+0  
 
0
37
2
avatar

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
avatar+20883 
+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
avatar+694 
+2

Rectangles ABCD and ANME are similar.

 

AE is 1/2 of AD

 

AG must be 1/2 of AE    indecision

 

 May 19, 2020
edited by Dragan  May 19, 2020

21 Online Users

avatar
avatar
avatar
avatar