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Rafael wrote a proof to prove that β–³QRSβˆΌβ–³TUV β–³ 𝑄 𝑅 𝑆 ∼ β–³ 𝑇 π‘ˆ 𝑉 . Drag and drop the missing pieces of information to complete his proof. (2 points) Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse. First, I will find the lengths of corresponding sides QRΒ―Β―Β―Β―Β―Β―Β―Β― 𝑄 𝑅 Β― and TUΒ―Β―Β―Β―Β―Β―Β― 𝑇 π‘ˆ Β― . Since they are vertical and horizontal, I can count the units. The length of QRΒ―Β―Β―Β―Β―Β―Β―Β― 𝑄 𝑅 Β― is 4. The length of TUΒ―Β―Β―Β―Β―Β―Β― 𝑇 π‘ˆ Β― is 7. I will set up a ratio to find the scale factor from QRΒ―Β―Β―Β―Β―Β―Β―Β― 𝑄 𝑅 Β― to TUΒ―Β―Β―Β―Β―Β―Β― 𝑇 π‘ˆ Β― . Response area Second, I will use the Response area to find the lengths of corresponding sides RSΒ―Β―Β―Β―Β―Β―Β― 𝑅 𝑆 Β― and UVΒ―Β―Β―Β―Β―Β―Β―Β― π‘ˆ 𝑉 Β― . RS=(5βˆ’1)2+(7βˆ’5)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑅 𝑆 = ( 5 βˆ’ 1 ) 2 + ( 7 βˆ’ 5 ) 2 RS=(4)2+(2)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑅 𝑆 = ( 4 ) 2 + ( 2 ) 2 RS=16+4βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑅 𝑆 = 16 + 4 RS=20βˆ’βˆ’βˆš 𝑅 𝑆 = 20 RS=25β€“βˆš 𝑅 𝑆 = 2 5 UV=(14.5βˆ’11)2+(5βˆ’(βˆ’2))2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = ( 14.5 βˆ’ 11 ) 2 + ( 5 βˆ’ ( βˆ’ 2 ) ) 2 UV=(3.5)2+(7)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = ( 3.5 ) 2 + ( 7 ) 2 UV=12.25+49βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = 12.25 + 49 UV=61.25βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = 61.25 UV=3.55β€“βˆš π‘ˆ 𝑉 = 3.5 5 I will set up a ratio to find the scale factor from Response area. 3.55√25√=3.52=1.75 3.5 5 2 5 = 3.5 2 = 1.75 Finally, I will use the distance formula to find the lengths of corresponding sides SQΒ―Β―Β―Β―Β―Β―Β― 𝑆 𝑄 Β― and VTΒ―Β―Β―Β―Β―Β―Β― 𝑉 𝑇 Β― . SQ= 𝑆 𝑄 = Response area SQ=(βˆ’4)2+(βˆ’6)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑆 𝑄 = ( βˆ’ 4 ) 2 + ( βˆ’ 6 ) 2 SQ=16+36βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑆 𝑄 = 16 + 36 SQ=52βˆ’βˆ’βˆš 𝑆 𝑄 = 52 SQ=213βˆ’βˆ’βˆš 𝑆 𝑄 = 2 13 VT=(4βˆ’14.5)2+(βˆ’2βˆ’5)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = ( 4 βˆ’ 14.5 ) 2 + ( βˆ’ 2 βˆ’ 5 ) 2 VT=(βˆ’10.5)2+(βˆ’7)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = ( βˆ’ 10.5 ) 2 + ( βˆ’ 7 ) 2 VT=110.25+49βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = 110.25 + 49 VT=159.25βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = 159.25 VT=3.513βˆ’βˆ’βˆš 𝑉 𝑇 = 3.5 13 I will set up a ratio to find the scale factor from SQΒ―Β―Β―Β―Β―Β―Β― 𝑆 𝑄 Β― to VTΒ―Β―Β―Β―Β―Β―Β― 𝑉 𝑇 Β― . 3.513√213√=3.52=1.75 3.5 13 2 13 = 3.5 2 = 1.75 I have proven that all three pairs of corresponding sides are Response area in length because for each pair of sides the scale factor is 74=1.75 7 4 = 1.75 . Therefore, β–³QRSβˆΌβ–³TUV β–³ 𝑄 𝑅 𝑆 ∼ β–³ 𝑇 π‘ˆ 𝑉 by the Response area. congruentSide-Side-Side Similarity Theoremdistance formula(1βˆ’5)2+(1βˆ’7)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš ( 1 βˆ’ 5 ) 2 + ( 1 βˆ’ 7 ) 2 (1βˆ’1)2+(5βˆ’7)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš ( 1 βˆ’ 1 ) 2 + ( 5 βˆ’ 7 ) 2 proportionalRSΒ―Β―Β―Β―Β―Β―Β― 𝑅 𝑆 Β― to UVΒ―Β―Β―Β―Β―Β―Β―Β― π‘ˆ 𝑉 Β― 74=1.75 7 4 = 1.75 slope formula47β‰ˆ0.57 4 7 β‰ˆ 0.57 UVΒ―Β―Β―Β―Β―Β―Β―Β― π‘ˆ 𝑉 Β― to RSΒ―Β―Β―Β―Β―Β―Β― 𝑅 𝑆 Β― Side-Angle-Side Similarity Theorem Skip to navigation Item 1 , Fully Attempted. Item 2 , Fully Attempted. Item 3 , Unattempted. Item 4 , Unattempted. Item 5 , Unattempted. Previous Next

7 answers

To complete Rafael's proof for proving that \( \triangle QRS \sim \triangle TUV \), the missing pieces of information can be filled in as follows:

  1. For the ratio setup between \( QR \) and \( TU \):

    • Response Area:
    • I'll set up a ratio to find the scale factor from \( QR \) to \( TU \).
    • \(\frac{QR}{TU} = \frac{4}{7} \)

  2. To find the lengths of \( RS \) and \( UV \):

    • Response Area:
    • I will set up a ratio to find the scale factor from \( RS \) to \( UV \).
    • \(\frac{RS}{UV} = \frac{\sqrt{20}}{\sqrt{61.25}} = \frac{2\sqrt{5}}{\sqrt{61.25}}\)

  3. For the last ratio setup between \( SQ \) and \( VT \):

    • Response Area:
    • I will set up a ratio to find the scale factor from \( SQ \) to \( VT \).
    • \(\frac{SQ}{VT} = \frac{\sqrt{52}}{\sqrt{159.25}} = \frac{2\sqrt{13}}{\sqrt{159.25}}\)

  4. To finalize the proof that all three pairs of corresponding sides are proportional:

    • Response Area:
    • I have proven that all three pairs of corresponding sides are proportional in length because for each pair of sides, the scale factor is \( \frac{7}{4} = 1.75 \).

  5. For the concluding statement of the proof:

    • Response Area:
    • Therefore, \( \triangle QRS \sim \triangle TUV \) by the Side-Angle-Side Similarity Theorem.

By filling in these specific responses in the appropriate areas, the proof will be complete and logically structured.

the options are 1 congruent 2 sss similarity theorem 3 distance formula 4 sqrt (1-5)^2 + (1-7)^2. 5 sqrt (1-1)^2 + (5-7)^2. 6 proportional 7 RS to UV 8 7/4 = 1.75 9 slope formula 10 4/7 β‰ˆ 0.57 11 UV to RS 12 SAS similarity theorem

Based on the given options and the context of the proof for \( \triangle QRS \sim \triangle TUV \), here’s how to complete Rafael’s proof by choosing the correct options:

  1. For the ratio setup between \( QR \) and \( TU \):

    • Response Area:
    • I will set up a ratio to find the scale factor from \( QR \) to \( TU \).
    • Selected Response: \( \frac{4}{7} \approx 0.57 \)

  2. To find the lengths of \( RS \) and \( UV \):

    • Response Area:
    • I will set up a ratio to find the scale factor from \( RS \) to \( UV \).
    • Selected Response: RS to UV.

  3. For the last ratio setup between \( SQ \) and \( VT \):

    • Response Area:
    • I will set up a ratio to find the scale factor from \( SQ \) to \( VT \).
    • Selected Response: UV to RS.

  4. To finalize the proof that all pairs of corresponding sides are proportional:

    • Response Area:
    • I have proven that all three pairs of corresponding sides are proportional in length because for each pair of sides, the scale factor is \( \frac{7}{4} = 1.75 \).
    • Selected Response: Proportional.

  5. For the concluding statement of the proof:

    • Response Area:
    • Therefore, \( \triangle QRS \sim \triangle TUV \) by the SAS similarity theorem.
    • Selected Response: SAS similarity theorem.

Putting it all together, the completed proof will look like this:

  1. I will set up a ratio to find the scale factor from \( QR \) to \( TU \).

    • Response: 4/7 β‰ˆ 0.57

  2. I will set up a ratio to find the scale factor from \( RS \) to \( UV \).

    • Response: RS to UV

  3. I will set up a ratio to find the scale factor from \( SQ \) to \( VT \).

    • Response: UV to RS

  4. I have proven that all three pairs of corresponding sides are proportional in length because for each pair of sides the scale factor is \( 7/4 = 1.75 \).

    • Response: Proportional

  5. Therefore, \( \triangle QRS \sim \triangle TUV \) by the SAS similarity theorem.

    • Response: SAS similarity theorem

This sequence provides a clear and coherent proof of similarity between the two triangles.

First, I will find the lengths of corresponding sides QRΒ―Β―Β―Β―Β―Β―Β―Β― 𝑄 𝑅 Β― and TUΒ―Β―Β―Β―Β―Β―Β― 𝑇 π‘ˆ Β― . Since they are vertical and horizontal, I can count the units. The length of QRΒ―Β―Β―Β―Β―Β―Β―Β― 𝑄 𝑅 Β― is 4. The length of TUΒ―Β―Β―Β―Β―Β―Β― 𝑇 π‘ˆ Β― is 7. I will set up a ratio to find the scale factor from QRΒ―Β―Β―Β―Β―Β―Β―Β― 𝑄 𝑅 Β― to TUΒ―Β―Β―Β―Β―Β―Β― 𝑇 π‘ˆ Β― . 47β‰ˆ0.57 4 7 β‰ˆ 0.57 Second, I will use the Response area to find the lengths of corresponding sides RSΒ―Β―Β―Β―Β―Β―Β― 𝑅 𝑆 Β― and UVΒ―Β―Β―Β―Β―Β―Β―Β― π‘ˆ 𝑉 Β― . RS=(5βˆ’1)2+(7βˆ’5)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑅 𝑆 = ( 5 βˆ’ 1 ) 2 + ( 7 βˆ’ 5 ) 2 RS=(4)2+(2)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑅 𝑆 = ( 4 ) 2 + ( 2 ) 2 RS=16+4βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑅 𝑆 = 16 + 4 RS=20βˆ’βˆ’βˆš 𝑅 𝑆 = 20 RS=25β€“βˆš 𝑅 𝑆 = 2 5 UV=(14.5βˆ’11)2+(5βˆ’(βˆ’2))2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = ( 14.5 βˆ’ 11 ) 2 + ( 5 βˆ’ ( βˆ’ 2 ) ) 2 UV=(3.5)2+(7)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = ( 3.5 ) 2 + ( 7 ) 2 UV=12.25+49βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = 12.25 + 49 UV=61.25βˆ’βˆ’βˆ’βˆ’βˆš π‘ˆ 𝑉 = 61.25 UV=3.55β€“βˆš π‘ˆ 𝑉 = 3.5 5 I will set up a ratio to find the scale factor from Response area. 3.55√25√=3.52=1.75 3.5 5 2 5 = 3.5 2 = 1.75 Finally, I will use the distance formula to find the lengths of corresponding sides SQΒ―Β―Β―Β―Β―Β―Β― 𝑆 𝑄 Β― and VTΒ―Β―Β―Β―Β―Β―Β― 𝑉 𝑇 Β― . SQ= 𝑆 𝑄 = Response area SQ=(βˆ’4)2+(βˆ’6)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑆 𝑄 = ( βˆ’ 4 ) 2 + ( βˆ’ 6 ) 2 SQ=16+36βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑆 𝑄 = 16 + 36 SQ=52βˆ’βˆ’βˆš 𝑆 𝑄 = 52 SQ=213βˆ’βˆ’βˆš 𝑆 𝑄 = 2 13 VT=(4βˆ’14.5)2+(βˆ’2βˆ’5)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = ( 4 βˆ’ 14.5 ) 2 + ( βˆ’ 2 βˆ’ 5 ) 2 VT=(βˆ’10.5)2+(βˆ’7)2βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = ( βˆ’ 10.5 ) 2 + ( βˆ’ 7 ) 2 VT=110.25+49βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = 110.25 + 49 VT=159.25βˆ’βˆ’βˆ’βˆ’βˆ’βˆš 𝑉 𝑇 = 159.25 VT=3.513βˆ’βˆ’βˆš 𝑉 𝑇 = 3.5 13 I will set up a ratio to find the scale factor from SQΒ―Β―Β―Β―Β―Β―Β― 𝑆 𝑄 Β― to VTΒ―Β―Β―Β―Β―Β―Β― 𝑉 𝑇 Β― . 3.513√213√=3.52=1.75 3.5 13 2 13 = 3.5 2 = 1.75 I have proven that all three pairs of corresponding sides are Response area in length because for each pair of sides the scale factor is 74=1.75 7 4 = 1.75 . Therefore, β–³QRSβˆΌβ–³TUV β–³ 𝑄 𝑅 𝑆 ∼ β–³ 𝑇 π‘ˆ 𝑉 by the Response area. fill in the blanks

To complete your proof, based on the context you've provided, I'll fill in the blanks with the appropriate responses. Here's how you can fill in the missing pieces:

  1. To find the lengths of corresponding sides \( QR \) and \( TU \):

    • First, I will find the lengths of corresponding sides \( \overline{QR} \) and \( \overline{TU} \). Since they are vertical and horizontal, I can count the units. The length of \( \overline{QR} \) is 4. The length of \( \overline{TU} \) is 7. I will set up a ratio to find the scale factor from \( \overline{QR} \) to \( \overline{TU} \).
    • Fill in the Blank: \( \frac{4}{7} \approx 0.57 \)

  2. To find the lengths of corresponding sides \( RS \) and \( UV \):

    • Second, I will use the distance formula to find the lengths of corresponding sides \( \overline{RS} \) and \( \overline{UV} \).

    • Fill in the Blank: distance formula

    • \( RS = (5-1)^2 + (7-5)^2\)

    • \[ RS = (5 - 1)^2 + (7 - 5)^2 = (4)^2 + (2)^2 = 16 + 4 = \sqrt{20} \]

    • \( UV = (14.5 - 11)^2 + (5 - (-2))^2\)

    • \[ UV = (3.5)^2 + (7)^2 = 12.25 + 49 = \sqrt{61.25} \]

  3. To set up the corresponding ratio:

    • I will set up a ratio to find the scale factor from \( RS \) to \( UV \).
    • Fill in the Blank: Response area could be: \( \frac{RS}{UV} \) or you can use values \( \frac{\sqrt{20}}{\sqrt{61.25}} = \frac{2\sqrt{5}}{\sqrt{61.25}} \)

  4. Calculating the scale factor:

    • I will set up a ratio to find the scale factor from \( \overline{SQ} \) to \( \overline{VT} \).
    • Fill in the Blank: proportional

  5. Lengths of corresponding sides \( SQ \) and \( VT \):

    • Finally, I will use the distance formula to find the lengths of corresponding sides \( \overline{SQ} \) and \( \overline{VT} \).
    • Fill in the Blank: distance formula (for the lengths computed)

  6. Fill in the last part:

    • I have proven that all three pairs of corresponding sides are proportional in length because for each pair of sides the scale factor is \( \frac{7}{4} = 1.75 \).
    • Fill in the Blank: proportional

  7. Conclusion of the proof:

    • Therefore, \( \triangle QRS \sim \triangle TUV \) by the SAS similarity theorem.
    • Fill in the Blank: SAS similarity theorem

Putting all these together, the missing pieces filled in your proof would look like this:

  1. \( \frac{4}{7} \approx 0.57 \)
  2. distance formula
  3. proportional
  4. distance formula
  5. proportional
  6. SAS similarity theorem

This should complete the proof showing that \( \triangle QRS \sim \triangle TUV \) based on the distances and ratios of the corresponding sides you calculated.

how is it sas when it says that all three pairs of coresponfing sides are either A proportional or B congruent

The confusion seems to stem from the difference between "congruent" and "proportional" in terms of triangle similarity. Let's clarify these terms:

  1. Proportional: When discussing triangle similarity, two triangles are similar if their corresponding sides are in proportion. This means the ratios of the lengths of their corresponding sides are equal. For example, if you have triangles ABC and DEF, then if \(\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}\), the triangles are similar (\( \triangle ABC \sim \triangle DEF\)).

  2. Congruent: Congruent triangles are those that are exactly the same in shape and size, which means all their corresponding sides are equal in length (\(AB = DE\), \(BC = EF\), \(AC = DF\)).

Why SAS Similarity Theorem Applies

The Side-Angle-Side (SAS) Similarity Theorem states that if two sides of one triangle are proportional to two sides of another triangle, and the included angles are congruent, then the two triangles are similar.

Your Situation

In your proof, since you are establishing that all corresponding pairs of sides are either proportional or congruent, you need to assert that:

  • If all three pairs of sides are determined to be proportional (which means you have ratios like \( \frac{QR}{TU} \), \( \frac{RS}{UV} \), and \( \frac{SQ}{VT} \) that are equal), you conclude the triangles are similar based on the definition of similarity.
  • If you were to find all three pairs of sides equal (congruent), then the triangles would be congruent.

Conclusion

So, in your proof, if you indeed found that all corresponding sides are proportional, then you conclude the triangles are similar by the criteria for similarity (not by SAS specifically if only the sides are compared). The conclusion should reinforce that:

  • If all three corresponding side pairs are proportional, then you would conclude \( \triangle QRS \sim \triangle TUV\) due to proportional sides.
  • If they were equal (like \( AB = DE \)), it would imply congruence.

So your final response should state:

I have proven that all three pairs of corresponding sides are proportional in length because for each pair of sides, the scale factor is \( \frac{7}{4} = 1.75\). Therefore, \( \triangle QRS \sim \triangle TUV \) by the Side-Side-Side (SSS) similarity theorem since all sides are proportional.

If one of those side pairs had been proven congruent instead of just proportional, then SAS or SSS could apply based on the information given. In any case, focus on whether your proof hinges on proportions or congruences when drawing your conclusion.

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