\type ost
\title Ostros¬up
\begin exercise
\begin text
\cont
Oblicz obj«to¯ª ostros¬upa prawid¬owego czworok©tnego, w kt®rym wysoko¯ª
¯ciany bocznej o d¬ugo¯ci &1 cm jest nachylona do p¬aszczyzny podstawy
pod k©tem 30².
\end text
\begin params
4x4
6
108
3
6¸(3)
3
27/2
3/2
3¸(3)
4
32
2
4¸(3)
2
4
1
2¸(3)
\end params
\begin question
\title Ostros¬up
\begin text
\cont
Oblicz obj«to¯ª ostros¬upa prawid¬owego czworok©tnego, w kt®rym wysoko¯ª
¯ciany bocznej o d¬ugo¯ci &1 cm jest nachylona do p¬aszczyzny podstawy
pod k©tem 30².
\end text
\points 16
\begin answer
\length 5
\evaluate
\correct &2
\wrong 4
\end answer
\begin help
\window 1,2,21,8
\scale 0,0,1,1
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\text 2.71,0.5,M
\text 2.45,0.2,w
\text 2.55,0.38,¶
\window 1,11,79,23
\lose 5
\next
\lose 2
Niech a oznacza d¬ugo¯ª kraw«dzi podstawy ostros¬upa 
     i niech SM b«dzie wysoko¯ci© ¯ciany bocznej.
Mamy dane ÏSMÏ = w = &1 oraz k©t S'MS = ¶ = 30².
Chcemy obliczyª obj«to¯ª V = Âa¾h.
\next
\lose 4
Rozwa±my tr®jk©t MSS'.  Poniewa± ¶ = 30², zatem tr®jk©t SS'M 
jest "po¬ow©" tr®jk©ta r®wnobocznego SNM o boku w. 
St©d h = Àw = &3.
\next
\lose 3
Ponadto, ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, mamy 
     ÏS'MÏ = Àa = Àw¸(3). 
Wynika st©d, ±e a = w¸(3) = &4.
\next
\lose 1
Ostatecznie, V = Âa¾h = ?
\end help
\end question
\end exercise
\type ost
\title Ostros¬up
\begin exercise
\begin text
\cont
W ostros¬upie prawid¬owym czworok©tnym 
kraw«d° podstawy ma d¬ugo¯ª &1 cm. 
K©t nachylenia ¯ciany bocznej 
do p¬aszczyzny podstawy ma miar« 60².
Oblicz pole powierzchni ca¬kowitej ostros¬upa.
\end text
\begin params
4x2
12
432
4
48
5
75
6
108
\end params
\begin question
\title Ostros¬up
\begin text
\cont
W ostros¬upie prawid¬owym czworok©tnym kraw«d° podstawy ma d¬ugo¯ª
&1 cm. K©t nachylenia ¯ciany bocznej do p¬aszczyzny podstawy ma miar« 60².
Oblicz pole powierzchni ca¬kowitej ostros¬upa.
\end text
\points 12
\begin answer
\evaluate
\length 5
\correct &2
\wrong 3
\end answer
\begin help
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\window 1,11,79,23
\lose 4
\next
\lose 3
Niech SM oznacza wysoko¯ª ¯ciany bocznej ostro¬upa: w = ÏSMÏ.
Mamy dan© kraw«d° podstawy a = &1 oraz k©t S'MS = ¶ = 60².
Chcemy obliczyª pole powierzchni ca¬kowitej 
     Pc = a¾ + 4(ÀaÔw) = a¾ + 2aÔw.
\next
\lose 2
Rozwa±my tr®jk©t MSS'. Oczywi¯cie ÏMS'Ï = Àa.
Poniewa± ¶ = 60², zatem w = a.
\next
\lose 1
Ostatecznie, Pc = a¾ + 2aÔw = a¾ + 2a¾ = 3a¾ = ?
\end help
\end question
\end exercise
\type ost
\title Ostros¬up
\begin exercise
\begin text
\cont
W ostros¬upie prawid¬owym czworok©tnym kraw«d° podstawy ma d¬ugo¯ª
&1 cm. Kraw«d° boczna ostros¬upa tworzy z p¬aszczyzn© podstawy k©t 60².
Oblicz obj«to¯ª ostros¬upa.
\end text
\begin params
4x4
12
288s(6)
12¸(2)
6¸(6)
6
36s(6)
6¸(2)
3¸(6)
¸(6)
6
2¸(3)
3
¸(2)
s(12)/3
2
¸(3)
\end params
\begin question
\title Ostros¬up
\begin text
\cont
W ostros¬upie prawid¬owym czworok©tnym kraw«d° podstawy ma d¬ugo¯ª
&1 cm. Kraw«d° boczna ostros¬upa tworzy z p¬aszczyzn© podstawy k©t 60².
Oblicz obj«to¯ª ostros¬upa.
\end text
\points 12
\begin answer
\evaluate
\length 8
\correct &2
\wrong 4
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
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\window 1,11,79,23
\lose 3
\next
\lose 1
Mamy dan© kraw«d° podstawy a = &1 oraz k©t S'AS = ¶ = 60².
Szukamy obj«to¯ci ostros¬upa V = Âa¾h. Trzeba znale°ª h.
\next
\lose 3
Rozwa±my tr®jk©t ACS.
Oczywi¯cie AC jest przek©tn© kwadratu o boku a, zatem 
     ÏACÏ = a¸(2) = &3.
\next
\lose 2
Poniewa± ¶ = 60², zatem tr®jk©t ACS jest tr®jk©tem r®wnobocznym o boku ÏACÏ.
St©d h = ÀÏACϸ(3) = &4.
\next
\lose 1
Ostatecznie, V = Âa¾h = ?
\end help
\end question
\end exercise
\type ost
\title Ostros¬up
\begin exercise
\begin text
\cont
Oblicz pole powierzchni ca¬kowitej ostros¬upa prawid¬owego czworok©tnego,
kt®rego kraw«d° podstawy wynosi &1 cm, a k©t nachylenia kraw«dzi bocznej
do p¬aszczyzny podstawy jest r®wny 45².
\end text
\begin params
4x5
8
64(1+s(3))
4¸(2)
48
4¸(3)
4
16(1+s(3))
2¸(2)
12
2¸(3)
1
1+s(3)
¸(2)/2
3/4
¸(3)/2
3
9(1+s(3))
3¸(2)/2
27/4
3¸(3)/2
\end params
\begin question
\title Ostros¬up
\begin text
\cont
Oblicz pole powierzchni ca¬kowitej ostros¬upa prawid¬owego czworok©tnego,
kt®rego kraw«d° podstawy wynosi &1 cm, a k©t nachylenia kraw«dzi bocznej
do p¬aszczyzny podstawy jest r®wny 45².
\end text
\points 18
\begin answer
\evaluate
\length 15
\correct &2
\wrong 5
\end answer
\begin help
\lose 5
\window 1,2,21,10
\scale 0,0,1,1
\dotted
\draw 0.1,0.9,0.55,0.75
\solid
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\window 1,11,79,23
\next
\lose 1
Niech SM oznacza wysoko¯ª ¯ciany bocznej ostros¬upa; w = ÏSMÏ.
Mamy dan© kraw«d° podstawy ostros¬upa a = &1  oraz k©t S'AS = ¶ = 45².
Mamy obliczyª pole powierzchni ca¬kowitej Pc.
\next
\lose 3
Przypomnijmy, ±e Pc = a¾ + 4(ÀaÔw) = a¾ + 2aÔw.
\next
\lose 2
Poniewa± k©t S'AS jest r®wny 45², zatem tr®jk©t AS'S jest 
r®wnoramiennym tr®jk©tem prostok©tnym. St©d h = ÏAS'Ï.
\next
\lose 2
Zauwa±my, ±e AS' jest po¬ow© przek©tnej kwadratu o boku a.
St©d h = Àa¸(2) = &3.
\next
\lose 3
Rozwa±my teraz tr®jk©t S'MS. Z twierdzenia Pitagorasa mamy:
     w¾ = ÏSS'Ͼ + ÏS'MϾ = h¾ + (Àa)¾ = &4.
\next
\lose 1
St©d w = &5. Ostatecznie, Pc = a¾ + 2aÔw = ?
\end help
\end question
\end exercise
\type ost
\title Ostros¬up
\begin exercise
\begin text
\cont
Pole powierzchni ca¬kowitej ostros¬upa prawid¬owego czworok©tnego
o kraw«dzi podstawy &1 dm wynosi &2 dm¾. Oblicz obj«to¯ª tego ostros¬upa.
\end text
\begin params
3x6
1.6
8
1.7
0.8
1.5
1.28
2
12
2
1
¸(3)
4s(3)/3
6
84
4
3
¸(7)
12s(7)
\end params
\begin question
\title Ostros¬up
\begin text
\cont
Pole powierzchni ca¬kowitej ostros¬upa prawid¬owego czworok©tnego
o kraw«dzi podstawy &1 dm wynosi &2 dm¾. Oblicz obj«to¯ª tego ostros¬upa.
\end text
\points 16
\begin answer
\evaluate
\length 10
\correct &6
\wrong 4
\end answer
\begin help
\lose 5
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.55,0.1
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\window 1,11,79,23
\next
\lose 2
Niech SM oznacza wysoko¯ª ¯ciany bocznej ostros¬upa; w = ÏSMÏ.
Mamy dan© kraw«d° podstawy ostros¬upa a = &1
oraz pole powierzchni ca¬kowitej Pc = &2.
Chcemy obliczyª obj«to¯ª V = Âa¾h.
\next
\lose 4
Przypomnijmy, ±e Pc = a¾ + 4(ÀaÔw) = a¾ + 2aÔw,
czyli &2 = &1¾ + 2Ô&1Ôw, sk©d w = &3.
\next
\lose 3
Rozwa±my tr®jk©t MSS'.
Oczywi¯cie ÏS'MÏ = Àa.
Z twierdzenia Pitagorasa mamy h = ¸(w¾ - (Àa)¾) = ¸(&3¾ - &4¾) = &5.
\next
\lose 1
Ostatecznie, V = Âa¾h = ?
\end help
\end question
\end exercise
\type ost
\title Ostros¬up
\begin exercise
\begin text
\cont
Podstaw© ostros¬upa jest prostok©t, kt®rego pole wynosi &1 cm¾,
a stosunek bok®w wynosi 3:2. Kraw«dzie boczne ostros¬upa tworz©
z p¬aszczyzn© podstawy k©ty po 30². Obliczyª obj«to¯ª tego ostros¬upa.
\end text
\begin params
3x6
54
9
6
3¸(13)/2
À¸(39)
9s(39)
6
3
2
À¸(13)
¸(39)/6
s(39)/3
24
6
4
¸(13)
¸(39)/3
8s(39)/3
\end params
\begin question
\title Ostros¬up
\begin text
\cont
Podstaw© ostros¬upa jest prostok©t, kt®rego pole wynosi &1 cm¾,
a stosunek bok®w wynosi 3:2. Kraw«dzie boczne ostros¬upa tworz©
z p¬aszczyzn© podstawy k©ty po 30². Obliczyª obj«to¯ª tego ostros¬upa.
\end text
\points 19
\begin answer
\evaluate
\length 12
\correct &6
\wrong 5
\end answer
\begin help
\lose 5
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\cont
\next
\lose 2
Niech a i b oznaczaj© d¬ugo¯ci kraw«dzi podstawy i niech h b«dzie 
wysoko¯ci© ostros¬upa. Mamy dane pole podstawy ostros¬upa Pp = &1,
wiemy, ±e a/b = 3/2 oraz znamy k©t S'AS = ¶ = 30².
Mamy obliczyª obj«to¯ª ostros¬upa V = Âa¾h = ÂPpÔh.
\next
\lose 4
Rozpatrzmy tr®jk©t prostok©tny AS'S. Poniewa± ¶ = 30²,zatem tr®jk©t 
ten jest "po¬ow©" tr®jk©ta r®wnobocznego MSA' o boku 2h. St©d ÏAS'Ï = h¸(3).
\next
\lose 2
AS' to po¬owa przek©tnej prostok©ta ABCD, zatem ÏAS'Ï = À¸(a¾ + b¾).
\next
\lose 3
Mamy uk¬ad r®wna­ aÔb = &1, 2a = 3b.
Rozwi©zuj©c ten uk¬ad dostajemy a = &2, b = &3.
\next
\lose 2
ÏAS'Ï = À¸(&2¾ + &3¾) = &4. h = ÏAS'Ï/¸(3) = (&4)/¸(3) = &5.
Ostatecznie V = ÂPpÔh = ?
\end help
\end question
\end exercise