I really need help if I fail this test I fail my grade so please help!!?

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#GA # is the diameter of the circle #T#.

Therefore#TA, TB, TD, TF, TG# are all radii of this circle and have a length of #5cm# as given in the question.

It is also given that #TD# (radius) is perpendicular to #BF# (chord). This automatically means that #BF# is bisected since it is perpendicular to the radius.

So as a result, #BH = FH = 1/2 xx BF = 4#.

Also, since #TD# is perpendicular to #BF#, that means we can look at triangle #THB#, where angle #THB = 90°#.

Apply Pythagorean theorem:

(#BH)^2 + (TH)^2 = (TB)^2#

#4^2 + (TH)^2 = 5^2#

#(TH)^2 = 25-16=9#

#TH = sqrt9=3#

You can solve for #TH#, or recognize this is a 3-4-5 triangle.

#TH = 3#

Where your question read 'BE' I think that should be 'BF'

#TA=5# and is a radius, #TB =5# as it too is a radius.

When a line from the centre meets a chord at #90°#, it bisects the chord. So when #TH# meets #BF# it bisects it making #BH=4#

#TBH# is a right angled triangle where the hypotenuse is #5#, one side is #4# so use Pythagoras to find the missing side #TH#

#5^2-4^2=TH^2#
#25-16=TH^2#
#TH=sqrt9 =3#

On the second question, #TL# is a radius and #ML# is a tangent so they meet at #90°#.

So #TML# is a right angled triangle so again use Pythagoras;

#12^2-9^2=TL^2#
#144-81=TL^2#
#TL=sqrt63#
#TL=7.937253933#