\type pro
\title Prostopad¬o¯cian
\begin exercise
\begin text
\cont
Oblicz pole powierzchni ca¬kowitej prostopad¬o¯cianu, kt®rego
podstaw© jest kwadrat o polu r®wnym &1 cm¾, a k©t mi«dzy przek©tn©
¯ciany bocznej i kraw«dzi© postawy wynosi 60².
\end text
\begin params
4x4
81
162(1+2s(3))
9
9¸(3)
64
128(1+2s(3))
8
8¸(3)
3
6(1+2s(3))
¸(3)
3
6
12(1+s(3))
¸(6)
3¸(2)
\end params
\begin question
\title Prostopad¬o¯cian
\begin text
\cont
Oblicz pole powierzchni ca¬kowitej prostopad¬o¯cianu, kt®rego
podstaw© jest kwadrat o polu r®wnym &1 cm¾, a k©t mi«dzy przek©tn©
¯ciany bocznej i kraw«dzi© postawy wynosi 60².
\end text
\points 15
\begin answer
\evaluate
\length 15
\correct &2
\wrong 4
\end answer
\begin help
\lose 4
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.2,0.3,0.5,0.3
\draw 0.4,0.1,0.2,0.3
\draw 0.4,0.1,0.7,0.1
\draw 0.7,0.1,0.5,0.3
\draw 0.2,0.3,0.2,0.9
\draw 0.5,0.3,0.5,0.9
\dotted
\draw 0.4,0.1,0.4,0.7
\solid
\draw 0.7,0.1,0.7,0.7
\draw 0.2,0.9,0.5,0.9
\dotted
\draw 0.4,0.7,0.7,0.7
\draw 0.2,0.9,0.4,0.7
\solid
\draw 0.5,0.9,0.7,0.7
\draw 0.5,0.9,0.2,0.3
\text 0.11,0.3,A'
\text 0.51,0.9,B
\text 0.15,0.9,A
\text 0.4,0.8,¶
\text 0.3,0.9,a
\text 0.65,0.75,a
\text 0.15,0.55,h
\solid
\draw 2.0,0.9,2.3,0.9
\dotted
\draw 1.7,0.9,2.0,0.9
\solid
\draw 2.0,0.9,2.0,0.3
\dotted
\draw 1.7,0.9,2.0,0.3
\solid
\draw 2.3,0.9,2.0,0.3
\text 2.0,0.2,A'
\text 1.7,0.91,M
\text 2.3,0.91,B
\text 2.0,0.91,A
\text 2.2,0.8,¶
\text 1.85,0.90,a
\text 2.01,0.6,h
\text 2.15,0.90,a
\window 1,11,79,23
\next
\lose 1
Mamy dane pole podstawy P = &1 oraz k©t ABA' = ¶ = 60². 
Niech a oznacza d¬ugo¯ª kraw«dzi podstawy prostopad¬o¯cianu, 
    za¯ h - jego wysoko¯ª. 
Chcemy obliczyª pole powierzchni ca¬kowitej Pc.
\next
\lose 3
Przypomnijmy, ±e Pc = 2a¾+4aÔh = 2P+4aÔh.
Zauwa±my, ±e a = &3.
\next
\lose 4
Rozwa±my tr®jk©t ABA'. Poniewa± ¶ = 60², zatem tr®jk©t ten 
jest "po¬ow©" tr®jk©ta r®wnobocznego MBA' o kraw«dzi 2a. 
St©d, korzystaj©c ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, mamy 
     h = À(2a)¸(3) = &4.
\next
\lose 2
Podstawiaj©c ten zwi©zek do wzoru na Pc, dostajemy: 
     Pc = 2P+4aÔh = ?
\end help
\end question
\end exercise
\type pro
\title Prostopad¬o¯cian
\begin exercise
\begin text
\cont
Podstaw© prostopad¬o¯cianu jest prostok©t o wymiarach a = &1 cm, b = &2
cm. Przek©tna tego prostopad¬o¯cianu nachylona jest do podstawy pod
k©tem 45². Oblicz pole powierzchni prostopad¬o¯cianu.
\end text
\begin params
3x6
7
5
35
12
¸(74)
70+24s(74)
4
3
12
7
5
94
4
5
20
9
¸(41)
40+18s(41)
\end params
\begin question
\title Prostopad¬o¯cian
\begin text
\cont
Podstaw© prostopad¬o¯cianu jest prostok©t 
o wymiarach a = &1 cm, b = &2 cm. 
Przek©tna tego prostopad¬o¯cianu nachylona jest do podstawy pod k©tem 45². 
Oblicz pole powierzchni prostopad¬o¯cianu.
\end text
\points 16
\begin answer
\evaluate
\length 12
\correct &6
\wrong 4
\end answer
\begin help
\lose 5
\window 1,2,21,10
\scale 0,0,1,1
\draw 0.1,0.9,0.6,0.9
\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
\draw 0.1,0.9,0.9,0.7
\draw 0.1,0.9,0.9,0.1
\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 0.2,0.77,¶
\solid
\draw 2.0,0.9,2.6,0.9
\draw 2.6,0.9,2.6,0.3
\draw 2.6,0.3,2.0,0.9
\text 2.0,0.9,A
\text 2.6,0.9,C
\text 2.6,0.2,C'
\text 2.1,0.8,¶
\text 2.62,0.6,h
\window 1,11,79,23
\next
\lose 1
Mamy dane ÏABÏ = a = &1, ÏBCÏ = b = &2 i k©t CAC' = ¶ = 45². 
Niech h oznacza wysoko¯ª prostopad¬o¯cianu. 
Chcemy obliczyª pole powierzchni ca¬kowitej Pc.
\next
\lose 3
Przypomnijmy, ±e Pc = 2(aÔb + aÔh + bÔh). 
St©d Pc = 2[&1Ô&2 + (&1 + &2)h] = 2(&3 + &4Ôh).
\next
\lose 3
Rozwa±my tr®jk©t C'AC i zauwa±my, ±e poniewa± ¶ = 45², 
     to jest on r®wnoramienny. W szczeg®lno¯ci h = ÏACÏ.
\next
\lose 2
Z drugiej strony, AC to przek©tna prostok©ta ABCD, 
a zatem (z twierdzenia Pitagorasa) ÏACÏ = ¸(a¾ + b¾) = ¸(&1¾ + &2¾) = &5. 
\next
\lose 1
Ostatecznie, Pc = 2(&3 + &4Ô&5) = ?
\end help
\end question
\end exercise
\type pro
\title Prostopad¬o¯cian
\begin exercise
\begin text
\cont
Oblicz obj«to¯ª prostopad¬o¯cianu, kt®rego podstaw© jest prostok©t o
bokach &1 cm i &2 cm, za¯ przek©tna prostopad¬o¯cianu jest nachylona do
p¬aszczyzny podstawy pod k©tem o mierze 45².
\end text
\begin params
4x5
8
6
48
10
480
4
3
12
5
60
4
5
20
¸(41)
20s(41)
3
5
15
¸(34)
15s(34)
\end params
\begin question
\title Prostopad¬o¯cian
\begin text
\cont
Oblicz obj«to¯ª prostopad¬o¯cianu, kt®rego podstaw© jest prostok©t o
bokach &1 cm i &2 cm, za¯ przek©tna prostopad¬o¯cianu jest nachylona do
p¬aszczyzny podstawy pod k©tem o mierze 45².
\end text
\points 16
\begin answer
\evaluate
\length 10
\correct &5
\wrong 4
\end answer
\begin help
\lose 5
\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
\draw 0.1,0.9,0.9,0.7
\draw 0.1,0.9,0.9,0.1
\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 0.2,0.77,¶
\solid
\draw 2.0,0.9,2.6,0.9
\draw 2.6,0.9,2.6,0.3
\draw 2.6,0.3,2.0,0.9
\text 2.0,0.9,A
\text 2.6,0.9,C
\text 2.6,0.2,C'
\text 2.1,0.8,¶
\text 2.62,0.6,h
\window 1,11,79,23
\next
\lose 1
Mamy dane ÏABÏ = a = &1, ÏBCÏ = b = &2 i k©t CAC' = ¶ = 45². 
Niech h oznacza wysoko¯ª prostopad¬o¯cianu. Chcemy obliczyª obj«to¯ª V. 
\next
\lose 3
Przypomnijmy, ±e V = aÔbÔh = &1Ô&2Ôh = &3Ôh. Trzeba znale°ª h.
\next
\lose 3
Rozwa±my tr®jk©t C'AC i zauwa±my, ±e poniewa± ¶ = 45², 
     to jest on r®wnoramienny. 
W szczeg®lno¯ci h = ÏACÏ.
\next
\lose 2
Z drugiej strony, AC to przek©tna prostok©ta ABCD, 
a zatem (z twierdzenia Pitagorasa) ÏACÏ = ¸(a¾ + b¾) = ¸(&1¾ + &2¾) = &4. 
\next
\lose 1
Ostatecznie, V = ?
\end help
\end question
\end exercise