i challenge you to answer this question! CHALLENGE YOU!!?
819602
If the bases are in the ratio of k:1, then we can say they have lengths of kx and x (kx and x have ratio k:1). Let's say the trapezoid has height h.\n\nThe mid-segment of the trapezoid cuts the trapezoid into two similar trapezoids, and has length (1/2)(kx + x)\n\nThe area of a trapezoid is (1/2)(height)(b1 + b2).\n\nThe top trapezoid has height h/2, and bases kx and (1/2)(kx + x).  It has an area \nA1 = (1/2)(h/2)(kx + (1/2)(kx + x))\n     = (h/4)(kx + kx/2 + x/2)\n     = (h/4)(3kx/2 + x/2)\n     = (hx/8)(3k + 1)\n\nThe bottom trapezoid has height h/2 and bases (1/2)(kx + x) and x.  It has an area\nA2 = (1/2)(h/2)((1/2)(kx + x) + x)\n     = (h/4)(kx/2 + x/2 + x)\n     = (h/4)(kx/2 + 3x/2)\n     = (hx/8)(k + 3)\n\nIf A1:A2 = 5:2, then \n[(hx/8)(3k + 1)]/[(hx/8)(k + 3)] = 5/2\n(3k+1)/(k+3) = 5/2\n2(3k+1) = 5(k+3)\n6k + 2 = 5k + 15\nk = 13\n(C)
1058752
Answering this question.
618405
knock u out, huh? i think u just answered ur own question!
1131739
ABSOLUTELY! That's an awesome question! Can I use that?
1046487
please repeat that question ???
349034
when you start asking yourself that question!
375608
This is a trick question...No fair!
688304
The other question is been answered!
848756
Re-phrase the question.
261688
This is a trick question..right?????
344538
Hey!  What is self challenge? I am curious possibly willing to join the challenge?
155623
With this question, you have to use your opinion!!!
465365
do your own homework!
997809
how many people have to ask this question?? no offence on u but so many people have asked this already!
629795
At least 6...I'm gonna challenge people on it!!!!!
769047
Answering your question!
610910
The person asking this question!