The image shows two quadrilaterals, labeled $ABCD$ and $EFGH$, and it specifies that they are similar, denoted by $ABCD \sim EFGH$. Some side lengths are given for both quadrilaterals, and you are asked to solve for $y$, which seems to be a missing side of the second quadrilateral $EFGH$.

Steps to solve for $y$:

Since the quadrilaterals are similar, corresponding sides of the two figures are proportional. This means we can set up a proportion using the known side lengths to solve for $y$.

From the image:

• $AB = 16$, $BC = 15$, and $CD = 18.7$ for quadrilateral $ABCD$.
• $EF = 8$, $FG = y$, and $GH = 9.4$ for quadrilateral $EFGH$.

Using the similarity of the two quadrilaterals, we can write the proportion: $\frac{AB}{EF} = \frac{BC}{FG} = \frac{CD}{GH}$ Substitute the known values: $\frac{16}{8} = \frac{15}{y} = \frac{18.7}{9.4}$ Now, solving for $y$, we can use the fact that $\frac{16}{8} = 2$, which gives us: $2 = \frac{15}{y}$ Multiplying both sides by $y$: $2y = 15$ Solving for $y$: $y = \frac{15}{2} = 7.5$

Thus, $y = 7.5$.

Would you like more details on this solution or have any other questions?

Related Questions:

1. What are the properties of similar figures?
2. How do you determine the scale factor between two similar polygons?
3. Can similarity be used to find angles in quadrilaterals?
4. How does the proportionality of corresponding sides extend to corresponding diagonals?
5. What is the difference between congruent and similar figures?

Tip: Always check for corresponding sides or angles when solving problems involving similar polygons!