Page 94 - Engineering
P. 94
ﺍﳌﺜﻠﺜﺎﺕ ٨١
] [Aust.MC 1983اﻟﻨﺴﺒﺔ x ′ : y′ : z ′ﺑﲔ اﻟﺰواﻳﺎ اﳋﺎرﺟﻴﺔ ﻟﻠﻤﺜﻠﺚ اﳌﺮﻓﻖ )(٥
ﻫﻲ . 4 : 5 : 6ﻣﺎ اﻟﻨﺴﺒﺔ ﺑﲔ اﻟﺰواﻳﺎ اﻟﺪاﺧﻠﻴﺔ x : y : z؟
)أ( ) 7 : 5 : 3ب( ) 3 : 2 : 1ج( ) 8 : 5 : 2د( 6 : 5 : 4
z′
z
x y y′
x′
اﻟﺤﻞ :اﻹﺟﺎﺑﺔ ﻫﻲ )أ( :ﻟﺪﻳﻨﺎ x + y + z = 180و
.(x + x ′) + (y + y′) + (z + z ′) = 3 × 180 = 540ﻣﻦ ذﻟﻚ ﳒﺪ أن
.x ′ + y′ + z ′ = 360°وﲟﺎ أن 4 + 5 + 6 = 15ﻓﺈن
،y′ = 5 × 360° = 120° ،x′ = 4 × 360° = 96°
15 15
.z ′ = 6 × 360° = 144°إذن،x = 180 − 96 = 84° ،
15
.z = 180 − 144 = 36° ،y = 180 − 120 = 60°وﺬا ﻓﺈن x : y : zﻫﻲ
. 84 : 60 : 36أي . 7 : 5 : 3
] [Aust.MC 1982ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ △RPQ ،ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ و ،ST PR )(٦
،ST = 4 ،PR = 8 ،PQ = 6ﻣﺴﺎﺣﺔ △RQTﺗﺴﺎوي:
)د( 16 )ج( 12 )ب( 10 )أ( 6