Geometry problem????

Question

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. Find XY.





Pls. answer it i need it now pls. show your solution pls. pls. i'm begging you pls. pls. pls. pls. i'm so very begging you pls.

Answer 1

You haven't provided any numbers but...

XY will be BC + (AD - BC) * 1/4

Normally the midpoints of the legs of a trapezoid are connected to form a line that is the average of the two bases. But here you are taking a line segment that is closer to the shorter base. It's actually the average of the midline and the shorter base.

For example, if BC were 10 and AD were 30, then XY would be 15
......B -------- 10 --------- C
....../ ....................... .... \
.....X ------------ 15 ------------- Y
..../ .................... ............ .... \
.../ .............. ........... ... ..... .......\
../ ........... ............... ........ ... ........\
./ ............ .................... ................ ...\
/ ............. ..................... ........... .... .....\
A ------------------------- 30 -------------------------- D

Here's another way to get the answer:
Average of AD and BC = (10 + 30) / 2 = 20.
Average of BC and 20 = (10 + 20) / 2 = 15

Answer 2

BC ll XY ll AD and
BX = 1/4 AB => 4BX = AB
then XY = BC + 1/4(AD -- BC)
=> XY = (3BC + AD)/4