Page 95 - Engineering

‫ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ(‬ ‫‪٨٢‬‬

‫‪Q‬‬

‫‪T 4S6‬‬

‫‪R‬‬ ‫‪P‬‬
‫‪8‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ج(‪ :‬ﻣﺪ ‪ TS‬ﻟﻴﻼﻗﻲ ‪ PQ‬ﰲ ‪ .S′‬ﻋﻨﺪﺋﺬ‪،‬‬

‫]‪[RQT ] = [TSQ] + [TSR‬‬
‫‪= 1 ×TS ×QS ′ + 1 ×TS × S ′P‬‬
‫‪22‬‬
‫)‪= 1 ×TS × (QS ′ + S ′P‬‬
‫‪2‬‬
‫‪= 1 ×TS ×QP‬‬
‫‪2‬‬
‫‪= 1 × 4 × 6 = 12‬‬
‫‪2‬‬

‫)‪ [Aust.MC 1978] (٧‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ AB = AC ،‬و ‪.KL = LM‬‬

‫ﻋﻨﺪﺋﺬ‪ ،‬اﻟﻨﺴﺒﺔ ‪ KB‬ﻫﻲ‬

‫‪LC‬‬

‫)أ( ‪) 1.5‬ب( ‪) 2‬ج( ‪) 2.5‬د( ‪3‬‬

‫‪A‬‬
‫‪K‬‬

‫‪L‬‬

‫‪B CM‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ب(‪ :‬أﻧﺸﺊ ‪ KD LC‬ﻛﻤﺎ ﻫﻮ ﻣﺒﲔ ﰲ اﻟﺸﻜﻞ‬