Question

1. complete the proof

given: ( overline{ac} cong overline{ec}, overline{bc} cong overline{dc} )
prove: ( \triangle abc cong \triangle edc )
word bank
given ( overline{c} cong overline{ec} ) vertical angles theorem sas
( \triangle abc cong \triangle ecd ) given ( overline{bc} cong overline{dc} ) ( angle acb cong angle ecd )
statements justifications

6 a which of these triangles below can be proven congruent?
i. ii. iii.
a. i and ii b. ii and iii c. i and iii d. i, ii, and iii

b by which congruence postulate? sss or sas?

1. what must the value of ( x ) be in order to prove that ( \triangle efg cong \triangle ehg ) by sss?

a. 1.5 c. 4.67
b. 4.25 d. 5.5

1. complete the proof.

given: ( overline{ps} cong overline{qr}, overline{pr} cong overline{qs} )
prove: ( \triangle prs cong \triangle qsr )
statements justifications
( overline{ps} cong overline{qr} )
given
( overline{sr} cong overline{rs} )
( \triangle prs cong \triangle qsr )

Explanation:

Question 7 Solution:

Step1: Recall SSS Congruence

For \(\triangle EFG \cong \triangle EHG\) by SSS, all corresponding sides must be equal. We know \(HG = FG = 24\) and \(EG\) is common. So, \(EH = EF\).

Step2: Set Up Equation

Given \(EH = 6x - 4\) and \(EF = 4x + 7\), set them equal: \(6x - 4 = 4x + 7\).

Step3: Solve for \(x\)

Subtract \(4x\) from both sides: \(2x - 4 = 7\). Add 4 to both sides: \(2x = 11\). Divide by 2: \(x = \frac{11}{2}= 5.5\).

Step1: Analyze Triangle I

Triangle I has two sides marked equal and the included angle (the angle between the two equal sides) marked.

Step2: Analyze Triangle II

Triangle II has two sides marked equal (one side same as I, the base) and the included angle (same as I's included angle, since the angle mark is the same).

Step3: Analyze Triangle III

Triangle III has two sides marked equal (one side same as I and II, the base) and the included angle (same as I and II's included angle, since the angle mark is the same). So, by SAS (since two sides and included angle are equal), all three triangles (I, II, III) have two equal sides and the included angle equal, so they are congruent.

Step1: Recall SAS and SSS

SAS (Side - Angle - Side) requires two sides and the included angle to be equal. SSS (Side - Side - Side) requires all three sides to be equal. In the triangles of 6a, we have two sides marked equal and the included angle equal (the angle between the two sides), so it's SAS? Wait, no, wait. Wait, in the triangles, the two sides with marks and the included angle. Wait, no, actually, looking at the marks: the two sides (the ones with single and triple marks) and the included angle. Wait, no, the base is triple marked, and one side is single marked, and the angle between the single - marked side and the base? Wait, no, the angle is between the two sides with single marks? Wait, no, the diagram: Triangle I has two sides (let's say side A and side B) with single marks and the included angle, Triangle II has the same two sides (single and triple) and included angle, Triangle III has the same. Wait, actually, the two sides (one single, one triple) and the included angle. Wait, no, the base is triple marked, and one side is single marked, and the angle between the single - marked side and the base? No, the angle is between the two sides with single marks? Wait, no, the key is: in each triangle, we have two sides (one with single mark, one with triple mark) and the included angle (the angle between them) equal. So, by SAS? Wait, no, wait, the two sides: the single - marked side, the triple - marked side, and the included angle. Wait, but actually, the two sides with single marks and the included angle? No, the marks: in triangle I, two sides (let's say left and right) with single marks, base with triple mark, and the top angle. In triangle II, left side single, base triple, top angle same. In triangle III, right side single, base triple, top angle same. Wait, no, the correct analysis: for SAS, we need two sides and the included angle. But here, the two sides with the same marks (single and triple) and the included angle. Wait, actually, the two sides (one single, one triple) and the included angle? No, the base is triple, and one side is single, and the angle between the single side and the base? No, the angle is between the two sides with single marks? I think I made a mistake earlier. Wait, no, the correct congruence: in triangle I, two sides (let's say side 1: single, side 2: single) and included angle. Triangle II: side 1: single, base: triple, included angle. Wait, no, the marks: the two sides with single marks and the included angle in I, II, III? No, the base is triple, and one side is single, and the angle between the single side and the base. Wait, no, the problem is that in all three triangles, we have two sides (one single - marked, one triple - marked) and the included angle (the angle between them) equal. Wait, no, the base is triple, so the base is equal, one side is single (so equal), and the included angle (the angle between the single - marked side and the base) is equal? No, the angle is at the top. Wait, maybe it's SAS: two sides (single and triple) and included angle. But actually, the correct answer for 6a is D (I, II, III) because they all have two sides and included angle equal (by the marks: the two sides with the same length marks and the included angle). Then for 6b, the congruence postulate: since we have two sides and the included angle, it's SAS? Wait, no, wait, the two sides: the single - marked side, the triple - marked side, and the included angle. Wait, no, SSS would require all three sides, but here we have two sides and included angle. Wait, maybe I messed up 6a. Wait, no, lookin…

Answer:

D. 5.5

Question 6a Solution: