Question

The figure below shows CB = 4, BE = 5, AB = 2x – 2, and DB = x + 2.



If ΔABC ~ ΔDBE, the value of x is _________

Answer 1

http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_Seg_Two_Exam_36/image0044e9d7cb1.jpg

Answer 2

PLEASE!!!!!

Answer 3

@panlac01

Answer 4

May someone please help?

Answer 5

Is it 4?

Answer 6

Since ΔABC ~ ΔDBE
\(\frac{AB}{DB} = \frac{BC}{BE}\) (corr. sides, ~Δs)
\[\frac{2x-2}{x+2} = \frac{4}{5}\]5(2x-2) = 4(x+2)
10x - 10 = 4x + 8
Solve x.

Answer 7

3?

Answer 8

I think so.

Answer 9

Now the question is 4 or 3......

Answer 10

hmm...

Answer 11

x = 4 is the ans.

Answer 12

Thank you!!!

Answer 13

i have found this without using the similarity condition

Answer 14

@manishsatywali May I know how you get the answer and where I did it wrong?

Answer 15

jst. find the intersection point of two lines, you will get your ans.

Answer 16

How?

Answer 17

I mean intersection point of two lines.

Answer 18

@manishsatywali ?

Answer 19

dear i m attaching drawing for u