Page 146 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪١٣٣‬‬

‫‪5 = A1B = A1B‬‬
‫‪3 A1C 4 − A1B‬‬

‫‪. A1C‬‬ ‫=‬ ‫‪3‬‬ ‫‪= PR‬‬ ‫َو‬ ‫‪A1B‬‬ ‫=‬ ‫‪5‬‬ ‫‪= PQ‬‬ ‫إذن‪،‬‬
‫‪2‬‬ ‫‪2‬‬

‫وﻣﻦ ﻣﱪﻫﻨﺔ ﻓﻴﺜﺎﻏﻮرس ﳒﺪ أن ‪.RQ = 25 − 9 = 4 = 2‬‬

‫‪44‬‬

‫ﲟﺎ أن ‪ AB = AC = BC = 2‬ﻓﺈن ‪.△ABC ∼ △PQR‬‬

‫‪PQ PR QR 1‬‬

‫و‪‬ﺬا ﻓﺈن ‪ . AA1 = 2‬وﻟﻜﻦ‬

‫‪PP1 1‬‬

‫= ‪AA1‬‬ ‫= ‪(AC )2 + (CA1 )2‬‬ ‫‪9+9 = 3 5‬‬
‫‪42‬‬

‫‪. PP1‬‬ ‫=‬ ‫‪1‬‬ ‫‪AA1‬‬ ‫=‬ ‫‪35‬‬ ‫إذن‪،‬‬
‫‪2‬‬ ‫‪4‬‬

‫)‪ (٧٤‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ △ABC ،‬ﻣﺘﺴﺎوي اﻷﺿﻼع‪،AM = MB ،‬‬
‫‪ .CP = PM ،CK = KS = SN = NB‬ﻣﺎ ﻗﻴﺎس اﻟﺰاوﻳﺔ ‪ BNM‬؟‬

‫‪C‬‬

‫‪K‬‬

‫‪PS‬‬
‫‪N‬‬

‫)د( ‪90°‬‬ ‫‪AM‬‬ ‫‪B‬‬ ‫)أ( ‪70°‬‬

‫)ج( ‪80°‬‬ ‫)ب( ‪75°‬‬