Question

yes
no
complete the similarity statement.
efgh ~
find the ratio of a side length in the first polygon to its corresponding si
second polygon.
simplify your answer and write it as a proper fraction, improper fraction,
number.

Explanation:

Part 1: Complete the similarity statement

To determine the similarity statement, we need to match the corresponding sides of the similar rectangles. For rectangle \(EFGH\), the sides are \(12\) (length) and \(6\) (width). For rectangle \(VWUT\) (let's check the labels), the sides are \(8\) (width) and \(16\) (length)? Wait, no, let's check the correspondence. Let's see the angles: all angles are right angles, so we need to match the sides. Let's find the ratio of sides. For \(EFGH\), length \(FG = 12\), width \(GH = 6\). For the other rectangle, let's see \(WT = 8\), \(VW = 16\)? Wait, no, maybe the correspondence is \(EFGH\) and \(WVTU\)? Wait, no, let's check the order of the vertices. Let's list the sides:

For \(EFGH\): \(FG = 12\), \(GH = 6\)

For the second rectangle: let's see the sides. Let's assume the sides are \(WT = 8\), \(VW = 16\). Wait, maybe the correct correspondence is \(EFGH \sim WVTU\)? Wait, no, let's check the ratios. Let's find the ratio of length to width for each rectangle.

For \(EFGH\): \(\frac{FG}{GH}=\frac{12}{6} = 2\)

For the second rectangle: let's take the sides. Let's say the sides are \(WT = 8\) and \(VW = 16\). Then \(\frac{VW}{WT}=\frac{16}{8}=2\). Wait, no, that's the same ratio. Wait, maybe the sides are \(GH = 6\) and \(WT = 8\)? No, that can't be. Wait, maybe I mixed up. Wait, the first rectangle: \(FG = 12\) (horizontal), \(GH = 6\) (vertical). The second rectangle: \(WT = 8\) (horizontal), \(VW = 16\) (vertical). Wait, no, the vertical side of the second rectangle is \(16\), horizontal is \(8\). So the ratio of horizontal to vertical for first rectangle: \(12/6 = 2\), for second: \(8/16 = 0.5\). Wait, that's the reciprocal. So maybe the correspondence is \(EFGH \sim UVWT\)? Wait, no, let's check the vertices. Let's list the rectangles:

Rectangle \(EFGH\): vertices \(E, F, G, H\) with right angles at each corner. So \(EF\) and \(FG\) are adjacent sides, \(FG\) and \(GH\) are adjacent, etc.

Rectangle \(VWUT\): vertices \(V, W, T, U\) with right angles. So \(VW\) and \(WT\) are adjacent, \(WT\) and \(TU\) are adjacent, etc.

Wait, maybe the correct correspondence is \(EFGH \sim WVTU\)? No, maybe I should look at the order of the sides. Let's find the ratio of corresponding sides. Let's take \(GH = 6\) (vertical side of \(EFGH\)) and \(WT = 8\) (vertical side of the second rectangle)? No, that doesn't make sense. Wait, maybe the second rectangle has sides \(8\) (horizontal) and \(16\) (vertical). So the horizontal side of \(EFGH\) is \(12\), vertical is \(6\). The horizontal side of the second rectangle is \(8\), vertical is \(16\). Wait, no, that would be a ratio of \(12/8 = 3/2\) and \(6/16 = 3/8\), which is not equal. Wait, I must have made a mistake. Wait, maybe the sides are \(GH = 6\) (vertical) and \(VW = 16\) (vertical)? No, that ratio is \(6/16 = 3/8\), and \(FG = 12\) (horizontal) and \(WT = 8\) (horizontal), ratio \(12/8 = 3/2\), which is not equal. Wait, that can't be. Wait, the problem says the rectangles are similar, so the ratios of corresponding sides must be equal. Wait, maybe I mixed up the sides. Let's check again.

Wait, the first rectangle: length \(12\), width \(6\). The second rectangle: length \(16\), width \(8\)? Wait, no, the second rectangle's vertical side is \(16\), horizontal is \(8\). So if we take the width of the first rectangle (\(6\)) and the length of the second (\(16\)), no. Wait, maybe the correct ratio is \(6/8 = 3/4\) and \(12/16 = 3/4\). Ah! There we go. So \(GH = 6\) (vertical side of first rectangle) corresponds to \(WT = 8\) (vertical side of second rectangle)? No, \(GH = 6\), \(VW = 16\)? N…

Step 1: Identify corresponding sides

We have rectangle \(EFGH\) with side \(FG = 12\) and \(GH = 6\), and rectangle \(UVWT\) (or the corresponding rectangle) with side \(VW = 16\) and \(WT = 8\). We found that the ratio of corresponding sides is \(\frac{6}{8} = \frac{3}{4}\) or \(\frac{12}{16} = \frac{3}{4}\). Let's confirm with \(GH = 6\) (side of first polygon) and \(WT = 8\) (corresponding side of second polygon), or \(FG = 12\) and \(VW = 16\).

Step 2: Calculate the ratio

Take \(FG = 12\) (side of \(EFGH\)) and its corresponding side \(VW = 16\) (side of the second polygon). The ratio is \(\frac{12}{16}\).

Step 3: Simplify the ratio

Simplify \(\frac{12}{16}\) by dividing numerator and denominator by their greatest common divisor, which is 4. So \(\frac{12 \div 4}{16 \div 4} = \frac{3}{4}\).

Answer:

Part 1: \(EFGH \sim UVWT\) (assuming the correct correspondence, but based on the ratio, the similarity statement is \(EFGH \sim WVTU\) or \(UVWT\); however, the most likely is \(WVTU\) or \(UVWT\), but based on the side ratios, the correct similarity statement is \(EFGH \sim WVTU\) (or \(UVWT\), but let's go with \(WVTU\) for now, but actually, the correct one is \(EFGH \sim WVTU\) or \(UVWT\); however, the key is the ratio is \(3/4\)).

Part 2: The ratio is \(\frac{3}{4}\)

Wait, but let's recheck. For \(EFGH\), side \(GH = 6\), corresponding side in the second rectangle is \(WT = 8\). So ratio \(6/8 = 3/4\). Or \(FG = 12\) and corresponding side \(VW = 16\), ratio \(12/16 = 3/4\). So the ratio is \(3/4\).

So the final answers:

Similarity statement: \(EFGH \sim WVTU\) (or \(UVWT\), but based on the diagram, the correct correspondence is \(EFGH \sim WVTU\))

Ratio: \(\frac{3}{4}\)

But let's confirm the similarity statement. Let's list the vertices in order. For \(EFGH\), the order is \(E, F, G, H\) (clockwise). For the second rectangle, let's say the order is \(W, V, T, U\) (clockwise)? No, maybe \(U, V, W, T\) (clockwise). So \(E\) corresponds to \(U\), \(F\) to \(V\), \(G\) to \(W\), \(H\) to \(T\). So \(EFGH \sim UVWT\). Yes, that makes sense. So \(EFGH \sim UVWT\).

And the ratio is \(3/4\).

So:

Part 1: \(EFGH \sim UVWT\)

Part 2: \(\frac{3}{4}\)