Page 140 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪١٢٧‬‬

‫)زواﻳﺎ اﳌﺜﻠﺚ(‬ ‫‪x + x + 110 = 180‬‬
‫‪2x = 70‬‬
‫‪x = 35°‬‬

‫)‪ [MAΘ 2010] (٦٦‬رﲰﻨﺎ ارﺗﻔﺎﻋﺎً ﻃﻮﻟﻪ ‪ 2 3‬إﱃ وﺗﺮ ﻣﺜﻠﺚ ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ‪ .‬إذا ﻛﺎن‬
‫ﻃﻮل إﺣﺪى ﻗﻄﻌﱵ اﻟﻮﺗﺮ ﻳﺴﺎوي ‪ 2‬ﻓﻤﺎ ﳏﻴﻂ اﳌﺜﻠﺚ ؟‬

‫)ب( ) ‪2(6 + 3‬‬ ‫)أ( ) ‪2(5 + 2 3‬‬

‫)د( ) ‪6(3 + 3‬‬ ‫)ج( ) ‪4(3 + 3‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ج(‪:‬‬

‫‪A‬‬

‫‪zy‬‬
‫‪23‬‬

‫‪Bx‬‬ ‫‪D2 C‬‬

‫‪( )y = 2 3 2 + 22 = 16 = 4‬‬

‫‪(x + 2)2 = y2 + z2 = 16 + z 2‬‬
‫‪x 2 + 4x + 4 = 16 + z2‬‬

‫وﻟﻜﻦ‪ .z2 = (2 )3 2 + x2 = 12 + x2 ،‬إذن‪،‬‬

‫‪x 2 + 4x + 4 = 16 + 12 + x2‬‬
‫‪4x = 24‬‬
‫‪x =6‬‬

‫و‪‬ﺬا ﻳﻜﻮن‪ .z = 12 + 36 = 48 = 4 3 ،‬وﻟﺬا ﻓﺈن ﳏﻴﻂ اﳌﺜﻠﺚ‪:‬‬