ABCD is a trapezoid where AB and CD are the legs while AD is the longest base and BC is the shortest base. XY is parallel to AD, BX = 1/4AB, BC=10 and AD=20. Find XY.
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2007-11-25 20:26:58 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I already gave you the method to get the answer in your other question...
BC = 10
AD = 20
Imagine there was a line that connected the midpoints of the two sides. I'll call this line EF. The length of EF would be the average of the two bases.
EF = AD + BC = (10 + 20)/2
EF equals 15. This would be the midline, but XY is the average of this midline (EF) and the top side (BC).
XY = (EF + BC) / 2
(10 + 15) / 2 = 12.5
XY has a length of 12.5
2007-11-25 20:32:20 · answer #1 · answered by Puzzling 7 · 1⤊ 0⤋
from the figure (tapezoid ABCD), drop a line from C normal or perpendicular to line AD (parallel to line AB).call that intersection point X..Line AX=BC=10, therefore line XD=10. use ratio and proportion and you should get distance XY=12.5
2007-11-25 20:59:31 · answer #2 · answered by dm_dizon 1 · 0⤊ 0⤋
as already you know Kokey,
XY = (3BC + AD)/4
= (3*10 + 20)/4
= 12.5
2007-11-25 20:39:29 · answer #3 · answered by sv 7 · 0⤊ 0⤋