HT bisects DG at O. If HO=3x+17, TO=4x-2, DO= 3x+10 and GO= 6x-20, how do you find the value of x?

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Answer 1 · 2017-07-20T18:32:06

$x = 10$ $x=10$

$H T$ $HT$ is the perpendicular bisector of $D G$ $DG$ which means that
$D O = O G$ $DO = OG$

$D G$ $DG$ is not automatically the bisector of $H T$ $HT$ , so you do NOT know that $H O = O T$ $HO= OT$

Find $x$ $x$ if $O G = D O$ $OG = DO$

$6 x - 20 = 3 x + 10$ $6x-20 = 3x+10$

$6 x - 3 x = 10 + 20$ $6x-3x=10+20$

$3 x = 30$ $3x =30$

$x = 10$ $x =10$