1. ‫ﺣﻞ‬‫ﺗﻤ‬‫ـ‬‫ﺎرﻳﻦ‬‫ا‬‫ﻷﻋــﺪاد‬‫واﻟﺤﺴﺎب‬)‫ك‬-‫م‬(‫اﻟﺠﺰء‬‫اﻟﺴﺎدس‬1AS
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‫ﺤل‬‫ﺤل‬‫ﺘﻤﺎﺭﻴﻥ‬‫ﺘﻤﺎﺭﻴﻥ‬‫ﺍ‬‫ﺍ‬‫ﻷﻋــﺩﺍﺩ‬‫ﻷﻋــﺩﺍﺩ‬‫ﻭﺍﻟﺤﺴﺎﺏ‬‫ﻭﺍﻟﺤﺴﺎﺏ‬
‫ﺘﻤﺎﺭﻴﻥ‬‫ﺘﻤﺎﺭﻴﻥ‬))‫ﺍﻟﻤﺩﺭﺴﻲ‬ ‫ﺍﻟﻜﺘﺎﺏ‬‫ﺍﻟﻤﺩﺭﺴﻲ‬ ‫ﺍﻟﻜﺘﺎﺏ‬((‫ﺍﻟﺠﺯﺀ‬‫ﺍﻟﺠﺯﺀ‬‫ﺍﻟﺴﺎﺩﺱ‬‫ﺍﻟﺴﺎﺩﺱ‬
‫ﺍﻟﻤﺴﺎﺌل‬:
‫ﻫﻭ‬909 526‫ﺭﻗﻤﺎ‬. 76(‫ﺃﺭﻗﺎﻡ‬ ‫ﻋﺩﺩ‬‫ﺍﻟﻌﺩﺩ‬p‫ﺤﻴﺙ‬3021377
p 2 1= −
77(ABC‫ﻀﻠﻌﻪ‬ ،‫ﺍﻷﻀﻼﻉ‬ ‫ﻤﺘﻘﺎﻴﺱ‬ ‫ﻤﺜﻠﺙ‬2.
‫ﻟﻴﻜﻥ‬[AO]‫ﺍﻻﺭﺘﻔﺎ‬‫ﻉ‬‫ﺒﺎﻟﻀﻠﻊ‬ ‫ﺍﻟﻤﺘﻌﻠﻕ‬[BC].
‫ﺃﻥ‬ ‫ﺒﻤﺎ‬‫ﺍﻟﻤﺜﻠﺙ‬ABC‫ﻓﺈﻥ‬ ‫ﺍﻷﻀﻼﻉ‬ ‫ﻤﺘﻘﺎﻴﺱ‬[AO]‫ﻭﺍﺭﺘﻔﺎﻉ‬ ‫ﻤﺤﻭﺭ‬ ‫ﻫﻭ‬
‫ﺍﻟﻭﻗﺕ‬ ‫ﻨﻔﺱ‬ ‫ﻓﻲ‬ ‫ﻭﻤﻨﺘﺼﻑ‬.
‫ﻓﻴﺘﺎﻏﻭﺭ‬ ‫ﻨﻅﺭﻴﺔ‬ ‫ﺒﺘﻁﺒﻴﻕ‬‫ﺱ‬‫ﺍﻟﻤﺜﻠﺙ‬ ‫ﻋﻠﻰ‬AOC‫ﻓﻲ‬ ‫ﺍﻟﻘﺎﺌﻡ‬O‫ﻟﺩﻴﻨﺎ‬ ‫ﻴﻜﻭﻥ‬ ،:
AO² AC² OC² 2² 1² 4 1
AO² 3 AO 3
3= − = − = −
= ⇔ =
=
0
39 3 13= ×
39 (a b) (a b) a b 3
39 3
a 8
b 513 a b 13
= − × + − =⎫ ⎧
⇔ ⇔⎬ ⎨
= ×
=⎧
⎨
=+ =⎭ ⎩ ⎩
39 (8 5) (8 5) 8² 5² 39 8² 5²= − × + = − ⇔ = −
‫ﻁﻭﻟﻬﺎ‬ ‫ﻗﻁﻌﺔ‬ ‫ﻹﻨﺸﺎﺀ‬ ‫ﺇﺫﻥ‬39‫ﻭﺘﺭﻩ‬ ‫ﻁﻭل‬ ‫ﻗﺎﺌﻡ‬ ‫ﻤﺜﻠﺙ‬ ‫ﺇﻨﺸﺎﺀ‬ ‫ﻴﺠﺏ‬8cm‫ﺃﻀ‬ ‫ﺃﺤﺩ‬ ‫ﻭﻁﻭل‬‫ﺍﻟﻘﺎﺌﻤﺔ‬ ‫ﺍﻟﺯﺍﻭﻴﺔ‬ ‫ﻼﻉ‬
5cm‫ﻁﻭل‬ ‫ﻴﻜﻭﻥ‬ ‫ﻋﻨﺩﻫﺎ‬ ،‫ﻫﻭ‬ ‫ﺍﻵﺨﺭ‬ ‫ﺍﻟﻀﻠﻊ‬39.
78(
I 3
39
α
β

2. ‫ﺣﻞ‬‫ﺗﻤ‬‫ـ‬‫ﺎرﻳﻦ‬‫ا‬‫ﻷﻋــﺪاد‬‫واﻟﺤﺴﺎب‬)‫ك‬-‫م‬(‫اﻟﺠﺰء‬‫اﻟﺴﺎدس‬1AS
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37,1m
AO
tan AO tan OC tan 30 64,3 0,577 64,3
OC
α = ⇔ = α× = × = × =° .
OB
tan OB tan OC tan 2,4 2,7 m5 64,3 0,042 64,3
OC
β = ⇔ = β× = × = × =° .
AC AO OB 37,1 2,7 3 , 08 4 m9= + = + ≈ .=
‫ﺤﻭﺍﻟﻲ‬ ‫ﻫﻭ‬ ‫ﺍﻟﻌﻤﺎﺭﺓ‬ ‫ﺍﺭﺘﻔﺎﻉ‬ ‫ﺇﺫﻥ‬40 m.
79(
‫ﺍﻟﻤﺴﺘﻁﻴل‬R1‫ﺒﻌﺩ‬‫ﻫﻤﺎ‬ ‫ﺍﻩ‬:‫ﺍﻟﻁﻭل‬L‫ﻭ‬‫ﺍﻟﻌﺭﺽ‬l.
‫ﺍﻟﻤﺴﺘﻁﻴل‬R2‫ﺒﻌﺩ‬‫ﻫﻤﺎ‬ ‫ﺍﻩ‬:‫ﺍﻟﻁﻭل‬l‫ﻭﺍﻟﻌﺭﺽ‬L-l.
‫ﺃﻥ‬ ‫ﺒﻤﺎ‬R2‫ﻨﻔﺱ‬ ‫ﻟﻪ‬‫ﺸﻜل‬R1‫ﻓﺈﻥ‬:
L l
l L
=
l−
.
L
c L c
l
= ⇔ = × l
L l c l l
l L l l c l l
(c l)(c l l) l²
c² l² c l² l²
l²(c
c² c 1 0− −
² c) l²
l²
c² c
l²
c² c 1
×
= ⇔ =
− × −
⇔ × × − =
⇔ × − × =
⇔ − =
⇔ − =
⇔ − = =⇔
2
1 5 1 1 5 1 5
c c² 2 c c² c 0 c² c 1 0
2 4 4 2 4 4 4
⎛ ⎞
− = ⇔ + − × × = ⇔ + − − = ⇔ − − = ⇔ =⎜ ⎟
⎝ ⎠
0 0
‫ﺇ‬‫ﻤﺤﻘﻘﺔ‬ ‫ﺍﻷﻭﻟﻰ‬ ‫ﺍﻟﻤﺴﺎﻭﺍﺓ‬ ‫ﻓﺈﻥ‬ ‫ﺫﻥ‬.
22 2
1 5 1 5 1 5 1 5
c c 0 c c
2 4 2 2 2 2 2 2
⎛ ⎞ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
− = ⇔ − − = ⇔ − − − + =⎜ ⎟ ⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦ ⎣ ⎦
0⎥
1 5 1 5 1 5 1c 0 c c c
2 2 2 2 2 2 2
1 5 1 5 1 5 1
c 0 c c c
2 2 2 2 2 2 2
⎧ ⎧ ⎧ ⎧⎛ ⎞ +− − =⎪ − = = + =⎜ ⎟ ⎪ ⎪ ⎪⎪⎝ ⎠ ⎪ ⎪ ⎪
⇔ ⇔ ⇔ ⇔⎨ ⎨ ⎨ ⎨
−⎛ ⎞⎪ ⎪ ⎪ ⎪− + = − = − = − =⎜ ⎟⎪ ⎪ ⎪ ⎪⎩ ⎩ ⎩⎝ ⎠⎩
5
5
1 5
0
2
+
>
:
. ‫ﺃﻥ‬ ‫ﻨﻌﻠﻡ‬
‫ﺃﺨﺭﻯ‬ ‫ﺠﻬﺔ‬ ‫ﻤﻥ‬‫ﻟﺩﻴﻨ‬‫ﺎ‬
1 5
0 1 5 0 1 5
2
−
< ⇔ − < ⇔ <.‫ﺍﻟﻭﺤﻴﺩ‬ ‫ﺍﻟﻤﻭﺠﺏ‬ ‫ﺍﻟﺤل‬ ‫ﺇﺫﻥ‬‫ﻫ‬‫ﻭ‬:
1 5
c
2
+
=.

3. ‫ﺣﻞ‬‫ﺗﻤ‬‫ـ‬‫ﺎرﻳﻦ‬‫ا‬‫ﻷﻋــﺪاد‬‫واﻟﺤﺴﺎب‬)‫ك‬-‫م‬(‫اﻟﺠﺰء‬‫اﻟﺴﺎدس‬1AS
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80(ABCD‫ﺤﻴﺙ‬ ‫ﻤﻨﺤﺭﻑ‬ ‫ﺸﺒﻪ‬CD = 4cm‫ﻭ‬D 6. 0= °
‫ﺍﻟﻤﺜﻠﺙ‬CDE‫ﺍﻷﻀﻼﻉ‬ ‫ﻤﺘﻘﺎﻴﺱ‬‫ﺃﻥ‬ ‫ﺃﻱ‬CD = DE = EC = 4cm.
‫ﻟﺘﻜﻥ‬O‫ﺘﻘﺎ‬ ‫ﻨﻘﻁﺔ‬‫ﻁﻊ‬‫ﻤﺤﻭ‬‫ﺭ‬[ED]‫ﻤﻊ‬[ED].
‫ﺃﻥ‬ ‫ﺒﻤﺎ‬CDE‫ﻓﺈﻥ‬ ‫ﺍﻷﻀﻼﻉ‬ ‫ﻤﺘﻘﺎﻴﺱ‬[CO]‫ﻫﻭ‬‫ﻤﺤﻭﺭ‬‫ﻭ‬‫ﻭﻤﻨﺘﺼﻑ‬ ‫ﻤﻨﺼﻑ‬‫ﺍﻟﻭﻗﺕ‬ ‫ﻨﻔﺱ‬ ‫ﻓﻲ‬ ‫ﻭﻤﺘﻭﺴﻁ‬‫ﻓﺈﻥ‬ ‫ﻭﻤﻨﻪ‬:
EO = OD = 2cm.
[CO]‫ﻤﺤﻭﺭ‬[ED]‫ﺍﻟﻤﺜﻠﺙ‬ ‫ﺇﺫﻥ‬COD‫ﻗﺎﺌﻡ‬‫ﻓ‬‫ﻲ‬O‫ﻨﻅﺭﻴﺔ‬ ‫ﻭﺒﺘﻁﺒﻴﻕ‬ ،‫ﻴﻠﻲ‬ ‫ﻤﺎ‬ ‫ﻋﻠﻰ‬ ‫ﻨﺘﺤﺼل‬ ‫ﻋﻠﻴﻪ‬ ‫ﻓﻴﺘﺎﻏﻭﺭﺱ‬:
CD²= CO² + OD²‫ﻭﻤﻨﻪ‬CO² = CD² - OD² = 4² - 2² = 16 – 4 = 12‫ﻭﻤﻨﻪ‬CO = 3,4 cm.
A B O 90= = =
= ⇔ = ⇔ + = ⇔ =
°‫ﺍﻟﺭﺒﺎﻋﻲ‬ ‫ﻓﺈﻥ‬ ‫ﻭﻤﻨﻪ‬ABCO‫ﻤﺴ‬‫ﺘﻁﻴل‬‫ﻭﻤﻨﻪ‬‫ﻓﺈﻥ‬:AB = CO = 3,4cm‫ﻭ‬BC = AO.
‫ﻟﻴﻜﻥ‬P1‫ﺍﻟﻤﺜﻠﺙ‬ ‫ﻤﺤﻴﻁ‬CDE‫ﻓﺈﻥ‬ ‫ﻭﻤﻨﻪ‬P1 = CD+DE+EC = 12cm.
‫ﻟﻴﻜﻥ‬P2‫ﺍﻟﻤﻨﺤﺭﻑ‬ ‫ﺸﺒﻪ‬ ‫ﻤﺤﻴﻁ‬ABCE.
P2 AB BC CE AE= + + +
P2 AB AO CE AE
P2 3,4 AE 2 4 AE
P2 2AE 9,4
= + + +
= + + + +
= +
.P1 P2 P2 12 2AE 9,4 12 2AE 12 9,4 2,6 AE 1,3cm=− = ⇔
81(ABC‫ﻓﻲ‬ ‫ﻗﺎﺌﻡ‬ ‫ﻤﺜﻠﺙ‬A.‫ﺤﻴﺙ‬BC = a; AC = b; AB = c.
‫ﻟﺘﻜﻥ‬S1‫ﺍﻟﻤﺜﻠﺙ‬ ‫ﻤﺴﺎﺤﺔ‬ABC:
bc
2
.S1=
‫ﻟﺘﻜﻥ‬S2‫ﻤﺴﺎﺤﺔ‬‫ﻨﺼﻑ‬‫ﺍ‬‫ﻟﺩﺍﺌﺭﺓ‬‫ﻗﻁﺭﻫﺎ‬ ‫ﺍﻟﺘﻲ‬[AB]:
2
c c²
c²2 4S2
2 2
⎛ ⎞
π× π×⎜ ⎟
8
π×⎝ ⎠= = =.
‫ﻟﺘﻜﻥ‬S3‫ﻤﺴﺎﺤﺔ‬‫ﻨﺼﻑ‬‫ﺍ‬‫ﻟﺩﺍﺌﺭﺓ‬‫ﻗﻁﺭﻫﺎ‬ ‫ﺍﻟﺘﻲ‬[BC]:
2
a a²
a²2 4
2 2 8
⎛ ⎞
π× π×⎜ ⎟
S3
π×⎝ ⎠= = =.
‫ﻟﺘﻜﻥ‬S4‫ﻤﺴﺎﺤﺔ‬‫ﻨﺼﻑ‬‫ﺍ‬‫ﻟﺩﺍﺌﺭﺓ‬‫ﻗﻁﺭﻫﺎ‬ ‫ﺍﻟﺘﻲ‬[AC]:
2
b b²
b²2 4
2 2 8
⎛ ⎞
π× π×⎜ ⎟
S4
π×⎝ ⎠= = =.
‫ﻟﺘﻜﻥ‬S‫ﻤﺴﺎﺤﺔ‬‫ﺍ‬‫ﺍﻟﻤﻅﻠل‬ ‫ﻟﺠﺯﺀ‬:
bc c² b² a² 4bc (c² b²
S S1 S2 S4 S3
2 8 8 8 8
a²)π× π× π× + π + −
= + + − = + + − = .
ABC‫ﻓﻲ‬ ‫ﻗﺎﺌﻡ‬ ‫ﻤﺜﻠﺙ‬A‫ﻴﻠﻲ‬ ‫ﻤﺎ‬ ‫ﻋﻠﻰ‬ ‫ﻨﺘﺤﺼل‬ ‫ﻋﻠﻴﻪ‬ ‫ﻓﻴﺘﺎﻏﻭﺭﺱ‬ ‫ﻨﻅﺭﻴﺔ‬ ‫ﻭﺒﺘﻁﺒﻴﻕ‬ ،:
BC² AB² AC² a² c² b²= + ⇔ = +‫ﻭﻤﻨﻪ‬:
4bc (a² a²) 4bc bc
S
8 8
+ π −
= =
2
= .
‫ﺍﻟﻤﺜﻠﺙ‬ ‫ﻤﺴﺎﺤﺔ‬ ‫ﺘﺴﺎﻭﻱ‬ ‫ﺍﻟﻤﻅﻠل‬ ‫ﺍﻟﺠﺯﺀ‬ ‫ﻤﺴﺎﺤﺔ‬ ‫ﺃﻥ‬ ‫ﻨﺴﺘﻨﺘﺞ‬ABC.