Similarity Problem

AC is parallel to BD, and AF is parallel to BE.
Triangles PAC and PBD are similar
Triangles QEB and QFA are similar.
 
I do not know what a Spoiler is, would someone please tell me?

This problem is actually quite easy.
angle PAC and angle PBD are the same since the triangles planes are parallel

write EB/16 = FA / 28 and AC/9 = BD/21

solve for EB * BD in terms of FA *AC

area of a triangle = 1/2 A B sin c

the angle is the same in both triangles so the area of each is proportional to the 2 side lengths which is given above, turns out to be 96 like it should
 
PWC said:
I do not know what a Spoiler is, would someone please tell me?

This problem is actually quite easy.
angle PAC and angle PBD are the same since the triangles planes are parallel

write EB/16 = FA / 28 and AC/9 = BD/21

solve for EB * BD in terms of FA *AC

area of a triangle = 1/2 A B sin c

the angle is the same in both triangles so the area of each is proportional to the 2 side lengths which is given above, turns out to be 96 like it should
Click to expand...
It is a hidden text which you can see only by clicking on the spoiler icon.
I thought that some people might find the problem as an interesting challenge, so I did not make my hints automatically visible.
 
You must log in or register to reply here.
Top