Page 129 - Engineering
P. 129
ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ( ١١٦
7 + BC + CD = 19
2BC = 12
BC = 6
ﳏﻴﻂ △ABCﻳﺴﺎوي 20و .BC = 6إذن،
AB + 6 + 6 = 20
AB = 20 − 12
AB = 8
) [Cayley 2007] (٥٢ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ △ABC ،ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ،M .C = 90° ،
P ،Nﻣﻨﺼﻔﺎت اﻷﺿﻼع AB ،AC ،BCﻋﻠﻰ اﻟﺘﻮاﱄ .إذا ﻛﺎﻧﺖ
ﻣﺴﺎﺣﺔ اﳌﺜﻠﺚ △APNﺗﺴﺎوي 2ﻓﻤﺎ ﻣﺴﺎﺣﺔ اﳌﺜﻠﺚ △ABC؟
A
NP
C MB
)د( 16 )ج( 8 )ب( 6 )أ( 4
اﻟﺤﻞ :اﻹﺟﺎﺑﺔ ﻫﻲ )ج(:
اﳊﻞ اﻷول
AN = 1 = APواﻟﺰاوﻳﺔ Aﻣﺸﱰﻛﺔ ﰲ اﳌﺜﻠﺜﲔ △APNو .△ABCإذن،
AC 2 AB
.△APN ∼ △ABCﻣﻦ ذﻟﻚ ﻧﺮى أن
] [APN = 1 2 = 1
] [ABC 2 4