line xy is the midsegment of trapezoid abcd. find ab if line dc is 98 and line xy is 73.

Question

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Answer 1

Consider a triangle PDC obtained by extending the sloping sides of the trapezoid ABCD so that they meet at P.

Let h be the height of the trapezoid XYCD, while the height of trapezoid ABCD=2h. Let the height of triangle PAB which is atop the trapezoid ABCD be x. There are 3 similar triangles: PAB, PXY, PDC. We have DC=98 and XY=73.

Consider two similar triangles PAB and PXY: x/(x+h)=AB/73; now similar triangles PAB and PDC: x/(x+2h)=AB/98.

Therefore AB=73x/(x+h)=98x/(x+2h), therefore 73/(x+h)=98/(x+2h), 73(x+2h)=98(x+h), 73x+146h=98x+98h, 48h=25x. This relates x and h, h=25x/48. It also helps us to find AB=73x/(x+25x/48)=98x/(x+25x/24); AB=73/(1+25/48) and 98/(1+25/24)=48 (because the x's cancel out).