4¾ - &3¾)/&6 = &7.
\next
\lose 3
Doszli¯my do uk¬adu r®wna­: ÏAD'Ï + ÏC'BÏ = &6, ÏAD'Ï - ÏC'BÏ = &7. 
St©d ÏAD'Ï = À(&6 + &7) = &8.
\next
\lose 3
Z twierdzenia Pitagorasa mamy: h = ¸(d¾ - ÏAD'Ͼ) = ¸(&4¾ - &8¾) = &9. 
Ostatecznie, P = &5h = &5Ô&9 = ?
\end help
\end question
\end exercise
\type tra
\title Trapez
\begin exercise
\begin text
\cont
W trapezie o polu &1 cm¾ wysoko¯ª ma d¬ugo¯ª &2 cm, a r®±nica d¬ugo¯ci 
r®wnoleg¬ych bok®w wynosi &3 cm. 
Oblicz d¬ugo¯ª d¬u±szego z r®wnoleg¬ych bok®w.
\end text
\begin params
2x5
594
22
6
54
30
600
20
10
60
35
\end params
\begin question
\title Trapez
\begin text
\cont
W trapezie o polu &1 cm¾ wysoko¯ª ma d¬ugo¯ª &2 cm,
a r®±nica d¬ugo¯ci r®wnoleg¬ych bok®w wynosi &3 cm.
Oblicz d¬ugo¯ª d¬u±szego z r®wnoleg¬ych bok®w.
\end text
\points 12
\begin answer
\evaluate
\length 5
\correct &5
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,1.0,0.9
\draw 1.0,0.9,0.9,0.1
\draw 0.9,0.1,0.3,0.1
\draw 0.3,0.1,0.1,0.9
\dotted
\draw 0.3,0.1,0.3,0.9
\text 0.1,0.9,A
\text 1.0,0.9,B
\text 0.9,0.0,C
\text 0.3,0.0,D
\text 0.6,0.89,a
\text 0.6,0.0,b
\text 0.3,0.9,D'
\text 0.32,0.5,h
\window 1,11,79,23
\lose 3
\next
\lose 1
Niech ÏABÏ = a, ÏCDÏ = b. 
Mamy dane pole P = &1, ÏDD'Ï = h = &2 oraz wiemy, ±e a - b = &3. 
Chcemy obliczyª a.
\next
\lose 4
Poniewa± P = À(a + b)h, zatem a + b = 2P/h = 2Ô&1/&2 = &4.
\next
\lose 3
Dostali¯my uk¬ad r®wna­ a - b = &3, a + b = &4, z kt®rego a = À(&3 + &4) = ?
\end help
\end question
\end exercise