Triangle Congruence: SAS

Triangles A Q R and A K P share point A. Triangle A Q R is rotated up and to the right for form triangle A Q R. Which rigid transformation would map ΔAQR to ΔAKP? a rotation about point A a reflection across the line containing AR a reflection across the line containing AQ a rotation about point R

Triangles RQS and NTV have the following characteristics: • Right angles at ∠Q and ∠T • RQ ≅ NT Can it be concluded that ΔRQS ≅ ΔNTV by SAS? Why or why not? Yes, one set of corresponding sides and one corresponding angle are congruent. Yes, they are both right triangles. No, it is necessary to know that another set of corresponding sides is congruent. No, it is not possible for the triangles to be congruent.

Which of these triangle pairs can be mapped to each other using a single translation? A.Triangles C E D and C N P are congruent. Triangle C E D is rotated about point C and then reflected across a line to form triangle C N P. B.Triangles C N E and C E D are congruent. Triangle C N E is reflected across a line and then rotated slightly to form triangle C E D. C.Triangles C E D and M D P are congruent. Triangle M D P is rotated and shifted up to form triangle C E D. D.Triangles C E D and M P N are congruent. Triangle C E D is shifted to the right to form triangle M P N. Mark this and return

Triangles H J K and L M N are congruent. Triangle H J K is rotated about point H to form triangle L N M. Triangle L M N is higher than triangle H J K. Two rigid transformations are used to map ΔHJK to ΔLMN. The first is a translation of vertex H to vertex L. What is the second transformation? a reflection across the line containing HK a rotation about point H a reflection across the line containing HJ a rotation about point K

The proof that ΔEFG ≅ ΔJHG is shown. Given: G is the midpoint of HF, EF ∥ HJ, and EF ≅ HJ. Prove: ΔEFG ≅ ΔJHG Triangles E F G and J H G share common point G. Statement Reason 1. G is the midpoint of HF 1. given 2. FG ≅ HG 2. def. of midpoint 3. EF ∥ HJ 3. given 4. ? 4. alt. int. angles are congruent 5. EF ≅ HJ 5. given 6. ΔEFG ≅ ΔJHG 6. SAS What is the missing statement in the proof? ∠FEG ≅ ∠HJG ∠GFE ≅ ∠GHJ ∠EGF ≅ ∠JGH ∠GEF ≅ ∠JHG

Which of these triangle pairs can be mapped to each other using both a translation and a rotation about C? Triangles X Y C and A B C are shown. Both triangles are congruent and share common point C. Triangle A B C is slightly lower than triangle X Y C. Triangles X Y Z and A B C are shown. Both triangles are congruent. Triangle X Y Z is identical to triangle A B C but is slightly higher. Triangles X Y Z and A B C are shown. Both triangles are congruent. Triangle X Y Z is reflected across a line to form triangle A B C. Triangles X Y Z and A B C are shown. Both triangles are congruent. Triangle X Y Z is rotated down and to the left to form triangle A B C. It is also slightly higher than triangle A B C.

Triangles H J K and L M N are congruent. Triangle H J K is rotated about point H to form triangle L N M. Triangle L M N is higher than triangle H J K. How can a translation and a rotation be used to map ΔHJK to ΔLMN? Translate H to L and rotate about H until HK lies on the line containing LM. Translate K to M and rotate about K until HK lies on the line containing LM. Translate K to N and rotate about K until HK lies on the line containing LN. Translate H to N and rotate about H until HK lies on the line containing LN.

Triangles J K L and M N R are shown. In the diagram, KL ≅ NR and JL ≅ MR. What additional information is needed to show ΔJKL ≅ ΔMNR by SAS? ∠J ≅ ∠M ∠L ≅ ∠R ∠K ≅ ∠N ∠R ≅ ∠K

Triangles A B C and X Y Z is congruent. Triangle X Y Z is slightly higher and to the right of triangle A B C. Triangle A B C is reflected across a line to form triangle X Y Z. Two rigid transformations are used to map ΔABC to ΔXYZ. The first is a translation of vertex A to vertex X. What is the second transformation? a reflection across the line containing line AB a reflection across the line containing line AC a rotation about point A a rotation about point B

The proof that ΔMNS is congruent to ΔQNS is shown. Given: ΔMNQ is isosceles with base , and and bisect each other at S. Prove: Square M N Q R is shown with point S in the middle. Lines are drawn from each point of the square to point S to form 4 triangles. We know that ΔMNQ is isosceles with base. So, by the definition of isosceles triangle. The base angles of the isosceles triangle, and , are congruent by the isosceles triangle theorem. It is also given that and bisect each other at S. Segments _______ are therefore congruent by the definition of bisector. Thus, by SAS. NS and QS NS and RS MS and RS MS and QS