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ABCD is a rectangle and the lines ending at E, F, G are all parallel to AB, as shown. If AD = 12, what is AG?

May 18, 2020

#1
+21953
+1

There may be a direct route to the answer using similar triangles, but I couldn't find it,

so I used coordinate geometry.

Placing this in quadrant I, with D at the origin and the distance from D to C being a:

D = (0,0)     C = (a,0)     A = (0,12)     B = (a,12).

The equation of AC is:  y - 12  =  (12 - 0)/(0 - a) · (x - 0)     --->     y  =  (-12/a)·x + 12

Since AC and DB are diagonals, they intersect at the center of the rectangle = (a/2,6).

This makes E = (0,6).

The equation of BE:  y - 6  =  (12 - 6)/(a - 0) · (x - 0)     --->     y  =  (6/a)·x + 6

Where BE and AC intersect:  (6/a)·x + 6  =  (-12/a)·x + 12

(18/a)·x  =  6

x  =  (1/3)·a

Since   y  =  (6/a)·x + 6     --->     y  =  (6/a)·[(1/3)·a] + 6     --->     y  =  8

Combining, we have the point  (  (1/3)·a, 8 )

This makes point F = (0,8).

The equation of FB:  y - 8  =  (12 - 8)/(a - 0) · (x - 0)     --->     y  =  (4/a)·x + 8

Where FB and AC inersect:  (4/a)·x + 8  =  (-12/a)·x + 12

(16/a)·x  =  4

x  =  a/4

Since  y  =  (4/a)·x + 8     --->     y  =  (4/a)·(a/4) + 8     --->     y  =  9

This makes point G = (0,9)

So, the distance from A to G is from (0,12) to (0,9)  =  3

May 18, 2020
#2
+1326
+2

Rectangles ABCD and ANME are similar.