Page 86 - Engineering
P. 86
٧٣ ﺍﳌﺜﻠﺜﺎﺕ
A
D
B HC E KF
،اﻵن
[ABC ] = 1 BC × AH = BC × AH
2
[DEF ] 1 EF × DK EF × DK
2
△( وأنABC ∼ △DEF ) B = E △ ﻷنABH ∼ △DEK وﻟﻜﻦ
. AH = AB ، ﻣﻦ ذﻟﻚ ﳒﺪ أن.( )ﻛﻞ ﻣﻨﻬﻤﺎ ﻗﺎﺋﻤﺔAHB = DKE
DK DE
ﺬا ﳒﺪ و. AH = BC ، إذن.(△ABC ∼ △DEF ) AB = BC ،وﻟﻜﻦ
DK EF DE EF
أن
[ABC ] = BC × AH = BC × BC = (BC )2 .
[DEF ] EF DK EF EF (EF )2
،DE = 5 ،BC ﻳﻮازيDE ،△ اﳌﺒﲔ أدﻧﺎﻩABC ﰲ اﳌﺜﻠﺚ:(١٣) ﻣﺜﺎل
.BCED ﺟﺪ ﻣﺴﺎﺣﺔ اﻟﺸﻜﻞ اﻟﺮﺑﺎﻋﻲ.[ADE ] = 15 ،BC = 8
A
D 5E
B C
8
