PLEASE HELP Square ABCD has a side length of 1. Point E lies on the interior of ABCD and is on the line ←→AC such that the length of AE is 1. Find the shortest distance from point E to a side of square ABCD.

he shortest distance from the point E to a side of square ABCD is 0.293

Step-by-step explanation:

The question parameters are

Shape of figure ABCD = Square

Point E lies on the diagonal line AC

The length of the segment AE = 1

Therefore, we have;

Length of AC = √(AB² + CD²) = √(1² + 1²) = √2

Hence, the point E is closer to the point C and the closest distance to a side   from E is the perpendicular from the point E to BC at point E' or to CD at poit E'' which is found as follows;

AC is a bisector of ∠DAB, hence;

∠DAC = 45° = ∠CAE'

EE' = EC × cos(45°)

EC = AC - AE = √2 - 1

Therefore;

EE' = (√2 - 1) × cos(45°) = (√2 - 1) × (√2)/2 = 1 - (√2)/2 = 0.293

Hence, the shortest distance from the point E to a side of square ABCD = 0.293.

answer: plz put a pic in so i can answer it

step-by-step explanation: