AB is parallet to segment YZ

Nekkamath

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In the figure to the right, segment AB is parallel to segment YZ. If AZ = 42 units, BQ = 12 units and QY = 24 units, what is the length of segment QZ?

The picture is of a smaller upside down triangle on top of a larger right side up triangle. The top smaller triangle side points are A and Z the middle where they MEET is Q and the side points of the larger triangle is Y and Z.

I don't know where to start.
 
Nekkamath said:
In the figure to the right, segment AB is parallel to segment YZ. If AZ = 42 units, BQ = 12 units and QY = 24 units, what is the length of segment QZ?

The picture is of a smaller upside down triangle on top of a larger right side up triangle. The top smaller triangle side points are A and Z the middle where they MEET is Q and the side points of the larger triangle is Y and Z.

Where is B?

The triangles are similar - so their sides should be proportional.

I don't know where to start.
 
The B is to the right of the A. So the bottom of the upside down smaller triangle looks like this:

Code:
     A ____________ B
      \            /
       \          /
        \        /
         \      /
          \    /
           \  /
            Q   <- tip of both triangles
          /  \
        /      \
      /         \
    /             \
  /                \
Y___________________Z
 
I was trying to draw the side lines of the triangles. So imagine them starting at point A and B and meeting downward at Q. And the larger triangle starting at Y and Z going upward meeting at the Q in the middle.
 
Nekkamath said:
I was trying to draw the side lines of the triangles. So imagine them starting at point A and B and meeting downward at Q. And the larger triangle starting at Y and Z going upward meeting at the Q in the middle.

First prove that the triangles ABQ and YZQ are similar (i.e. their corresponding angles are equal)

Then use the law of proportionality - (I have given you this hint in the post above) - and then show your work indicating exactly where you are stuck.
 
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