Thursday
March 23, 2017

Post a New Question

Posted by on Thursday, September 19, 2013 at 2:21am.

Given: FG (SEG) bisects BT (SEG)
BO=7x-6
OT=5x+10

Prove: BO=50

  • geometry - , Thursday, September 19, 2013 at 3:08am

    There appears to be insufficient information. Where is O, with respect to BT and FG?

  • geometry - , Thursday, September 19, 2013 at 4:06am

    It appears to me that O is midway between B and T. If so, then

    7x-6 = 5x+10
    2x = 16
    x = 8

    So, BO = 5*8+10 = 50

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question