Page 180 - Engineering

‫ﺍﳌﻀﻠﻌﺎﺕ ‪١٦٧‬‬

‫‪B 10 C‬‬

‫‪8‬‬

‫‪A E 6D‬‬

‫اﻟﺤﻞ‪ :‬ﲟﺎ أن ‪ AD = 10‬ﻓﺈن ‪ .AE = 10 − 6 = 4‬وﻳﻜﻮن‬

‫‪[BEDC ] = [ABCD] − [ABE ] = 8 ×10 − 1 × 4 × 8 = 64 .‬‬
‫‪2‬‬

‫ﻣﺜﺎل )‪ :[AJHSME 1995] (٣‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ ‪ ABCD‬ﻣﺘﻮازي أﺿﻼع‪،‬‬

‫‪ DE ⊥ AB‬و ‪ .DF ⊥ BC‬إذا ﻛﺎن ‪DE = 6 ، EB = 4 ،DC = 12‬‬

‫‪D 12‬‬ ‫ﻓﺠﺪ ‪.DF‬‬

‫‪C‬‬

‫‪6‬‬

‫‪A E 4 BF‬‬

‫اﻟﺤﻞ‪.[ABCD] = AB × ED = DF × BC = 12 × 6 = 72 :‬‬
‫إذن‪ .DF × BC = 72 ،‬اﻵن‪ △AED ،‬ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ‪ .‬إذن‪،‬‬

‫‪(AE)2 + (ED)2 = (AD)2‬‬

‫‪(AD)2 = 64 + 36 = 100‬‬

‫‪. DF‬‬ ‫=‬ ‫‪72‬‬ ‫=‬ ‫‪72‬‬ ‫=‬ ‫ﻳﻜﻮن‪7.2 ،‬‬ ‫‪ .AD‬ﻣﻦ ذﻟﻚ‬ ‫‪= BC‬‬ ‫‪= 10‬‬ ‫ﻓﺈن‬ ‫وﻟﺬا‬
‫‪BC‬‬ ‫‪10‬‬

‫ﻣﺜﺎل )‪ :[Euclid 2000] (٤‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ ABCD ،‬ﻣﺘﻮازي أﺿﻼع‪ .‬ﻧﻘﺎط‬
‫ﺗﻘﺎﻃﻊ ﻣﻨﺼﻔﺎت اﻟﺰواﻳﺎ ﻫﻲ رؤوس اﻟﺮﺑﺎﻋﻲ ‪ .PQRS‬أﺛﺒﺖ أن‬