\type gra
\title Graniastos¬up
\begin exercise
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy 
ma d¬ugo¯ª 2 cm,a przek©tna graniastos¬upa tworzy z podstaw© k©t 60².
Oblicz obj«to¯ª graniastos¬upa.
\end text
\begin question
\title Graniastos¬up
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy
 ma d¬ugo¯ª 2 cm,a przek©tna graniastos¬upa tworzy z podstaw© k©t 60².
Oblicz obj«to¯ª graniastos¬upa.
\end text
\points 15
\begin answer
\evaluate
\length 8
\correct 8s(6)
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\draw 0.1,0.9,0.1,0.2
\draw 0.1,0.2,0.6,0.2
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\dotted
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\text 0.4,0.9,a
\text 0.8,0.8,a
\text 0.92,0.3,h
\text 0.2,0.77,¶
\solid
\draw 2.0,0.9,2.6,0.9
\draw 2.6,0.9,2.6,0.2
\draw 2.6,0.2,2.0,0.9
\text 2.0,0.9,A
\text 2.6,0.9,C
\text 2.6,0.1,C'
\text 2.1,0.79,¶
\text 2.62,0.5,h
\text 3.2,0.9,M
\dotted
\draw 3.2,0.9,2.6,0.2
\draw 2.6,0.9,3.2,0.9
\window 1,11,79,23
\lose 5
\next
\lose 2
Mamy dan© kraw«d° podstawy a = 2 oraz k©t CAC' = ¶ = 60². 
Niech h oznacza wysoko¯ª graniastos¬upa. Mamy obliczyª obj«to¯ª V = a¾h.
\next
\lose 2
Rozwa±my tr®jk©t CAC'. Poniewa± ¶ = 60², 
zatem tr®jk©t ten jest "po¬ow©" tr®jk©ta r®wnobocznego AMC' o boku 2ÏACÏ. 
\next
\lose 2
Zauwa±my, ±e AC jest przek©tn© kwadratu o boku a, a zatem ÏACÏ = a¸(2). 
\next
\lose 2
Zatem, ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, dostajemy: 
     h = À(2ÏACÏ)¸(3) = ÏACϸ(3) = a¸(2)¸(3) = a¸(6). 
\next 
\lose 1
Ostatecznie, V = a¾h = a¿¸(6) = ?
\end help
\end question
\end exercise
\begin exercise
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy
 ma d¬ugo¯ª 5¸(2) cm, a przek©tna graniastos¬upa tworzy
 z podstaw© k©t 60². Oblicz obj«to¯ª 
graniastos¬upa.
\end text
\begin question
\title Graniastos¬up
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy
 ma d¬ugo¯ª 5¸(2) cm, a przek©tna graniastos¬upa tworzy
 z podstaw© k©t 60². Oblicz obj«to¯ª graniastos¬upa.
\end text
\points 15
\begin answer
\evaluate
\length 10
\correct 500s(3)
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\draw 0.1,0.9,0.1,0.2
\draw 0.1,0.2,0.6,0.2
\draw 0.6,0.2,0.6,0.9
\draw 0.6,0.9,0.9,0.7
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\dotted
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\text 0.6,0.9,B
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\text 0.91,0.0,C'
\text 0.4,0.9,a
\text 0.8,0.8,a
\text 0.92,0.3,h
\text 0.2,0.77,¶
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\draw 2.6,0.9,2.6,0.2
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\text 2.0,0.9,A
\text 2.6,0.9,C
\text 2.6,0.1,C'
\text 2.1,0.79,¶
\text 2.62,0.5,h
\text 3.2,0.9,M
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\dotted
\draw 2.6,0.9,3.2,0.9
\dotted
\window 1,11,79,23
\lose 5
\next
\lose 2
Mamy dan© kraw«d° podstawy a = 5¸(2) oraz k©t CAC' = ¶ = 60². 
Niech h oznacza wysoko¯ª graniastos¬upa. Mamy obliczyª obj«to¯ª V = a¾h.
\next
\lose 2
Rozwa±my tr®jk©t CAC'. Poniewa± ¶ = 60², 
zatem tr®jk©t ten jest "po¬ow©" tr®jk©ta r®wnobocznego AMC' o boku 2ÏACÏ. 
\next
\lose 2
Zauwa±my, ±e AC jest przek©tn© kwadratu o boku a, a zatem ÏACÏ = a¸(2). 
\next
\lose 2
Zatem, ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, dostajemy: 
     h = À(2ÏACÏ)¸(3) = ÏACϸ(3) = a¸(2)¸(3) = a¸(6). 
\next 
\lose 1
Ostatecznie, V = a¾h = a¿¸(6) = ?
\end help
\end question
\end exercise
\begin exercise
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy 
ma d¬ugo¯ª 3 cm, a przek©tna graniastos¬upa tworzy 
z podstaw© k©t 30². Oblicz obj«to¯ª graniastos¬upa.
\end text
\begin question
\title Graniastos¬up
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy 
ma d¬ugo¯ª 3 cm, a przek©tna graniastos¬upa tworzy 
z podstaw© k©t 30². Oblicz obj«to¯ª graniastos¬upa.
\end text
\points 15
\begin answer
\evaluate
\length 8
\correct 9s(6)
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\draw 0.1,0.9,0.1,0.2
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\dotted
\draw 2.0,0.45,2.6,0.9
\draw 2.6,0.45,2.6,0.9
\window 1,11,79,23
\lose 5
\next
\lose 2
Mamy dan© kraw«d° podstawy a = 3 oraz k©t CAC' = ¶ = 30². 
Niech h oznacza wysoko¯ª graniastos¬upa. Mamy obliczyª obj«to¯ª V = a¾h.
\next
\lose 2
Rozwa±my tr®jk©t CAC'. Poniewa± ¶ = 30², zatem tr®jk©t ten jest 
"po¬ow©" tr®jk©ta r®wnobocznego AMC' o wysoko¯ci ÏACÏ i boku 2h 
\next
\lose 2
Zauwa±my, ±e AC jest przek©tn© kwadratu o boku a, a zatem ÏACÏ = a¸(2). 
\next
\lose 2
Zatem, ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, dostajemy:
     ÏACÏ = À(2h)¸(3), 
wi«c h = ÏACÏ/¸(3) = a¸(2)/¸(3) = a¸(6)/3 = ¸(6). 
\next 
\lose 1
Ostatecznie, V = a¾h = ?
\end help
\end question
\end exercise
\begin exercise
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy
 ma d¬ugo¯ª ¸(3) cm, a przek©tna graniastos¬upa tworzy
 z podstaw© k©t 30². Oblicz obj«to¯ª graniastos¬upa.
\end text
\begin question
\title Graniastos¬up
\begin text
\cont
W graniastos¬upie prawid¬owym czworok©tnym kraw«d° podstawy
 ma d¬ugo¯ª ¸(3) cm, a przek©tna graniastos¬upa tworzy
 z podstaw© k©t 30². Oblicz obj«to¯ª graniastos¬upa.
\end text
\points 15
\begin answer
\evaluate 
\length 10
\correct 3s(2)
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\draw 0.1,0.9,0.1,0.2
\draw 0.1,0.2,0.6,0.2
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\draw 0.6,0.9,0.9,0.7
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\draw 2.6,0.45,2.6,0.9
\window 1,11,79,23
\lose 5
\next
\lose 2
Mamy dan© kraw«d° podstawy a = ¸(3) oraz k©t CAC' = ¶ = 30². 
Niech h oznacza wysoko¯ª graniastos¬upa. Mamy obliczyª obj«to¯ª V = a¾h.
\next
\lose 2
Rozwa±my tr®jk©t CAC'. Poniewa± ¶ = 30², zatem tr®jk©t ten jest 
"po¬ow©" tr®jk©ta r®wnobocznego AMC' o wysoko¯ci ÏACÏ i boku 2h 
\next
\lose 2
Zauwa±my, ±e AC jest przek©tn© kwadratu o boku a, a zatem ÏACÏ = a¸(2). 
\next
\lose 2
Zatem, ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, dostajemy:
     ÏACÏ = À(2h)¸(3), 
wi«c h = ÏACÏ/¸(3) = a¸(2)/¸(3) = ¸(2). 
\next 
\lose 1
Ostatecznie, V = a¾h = ?
\end help
\end question
\end exercise
\type gra
\title Graniastos¬up
\begin exercise
\begin text
\cont
Podstaw© graniastos¬upa prostego jest romb o d¬ugo¯ci boku &1 cm 
i k©cie ostrym ¶ = 60². Wysoko¯ª graniastos¬upa ma d¬ugo¯ª &2.
Obliczyª obj«to¯ª tego graniastos¬upa.
\end text
\begin params
4x4
2
5¸(3)
30
¸(3)
1
¸(3)
3/2
¸(3)/2
2
¸(2)
2s(6)
¸(3)
¸(3)
2¸(3)
9
3/2
\end params
\begin question
\title Graniastos¬up
\begin text
\cont
Podstaw© graniastos¬upa prostego jest romb o d¬ugo¯ci boku &1 cm 
i k©cie ostrym ¶ = 60². Wysoko¯ª graniastos¬upa ma d¬ugo¯ª &2.
Obliczyª obj«to¯ª tego graniastos¬upa.
\end text
\points 12
\begin answer
\evaluate
\length 8
\correct &3
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\draw 0.1,0.9,0.1,0.3
\draw 0.1,0.3,0.6,0.3
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\dotted
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\text 3.6,0.2,C
\text 3.0,0.9,B
\window 1,11,79,23
\lose 4
\next
\lose 1
Mamy dany bok podstawy a = &1, k©t BAD = ¶ = 60² oraz wysoko¯ª
ÏCC'Ï = h = &2. 
Mamy obliczyª obj«to¯ª graniastos¬upa V = PpÔh, gdzie Pp oznacza
pole podstawy.
\next
\lose 2
Niech w oznacza wysoko¯ª rombu ABCD. Mamy Pp = aÔw.
\next
\lose 3
Poniewa± ¶ = 60², zatem w jest wysoko¯ci© tr®jk©ta r®wnobocznego ABD o boku a. 
Wynika st©d, ±e w = Àa¸(3) = &4.
\next
\lose 1
Ostatecznie, V = aÔwÔh = ?
\end help
\end question
\end exercise
\type gra
\title Graniastos¬up
\begin exercise
\begin text
\cont
Podstaw© graniastos¬upa prostego jest r®wnoleg¬obok 
o bokach d¬ugo¯ci &1 cm i &2 cm i k©cie 30². 
Oblicz obj«to¯ª tego graniastos¬upa wiedz©c,
±e jego pole powierzchni ca¬kowitej wynosi &3 cm¾.
\end text
\begin params
4x5
6
4
72
12/5
144/5
3
2
18
6/5
3.6
4
2
24
4/3
16/3
5
3
39
3/2
45/4
\end params
\begin question
\title Graniastos¬up
\begin text
\cont
Podstaw© graniastos¬upa prostego jest r®wnoleg¬obok o bokach d¬ugo¯ci
&1 cm i &2 cm i k©cie 30². 
Oblicz obj«to¯ª tego graniastos¬upa wiedz©c, ±e jego 
pole powierzchni ca¬kowitej wynosci &3 cm¾.
\end text
\points 15
\begin answer
\length 8
\evaluate
\correct &5
\wrong 4
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\dotted
\draw 0.4,0.8,0.9,0.8
\solid
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\draw 0.1,0.9,0.1,0.3
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\draw 0.9,0.8,0.9,0.2
\dotted
\draw 0.4,0.8,0.4,0.2
\solid
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\draw 0.6,0.3,0.9,0.2
\dotted
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\solid
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\text 0.1,0.9,A
\text 0.6,0.9,B
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\text 0.92,0.5,h
\text 0.9,0.1,C'
\text 0.25,0.8,¶
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\dotted
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\solid
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\text 2.5,0.9,M
\text 2.5,0.4,D
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\text 2.52,0.7,w
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\text 3.5,0.4,C
\text 2.6,0.9,B
\window 1,11,79,23
\lose 5
\next
\cont
Mamy dane ÏABÏ = a = &1, ÏADÏ = b = &2, k©t BAD = ¶ = 30² 
oraz pole powierzchni ca¬kowitej graniastos¬upa Pc = &3. 
Niech h oznacza wysoko¯ª graniastos¬upa. 
Mamy obliczyª obj«to¯ª V = PpÔh, gdzie Pp oznacza pole podstawy.
\lose 2
Niech DM oznacza wysoko¯ª r®wnoleg¬oboku b«d©cego podstaw© graniastos¬upa; 
w = ÏDMÏ. Mamy Pp = aÔw.
\next
\lose 3
Rozwa±my tr®jk©t DAM. 
Poniewa± ¶ = 30², zatem w = Àb. 
Tak wi«c V = PpÔh = (aÔw)h = a(Àb)h = ÀaÔbÔh. Pozostaje obliczyª h.
\next
\lose 3
Zauwa±my, ±e Pc = 2(aÔw + aÔh + bÔh) = 2(ÀaÔb + aÔh + bÔh), 
sk©d h = À(Pc - aÔb)/(a + b) = À(&3 - &1Ô&2)/(&1 + &2) = &4. 
\next
\lose 1
Ostatecznie, V = ÀaÔbÔh = ?
\end help
\end question
\end exercise
\type gra
\title Graniastos¬up
\begin exercise
\begin text
\cont
Podstaw© graniastos¬upa prostego jest r®wnoleg¬obok
o bokach &1 cm i &2 cm oraz k©cie mi«dzy nimi 120². 
Kraw«d° boczna tego graniastos¬upa ma d¬ugo¯ª &3 cm. 
Oblicz obj«to¯ª graniastos¬upa.
\end text
\begin params
4x5
15
12
30
2700s(3)
6¸(3)
4
2
6
24s(3)
¸(3)
6
8
24
576s(3)
4¸(3)
1
6
18
54s(3)
3¸(3)
\end params
\begin question
\title Graniastos¬up
\begin text
Podstaw© graniastos¬upa prostego jest r®wnoleg¬obok 
o bokach &1 cm i &2 cm oraz k©cie mi«dzy nimi 120². 
Kraw«d° boczna tego graniastos¬upa ma d¬ugo¯ª &3 cm. 
\cr
Oblicz obj«to¯ª graniastos¬upa.
\end text
\points 12
\begin answer
\evaluate
\length 8
\correct &4
\wrong 3
\end answer
\begin help
\lose 4
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\draw 0.1,0.9,0.1,0.3
\draw 0.1,0.3,0.6,0.3
\draw 0.6,0.3,0.6,0.9
\draw 0.6,0.9,0.9,0.7
\draw 0.9,0.7,0.9,0.1
\draw 0.9,0.1,0.4,0.1
\draw 0.4,0.1,0.1,0.3
\draw 0.6,0.3,0.9,0.1
\dotted
\draw 0.1,0.9,0.4,0.7
\draw 0.4,0.7,0.4,0.1
\draw 0.4,0.7,0.9,0.7
\text 0.1,0.9,A
\text 0.6,0.9,B
\text 0.9,0.7,C
\text 0.4,0.7,D
\text 0.9,0.01,C'
\text 0.4,0.9,a
\text 0.8,0.8,b
\text 0.92,0.4,h
\text 0.55,0.8,¶
\solid
\draw 2.0,0.9,3.0,0.9
\dotted
\draw 2.5,0.9,2.5,0.3
\solid
\draw 2.5,0.3,2.0,0.9
\text 2.0,0.9,A
\text 2.5,0.9,M
\text 2.5,0.2,D
\text 2.92,0.8,¶
\text 2.52,0.6,w
\draw 2.5,0.3,3.6,0.3
\draw 3.6,0.3,3.0,0.9
\text 2.1,0.8,·
\text 3.6,0.2,C
\text 3.0,0.9,B
\window 1,11,79,23
\next
\lose 1
     Mamy dane ÏABÏ = a = &1, ÏADÏ = b = &2, k©t ABC = ¶ = 120² 
oraz wysoko¯ª ÏCC'Ï = h = &3. 
     Chcemy obliczyª obj«to¯ª graniastos¬upa V = PpÔh, 
gdzie Pp oznacza pole podstawy.
\next
\lose 3
Niech DM oznacza wysoko¯ª r®wnoleg¬oboku b«d©cego podstaw© graniastos¬upa. 
Mamy Pp = aÔw.
\next
\lose 2
Rozwa±my tr®jk©t DAM. Zauwa±my, ±e · = 180² - ¶ = 180² - 120² = 60².
\next
\lose 1
Zatem tr®jk©t AMD jest po¬ow© tr®jk©ta r®wnobocznego o boku b. 
St©d w = Àb¸(3) = &5.
\next
Ostatecznie, V = aÔwÔh = ?
\end help
\end question
\end exercise
\type gra
\title Graniastos¬up
\begin exercise
\begin text
\cont
Oblicz obj«to¯ª graniastos¬upa prawid¬owego tr®jk©tnego, 
kt®rego kraw«d° podstawy ma d¬ugo¯c &1 cm 
i tworzy z przek©tn© ¯ciany bocznej k©t 60².
\end text
\begin params
4x4
8
384
16¸(3)
8¸(3)
2
6
¸(3)
2¸(3)
4
48
4¸(3)
4¸(3)
6
162
9¸(3)
6¸(3)
\end params
\begin question
\title Graniastos¬up
\begin text
\cont
Oblicz obj«to¯ª graniastos¬upa prawid¬owego tr®jk©tnego, 
kt®rego kraw«d° podstawy ma d¬ugo¯c &1 cm 
i tworzy z przek©tn© ¯ciany bocznej k©t 60².
\end text
\points 15
\begin answer
\evaluate
\length 5
\correct &2
\wrong 4
\end answer
\begin help
\lose 5
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.5,0.9
\text 0.1,0.9,A
\text 0.5,0.9,B
\draw 0.1,0.3,0.5,0.3
\text 0.5,0.2,B'
\draw 0.1,0.9,0.1,0.3
\draw 0.5,0.9,0.5,0.3
\draw 0.1,0.9,0.5,0.3
\draw 0.1,0.3,0.4,0.1
\draw 0.5,0.3,0.4,0.1
\dotted
\draw 0.4,0.1,0.4,0.7
\draw 0.1,0.9,0.4,0.7
\draw 0.5,0.9,0.4,0.7
\text 0.18,0.78,¶
\text 0.52,0.5,h
\text 0.42,0.65,C
\text 0.3,0.88,a
\solid
\draw 1.6,0.9,2.0,0.9
\dotted
\draw 2.0,0.9,2.4,0.9
\solid
\draw 2.0,0.9,2.0,0.3
\draw 1.6,0.9,2.0,0.3
\dotted
\draw 2.4,0.9,2.0,0.3
\text 1.6,0.9,A
\text 2.0,0.9,B
\text 2.4,0.9,M
\text 1.8,0.88,a
\text 2.02,0.5,h
\text 2.0,0.2,B'
\text 1.68,0.8,¶
\window 1,11,79,23
\next
\lose 1
Mamy dan© d¬ugo¯ª kraw«dzi podstawy a = &1 oraz k©t BAB' = ¶ = 60². 
     Chcemy obliczyª obj«to¯ª V = PpÔh, 
gdzie Pp oznacza pole podstawy, za¯ h - wysoko¯ª graniastos¬upa.
\next
\lose 3
Korzystaj©c ze wzoru na pole tr®jk©ta r®wnobocznego dostajemy 
     Pp = a¾¸(3)/4 = &3. 
Pozostaje obliczyª h.
\next
\lose 4
Rozwa±my tr®jk©t ABB'. Poniewa± ¶ = 60², 
zatem tr®jk©t ten jest po¬ow© tr®jk©ta r®wnobocznego AMB' o boku 2a. 
St©d h = À(2a)¸(3) = a¸(3) = &4. 
\next
\lose 1
Ostatecznie, V = PpÔh =  ?
\end help
\end question
\end exercise