Page 119 - Engineering

‫ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ(‬ ‫‪١٠٦‬‬

‫‪XY = 4x 2 + 4y2 = 2 x 2 + y2 = 2 169 = 26 .‬‬

‫)‪ [AMC10B 2004] (٤٠‬اﳌﺜﻠﺚ ‪ △ACE‬ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ﻓﻴﻪ ‪،AC = 12‬‬
‫‪ .EA = 20 ،CE = 16‬رﲰﻨﺎ اﻟﻨﻘﺎط ‪ F ،D ،B‬ﻋﻠﻰ ‪،CE ،AC‬‬
‫‪ EA‬ﻋﻠﻰ اﻟﺘﻮاﱄ ﲝﻴﺚ ﻳﻜﻮن ‪ .EF = 5 ،CD = 4 ،AB = 3‬ﻣﺎ ﻧﺴﺒﺔ‬

‫ﻣﺴﺎﺣﺔ اﳌﺜﻠﺚ ‪ △DBF‬إﱃ ﻣﺴﺎﺣﺔ اﳌﺜﻠﺚ ‪ △ACE‬؟‬

‫)د( ‪7‬‬ ‫)ج( ‪11‬‬ ‫)ب( ‪3‬‬ ‫)أ( ‪1‬‬

‫‪16‬‬ ‫‪25‬‬ ‫‪8‬‬ ‫‪4‬‬
‫‪A‬‬
‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )د(‪:‬‬

‫‪3‬‬

‫‪B 15‬‬

‫‪9‬‬ ‫‪F‬‬
‫‪Gh‬‬ ‫‪5‬‬

‫‪C 4D‬‬ ‫‪12‬‬ ‫‪E‬‬

‫ﻻﺣﻆ أوﻻً أن‬

‫‪AB = CD = EF = 1‬‬
‫‪AC CE EA 4‬‬

‫‪BC = DE = FA = 3‬‬
‫‪AC CE EA 4‬‬

‫ارﺳﻢ اﻵن اﻻرﺗﻔﺎع ‪ h‬ﻣﻦ ‪ F‬إﱃ ‪ AC‬ﻟﻴﻼﻗﻲ ‪ AC‬ﰲ اﻟﻨﻘﻄﺔ ‪ .G‬ﻋﻨﺪﺋﺬ‪،‬‬

‫‪.(AA) △AFG ∼ △AEC‬‬