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zimovet [89]
2 years ago
12

7 Point Question Tyler was trying to prove that all rhombuses are similar. He thought, “If I draw 2 rhombuses, I can find dilati

ons that will take one rhombus exactly onto the other.” Then he wrote a proof. However, all rhombuses are not similar. Which steps in Tyler’s proof have flaws? Explain what is wrong with each step that is wrong. Step 1: Let ABCD and WXYZ be rhombuses. By the definition of a rhombus, AB≅BC≅CD≅DA and WX≅XY≅YZ≅ZW. Step 2: Dilate rhombus ABCD by the scale factor given by $$ WX AB​. Side A'B' will be the same length as WX because of how I chose the scale factor. Because all sides of a rhombus are congruent, B'C' will be the same length as WX and therefore the same length as XY, C'D' will be the same length as YZ, and A'D' will be the same length as WZ. That means all the corresponding sides of A'B'C'D' and WXYZ will be the same length. Step 3: If all the sides in 2 figures are proportional, then those 2 figures are similar, so there must be a sequence of transformations that take ABCD onto WXYZ using dilations and rigid motions. Step 4: Translate A'B'C'D' by the directed line segment A'W so that A″ and W coincide. Step 5: Rotate A″B″C″D″ by angle B″WX so that B‴ and X coincide. Now segments A‴B‴ and WX coincide. Step 6: If needed, reflect A‴B‴C‴D‴ over segment WX so that D⁗ and Z are on the same side of WX. Now segments A⁗D⁗ and WZ coincide. Step 7: Once 3 vertices of the rhombus are lined up, the other vertex has to line up as well or else the shapes wouldn’t be rhombuses. So, we have shown we can use rigid motions and dilations to line up any 2 rhombuses, and therefore, all rhombuses are similar. (Select all that apply.) Step 1 is incorrect. The definition of a rhombus does not say that all the sides are equal. Step 2 is incorrect. The scale factor is not $$ . Step 3 is incorrect. It is only true for triangles that if all the side lengths in a figure are proportional, the figures must be similar. Step 4 is incorrect. Translating by the directed line segment
Mathematics
1 answer:
Viefleur [7K]2 years ago
7 0

A rhombus can have the same shape and dimension as a square or a

parallelogram, therefore, all rhombuses are not similar.

  • A step in the proof that have flaws is <u>step 3</u>.

Reasons:

By reading through the given steps, we have, the first step that is incorrect

is step 3.

Step 3: The requirement for the similarity of two figures based on the

proportionality of the corresponding sides of the figures is related with

triangles as given by the triangle proportionality theorem.

This is so because, the lengths of the sides of a triangle depends on the

opposite interior angle, while the side of a rhombus can have a

combination of large length and and acute angle.

Two rhombuses that have proportional sides may be dissimilar,

because they can have different interior angles.

Therefore;

<u>The step in Tyler's proof that have a flaw is Step 3</u>

Learn more about similarity between geometric figures here:

brainly.com/question/10270676

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