\type rwb
\title R®wnoleg¬obok 
\begin exercise
\begin text
\cont
W r®wnoleg¬oboku dane s© d¬ugo¯ci bok®w a = &1 cm i b = &2 cm oraz
k©t ¶ = 30². Obliczyª d¬ugo¯ª kr®tszej przek©tnej.
\end text
\begin params
2x5
8
2¸(3)
¸(3)
5
2s(7)
4
2¸(3)
¸(3)
1
2
\end params
\begin question
\title R®wnoleg¬obok
\begin text
\cont
W r®wnoleg¬oboku dane s© d¬ugo¯ci bok®w a = &1 cm i b = &2 cm 
oraz k©t ¶ = 30². 
Obliczyª d¬ugo¯ª kr®tszej przek©tnej.
\end text
\points 18
\begin answer
\evaluate
\length 8
\correct &5
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\dotted
\draw 0.9,0.9,0.5,0.6
\text 0.8,0.7,d
\solid
\draw 0.1,0.9,0.9,0.9
\draw 0.9,0.9,1.3,0.6
\draw 1.3,0.6,0.5,0.6
\draw 0.5,0.6,0.1,0.9
\dotted
\draw 0.5,0.6,0.5,0.9
\text 0.1,0.9,A
\text 0.9,0.9,B
\text 1.3,0.5,C
\text 0.47,0.5,D
\text 0.6,0.89,a
\text 0.2,0.7,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
\dotted
\draw 1.9,0.5,1.9,0.8
\draw 1.9,0.8,1.5,0.5
\text 1.45,0.5,A
\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
\text 1.6,0.3,b
\window 1,11,79,23
\lose 4
\next
\lose 1
Mamy dane ÏABÏ = a = &1, ÏADÏ = b = &2 i k©t BAD = ¶ = 30². 
Chcemy obliczyª d = ÏBDÏ.
\next
\lose 3
Niech DD' oznacza wysoko¯ª r®wnoleg¬oboku; ÏDD'Ï = h. 
Rozwa±my tr®jk©t AD'D. Poniewa± ¶ = 30², 
zatem jest on "po¬ow©" tr®jk©ta r®wnobocznego MDA o boku b. 
St©d h = Àb = À&2 = &3.
\next
\lose 3
Ponadto, ze wzoru na wysoko¯ª w tr®jk©cie r®wnobocznym, mamy ÏAD'Ï = Àb¸(3).
\next
\lose 3
W takim razie ÏD'BÏ = ÏABÏ - ÏAD'Ï = a - Àb¸(3) = &1 - ÀÔ&2¸(3) = &4.
\next
\lose 3
Korzystaj©c z twierdzenia Pitgorasa w tr®jk©cie BDD', 
mamy d = ¸(ÏDD'Ͼ + ÏD'BϾ) = ¸(&3¾ + &4¾) = ?
\end help
\end question
\end exercise
\type rwb
\title R®wnoleg¬obok
\begin exercise
\begin text
\cont
Obw®d r®wnoleg¬oboku wynosi &1 dm. 
Stosunek d¬ugo¯ci bok®w tego r®wnoleg¬oboku jest r®wny 8:3. 
Obliczyª pole r®wnoleg¬oboku, je±eli k©t ostry wynosi 60².
\end text
\begin params
3x4
66
24
9
108s(3)
33
12
9/2
27s(3)
22
8
3
12s(3)
\end params
\begin question
\title R®wnoleg¬obok
\begin text
\cont
Obw®d r®wnoleg¬oboku wynosi &1 dm. 
Stosunek d¬ugo¯ci bok®w tego r®wnoleg¬oboku jest r®wny 8:3. 
Obliczyª pole r®wnoleg¬oboku, je±eli k©t ostry wynosi 60².
\end text
\points 15
\begin answer
\evaluate
\length 10
\correct &4
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\text 0.4,0.9,D'
\text 0.1,0.9,A
\text 0.9,0.9,B
\text 1.2,0.2,C
\text 0.4,0.2,D
\text 0.5,0.89,a
\text 0.18,0.5,b
\text 0.42,0.5,h
\text 0.16,0.8,¶
\draw 0.1,0.9,0.9,0.9
\draw 0.9,0.9,1.2,0.3
\draw 1.2,0.3,0.4,0.3
\draw 0.4,0.3,0.1,0.9
\dotted
\draw 0.4,0.3,0.4,0.9
\solid
\draw 1.5,0.9,1.8,0.9
\draw 1.8,0.9,1.8,0.3
\draw 1.8,0.3,1.5,0.9
\dotted
\draw 1.8,0.9,2.1,0.9
\draw 2.1,0.9,1.8,0.3
\text 1.5,0.9,A
\text 1.8,0.9,D'
\text 2.1,0.9,M
\text 1.8,0.2,D
\text 1.82,0.5,h
\text 1.6,0.5,b
\text 1.58,0.8,¶
\window 1,11,79,23
\lose 3
\next
\lose 2
Niech ÏABÏ = a, ÏADÏ = b. 
Wiemy, ±e 2(a + b) = &1, a/b = 8/3 oraz mamy dany k©t BAD = ¶ = 60². 
Niech DD' oznacza wysoko¯ª r®wnoleg¬oboku; h = ÏDD'Ï. 
Chcemy znale°ª pole r®wnoleg¬oboku P = aÔh.
\next
\lose 3
Poniewa± ¶ = 60², zatem h = Àb¸(3) 
(tr®jk©t AD'D jest "po¬ow©" tr®jk©ta r®wnobocznego AMD o boku b). 
Tak wi«c, P = aÔh = ÀaÔb¸(3).
\next
\lose 3
Mamy uk¬ad r®wna­: 3a = 8b, 2(a + b) = &1. 
Rozwi©zuj©c go dostajemy: a = 4Ô&1/11 = &2, b = 3Ô&1/22 = &3, 
\next
\lose 2
a st©d P = ÀaÔb¸(3) = ÀÔ&2Ô&3Ô¸(3) = ?
\end help
\end question
\end exercise
\type rwb
\title 
\begin exercise
\begin text
\cont
Przek©tne rombu maj© d¬ugo¯ª &1 i &2 cm. Oblicz jego obw®d.
\end text
\begin params
3x3
8
6
20
16
12
40
8
4
8s(5)
\end params
\begin question
\title R®wnoleg¬obok
\begin text
\cont
Przek©tne rombu maj© d¬ugo¯ª &1 i &2 cm. Oblicz jego obw®d.
\end text
\points 10
\begin answer
\evaluate
\length 8
\correct &3
\wrong 2
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\text 0.1,0.9,A
\text 0.9,0.9,B
\text 1.2,0.2,C
\text 0.4,0.2,D
\draw 0.1,0.9,0.9,0.9
\draw 0.9,0.9,1.2,0.3
\draw 1.2,0.3,0.4,0.3
\draw 0.4,0.3,0.1,0.9
\dotted
\draw 0.1,0.9,1.2,0.3
\draw 0.9,0.9,0.4,0.3
\text 0.6,0.5,E
\window 1,11,79,23
\lose 3
\next
\lose 1
Mamy dane ÏACÏ = &1 i ÏBDÏ = &2. 
Wystarczy obliczyª jeden bok rombu.
\next
\lose 2
Tr®jk©t ABE jest prostok©tny, 
a jego przyprostok©tne AE, EB, s© po¬owami przek©tnych.
\next
\lose 2
Z twierdzenia Pitagorasa mo±emy obliczyª d¬ugo¯ª boku:
\cr
     ÏABÏ = ¸(ÏAEϾ+ÏEBϾ) = ?
\cr
Obw®d = 4ÔÏAEÏ.
\end help
\end question
\end exercise
\type rwb
\title R®wnoleg¬obok
\begin exercise
\begin text
\cont
W rombie d¬u±sza przek©tna wynosi &1 cm,
a k©t rozwarty 120².
Obliczyª pole tego rombu.
\end text
\begin params
3x3
10
5
50/s(3)
6
3
18/s(3)
8
4
32/s(3)
\end params
\begin question
\title R®wnoleg¬obok
\begin text
\cont
W rombie d¬u±sza przek©tna wynosi &1 cm,
a k©t rozwarty 120².
Obliczyª pole tego rombu.
\end text
\points 14
\begin answer
\evaluate
\length 10
\correct &3
\wrong 3
\end answer
\begin help
\window 1,2,21,10
\scale 0,0,1,1
\text 0.1,0.9,A
\text 0.9,0.9,B
\text 1.2,0.2,C
\text 0.4,0.2,D
\draw 0.1,0.9,0.9,0.9
\draw 0.9,0.9,1.2,0.3
\draw 1.2,0.3,0.4,0.3
\draw 0.4,0.3,0.1,0.9
\dotted
\draw 0.1,0.9,1.2,0.3
\draw 0.9,0.9,0.4,0.3
\text 0.6,0.5,E
\window 1,11,79,23
\lose 3
\next
\lose 2
Wiemy, ±e ÏACÏ = &1, a k©t ABC = 120².
Wystarczy obliczyª pole tr®jk©ta ABD i pomno±yª je przez 2.
\next
\lose 2
K©t ABD ma 60², wi«c ten tr®jk©t jest r®wnoboczny. 
\next
\lose 2
Jego wysoko¯ª AE to po¬owa d¬u±szej przek©tnej, wi«c wynosi &2.
\next
\lose 2
Ze wzoru na wysoko¯ª tr®jk©ta r®wnobocznego mamy: ÏAEÏ = ÏBDϸ(3)/2.
Z tego mo±emy wyliczyª ÏBDÏ.
\next
\lose 1
Mamy ÏBDÏ = 2Ô&2/¸(3).
Podstawiamy do wzoru na pole rombu: 
     P = 2ÔÀÔÏAEÏÔÏBDÏ = ÏAEÏÔÏBDÏ = ?
\end help
\end question
\end exercise