Answer:
Part 9) [tex]x=5[/tex]
Part 10) [tex]x=-9[/tex]
Step-by-step explanation:
we know that
The Midpoint Theorem states that: The segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side
Part 9) we know that
Point Q is the midpoint segment XY (QY=QX)
Point R is the midpoint segment XW (RW=RX)
Applying the Midpoint Theorem
RQ is parallel to WY
and
[tex]RQ=\frac{1}{2}WY[/tex]
we have
[tex]RQ=2x-3\\WY=x+9[/tex]
substitute
[tex]2x-3=\frac{1}{2}(x+9)[/tex]
solve for x
[tex]4x-6=(x+9)[/tex]
[tex]4x-x=9+6[/tex]
[tex]3x=15[/tex]
[tex]x=5[/tex]
Part 10) we know that
Point B is the midpoint segment TS (BS=BT)
Point C is the midpoint segment RS (CS=CR)
Applying the Midpoint Theorem
BC is parallel to TR
and
[tex]BC=\frac{1}{2}TR[/tex]
we have
[tex]BC=x+19\\TR=x+29[/tex]
substitute
[tex]x+19=\frac{1}{2}(x+29)[/tex]
solve for x
[tex]2x+38=(x+29)[/tex]
[tex]2x-x=29-38[/tex]
[tex]x=-9[/tex]
