Can someone pretty please help me with #14 on page 312-Geometry? I looked at it & I see how they did it, but the whole process doesn't make since to me

312-geometry

#1

Can someone pretty please help me with #14 on page 312-Geometry? I looked at it & I see how they did it, but the whole process doesn’t make since to me.

Answer:

Since the two segments are parallel, then the two triangles XMN and XYZ are similar, and, according to the Triangle Proportionality Theorem, the sides intersected by MN are divided into proportional segments. This is very similar to comparing sides of similar triangles. But, in this case you are comparing pieces of the sides of one triangle (XYZ) that have been created by the parallel line (MN) cutting through it. The pieces on one side are XM and MY which together form the side XY. The pieces on the other side are XN and NZ which together form the side XZ. These pieces are proportional, keeping similar pieces in the numerator, and similar pieces in the denominator. So, if you keep the sides of the small triangle in the numerator, and the extra pieces between the parallel lines in the denominator, you end up with XM/MY = XN/NZ. You can then substitute what you know to find the missing piece, MY. But, the exercise doesn’t ask for the length of that piece, it asks for the length of the whole side, XY. So, you have to add the lengths of the pieces, XM and MY to get that total length.

You also could have done this without the Triangle Proportionality Theorem. Since, you have two similar triangles you could simply set up a proportion of their sides. XM/XY = XN/XZ. You would have to add the values you know, XN and NZ to get the value of XZ. Then substitute into the proportion to find the missing value, XY.