uestion 1

Question

A)△ABC has side lengths of 10, 16, and 23 units. Suppose you want to draw a congruent and smaller triangle within ΔABC. Investigate a pattern between the original triangle and smaller triangle and make a conjecture. What is a possible midsegment length of the smaller triangle? (1 point)
Responses

4 units
4 units

13 units
13 units

8 units
8 units

10 units
10 units
Question 2
A)
Use the image to answer the question.

A triangle marked clockwise from the left vertex as upper A upper B upper C. Congruent line segments and angles are marked.

In the figure, ΔDBE has the following side lengths: DB¯¯¯¯¯¯¯¯=12 units, BE¯¯¯¯¯¯¯¯=7 units, and DE¯¯¯¯¯¯¯¯=14 units. What is the length of AB¯¯¯¯¯¯¯¯?

(1 point)
Responses

24 units
24 units

28 units
28 units

12 units
12 units

14 units
14 units
Question 3
A) △ABC has side lengths of 10, 12, and 8 inches. Suppose you want to draw a congruent and smaller triangle within ΔABC. What is a possible midsegment length of the smaller triangle? (1 point)
Responses

8 inches
8 inches

10 inches
10 inches

6 inches
6 inches

16 inches
16 inches
Question 4
A)
Use the image to answer the question.

A triangle is marked clockwise from the lower left vertex as upper A upper B upper C. Point upper D is marked at the midpoint of side upper A upper B. Point upper F is marked at the midpoint of side upper A upper C.

Opal has already proved that △ADF∼△ABC. They are now attempting to prove that DF=12BC and DF¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯. Critique their reasoning. Which option should fill in the blank?

PROOF: Because corresponding sides of similar triangles are proportional, ________________. Multiply both sides of the equation by segment BC to show that DF=12BC. Because corresponding angles of similar triangles are congruent, ∠ADF≅∠ABC. If corresponding angles are congruent, then lines are parallel, so DF¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯.

(1 point)
Responses

BCDF=2
Start Fraction upper B upper C over upper D upper F End Fraction equals 2

DF=BC
upper D upper F equals upper B upper C

DFBC=12
Start Fraction upper D upper F over upper B upper C End Fraction equals Start Fraction 1 over 2 End Fraction

AD=DB
upper A upper D equals upper D upper B
Question 5
A)
Use the image to answer the question.

A triangle is labeled clockwise as upper F upper G upper H, with vertex F pointing left. Point upper I is marked at the midpoint of side upper F upper G. Point upper J is marked at the midpoint of side upper G upper H. A line connects upper I and upper J.

Sage has already proved that △IGJ∼△FGH. She is now attempting to prove that IJ¯¯¯¯¯¯∥FH¯¯¯¯¯¯¯¯. Help her to construct a viable argument. What is the missing statement and reason in her proof?

Statements Reasons
1. △IGJ∼△FGH 1. Given
2. 2.
3. IJ¯¯¯¯¯¯∥FH¯¯¯¯¯¯¯¯ 3. If corresponding angles are congruent, then lines are parallel
(1 point)
Responses

The missing statement is ∠JIF≅∠GFH and the missing reason is “CPCTC Theorem.”
The missing statement is angle upper J upper I upper F congruent to angle upper G upper F upper H and the missing reason is “CPCTC Theorem.”

The missing statement is ∠GIJ≅∠GFH and the missing reason is “CPCTC Theorem.”
The missing statement is angle upper G upper I upper J congruent to angle upper G upper F upper H and the missing reason is “CPCTC Theorem.”

The missing statement is ∠GJI≅∠GHF and the missing reason is “corresponding angles of similar triangles are congruent.”
The missing statement is angle upper G upper J upper I congruent to angle upper G upper H upper F and the missing reason is “corresponding angles of similar triangles are congruent.”

The missing statement is ∠JIF≅∠GFH and the missing reason is “corresponding angles of

Answer 1

Here are the answers to your questions:

Question 1: In triangle ΔABC with side lengths 10, 16, and 23 units, we can consider a triangle formed by the midsegments. The lengths of the sides of the smaller triangle formed by midsegments will be half the corresponding sides of the larger triangle. Thus, the lengths of the midsegments will be:

• Half of 10 = 5 units
• Half of 16 = 8 units
• Half of 23 = 11.5 units

From the options provided, the possible midsegment length of the smaller triangle is 8 units.

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Question 2: In triangle ΔDBE with side lengths DB = 12 units, BE = 7 units, and DE = 14 units, we can reason that triangle ΔDBE is similar to triangle ΔABC, so corresponding sides have the same ratio. The side lengths AB in triangle ABC can be found based on the proportions.

Calculating the ratio based on the given sides:

• Since DE is to AB as DB is to BC, and BE matches the BC, we consider DE as the proportional base to find AB.

If we assume the triangles are congruent under some vertical stretch, we can deduce the size of AB. It depends on how they maintain their proportions.

The answer here would be 24 units assuming a square scaling factor of 2 for dimension conservation based on provided lengths.

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Question 3: For triangle ABC with side lengths 10, 12, and 8 inches, the possible midsegment lengths can again be found by taking each side and halving them:

1. Half of 10 = 5 inches
2. Half of 12 = 6 inches
3. Half of 8 = 4 inches

From the options provided, the possible midsegment length of the smaller triangle is 6 inches.

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Question 4: In the proof, to show that DF = 1/2 BC, we rely on the properties of similar triangles. The missing statement should assert that segment DF corresponds to segment BC at half the proportional length based on the midpoint definition.

Therefore, the filled-in statement should be: "Start Fraction upper B upper C over upper D upper F End Fraction equals 2".

This clarifies the proportional relationship between the two lengths where segment DF is indeed half of segment BC.

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Question 5: To populate the proof that supports the claim IJ ∥ FH, we look to demonstrate that corresponding angles between similar triangles are congruent to deduce parallelism.

The missing statement should be: "The missing statement is ∠GJI≅∠GHF and the missing reason is 'corresponding angles of similar triangles are congruent.'"

By establishing this angle relationship, it concludes that the line segment IJ is parallel to segment FH.

If you have any further questions or need clarifications, feel free to ask!