急!!!!!!!!!!!!!!4題數學問題only~~(20分)－gbu94mz95e的部落格

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急!!!!!!!!!!!!!!4題數學問題only~~(20分)

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我最想知19題答案我想知以下問題的答案&點計 http://inlinethumb50.webshots.com/42737/2166068050101901781S600x600Q85.jpg http://inlinethumb24.webshots.com/43479/2935193280101901781S600x600Q85.jpg http://inlinethumb39.webshots.com/41190/2034711070101901781S600x600Q85.jpg

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Q19.) 3x > -2y 3x/y < -2 (Since y<0, the sign would reverse if both side is divided by y) x/y < -2/3 So the Answer is "D" Q20.) Let Y be the length of the Square ABCD If AD is the base of the Triangle ADE Then the base of the Triangle ADE is Y and the height of the Triangle ADE = Y/2 ( Since AE = BE) Area of AED = (Y)(Y/2)(1/2) = (Y^2)/4 Area of ABCD = Y^2 Area of AED / Area of ABCD = (Y^2/4) / Y^2 = 1 / 4 The answer is "D" Q28.) If AE is the base of Triangle ADE Then Triangle ADE will have the same height of parallelogram ABCD Let B be the base of Triangle ADE and H be the height of Triangle ADE Area of parallelogram ABCD = 2B x H = 2BH Area of Triangle ADE = (B x H)/2 = (BH) / 2 Similarly, Area of Triangle BCF = Area of Triangle ADE Area of Triangle BCF = (BH) / 2 Area of the polygon BFDE = Area of parallelogram ABCD - (Area of Triangle BCF and Triangle ADE) = 2BH - (BH)/2 - (BH)/2 = BH Since Area of QPEB = Area of QPFD Area of QPEB = (BH)/2 Since Area of parallelogram ABCD = 48 That is 2BH = 48 (BH)/2 = 48/4 = 12 Therefore Area of QPEB = 12 So the answer is "C" 2008-08-27 16:20:58 補充： Q32.) the Answer is "D"

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