問二題數學題~~ @ oob73zj92b的部落格

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問二題數學題~~

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3. 2x+3x = 180°(adj. ∠s on st. line) 5x = 180° x = 36° ∠APQ = 2x = 72° ∵∠APQ = ∠PQD ∴AB//CD (alt. ∠s eq.) 18. △ABD~△BCD (AAA) 6/10 = 5/x (corr. ∠s, ~△s) x = 25/3 6/10 = x/y (corr. ∠s, ~△s) 6/10 = (25/3)/y y = 125/9

其他解答:

1. ∵∠APY +∠BPY=180 ° ∴2X +3X =180° 5X=180° X=36° ∵ ∠XQD=∠CQY=72° ∠APY=2X=2*36°=72° ∴ AB ∥CD 2.∵ ∠BCD=∠ABD,∵ ∠BDC=∠ADB ∴ ⊿BCD ≌⊿ABD ∴AB/BC=AD/BD=BD/CD 6/10=5/X=X/Y ∴X=25/3 Y=125/9|||||1. 2x + 3x = 180 (adj. angle s on st. lie) 5x = 180 x = 36 angle APQ = 2x =72 angle XQD = 72 angle XQD = angle APQ so AB//CD (alt. angle s equal) 2. Consider triangle ABD and BCD. angle BCD = angle ABD (given) angle BDC = angle ADB (given) angle BCD + angle BDC + angle CBD = angle ABD + angle ADB angle BAD = 180 (angle sum of triangle) angle CBD = angle BAD triangle ABD ~ triangle BCD (AAA) AD/BD = AB/BC (corr. sides, similar triangle) 5/x = 6/10 x = 25/3 AB/BC = BD/CD (corr. sides, similar triangle) 6/10 = x/y 3/5 = (25/3)/y y = 125/9