F.2數 10分－飛行里程

標題:

F.2數 10分

發問:

求下列各未知數題目係呢個網入面 http://boxstr.com/files/4777313_7s6vt/dj.jpg

最佳解答:

4. PRS = P + Q (D外角) 54o = 27o + h h = 27o P = Q = 27o 所以 DRPQ 是等腰 D。 PR = QR (等腰 D 的兩腰) PR = 4.8 cm k = 4.8 5. ABD + DBE + EBC = 180o (直線同側鄰角) 58o + m + 79o = 180o m = 43o DBE = DEB = 43o 所以 DDBE 是等腰 D。 DB = DE (等腰 D 的兩腰) n = 3n - 6 2n = 6 n = 3 7. AB = AC = 4cm (己知) AD = AD (公共邊) BD = CD (己知) DABD DACD (SSS) ADB = ADC = a (全等三邊形對應角) BAD = CAD = b (全等三邊形對應角) CD = BD = 2 cm (全等三角形對應邊) ADB + ADC = 180o (直線同側鄰角) a + a = 180o a = 90o BC = (2 + 2) cm = 4 cm 所以，AB = BC = CA = 4 cm DABC 是等邊D。 BAC = 60o b + b = 60o b = 30o 8. 右上角的三角形是等腰三角形，兩底角均等於 x。 100o + x + x = 180o (D內角和) x = 40o 下方的三角形為等邊三角形。每一個內角 = 60o y = 60o + x y = 100o =

其他解答:

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4. h + 27 degrees = 54 degrees (ext. ∠ of triangle) h = 54 degrees - 27 degrees h = 27 degrees ∠Q = ∠P (from above) so, QR = PR (opp. sides, = ∠s) k cm = 4.8 cm k = 4.8 5. ∠ABD + ∠DBE + ∠CBE = 180 degrees (adj. ∠s on st. line) 58 degrees + m + 79 degrees = 180 degrees m = 180 degrees - 58 degrees - 79 degrees m = 43 degrees ∠DBE = ∠BED (from above) so, DB = DE (opp. sides, = ∠s) n = 3n-6 6 = 3n-n 6 = 2n n = 3 7. because AB = AC and BD = DC so triangle ABC is a isos. triangle a = 90 degrees (prop. of isos. triangle) BD = DC so, BC = (2+2)cm = 4cm = AB = CA so triangle ABC is also an equil. triangle 2b = 60 degrees (prop. of equil. triangle) b = 30 degrees 8. Let B & C be the pts on the left corner and the right one such that triangle ABC is an equil. triangle & triangle ACD is an isos. triangle. x = ∠ACD (base. ∠s, isos. triangle) 2x + 100 degrees = 180 degrees (∠ sum of triangle) 2x = 80 degrees x = 40 degrees y = ∠ACD + ∠ACB & ∠ACD = x = 40 degrees (from above) ∠ACB = 60 degrees (prop. of equil. triangle) so, y = 40 degrees + 60 degrees = 100 degrees