Triangle RTS is congruent to RQS by AAS postulate of congruent
Step-by-step explanation:
Let us revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles
and one side in the 2nd Δ
- HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ
∵ SR bisects angle TSQ ⇒ given
∴ ∠TSR ≅ ∠QSR
∴ m∠TSR ≅ m∠QSR
∵ ∠T ≅ ∠Q ⇒ given
∴ m∠T ≅ m∠Q
In two triangles RTS and RQS
∵ m∠T ≅ m∠Q
∵ m∠TSR ≅ m∠QSR
∵ RS is a common side in the two triangle
- By using the 4th case above
∴ Δ RTS ≅ ΔRQS ⇒ AAS postulate
Triangle RTS is congruent to RQS by AAS postulate of congruent
Learn more:
You can learn more about the congruent in brainly.com/question/3202836
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Answer
58
Step-by-step explanation:
So you would plug in 4 for h and 6 for g to get 4+9(6), then you multiply 9x6 to get 54 then you would add 4 to get 58
Answer:
question 1: answer is 4
question 2: answer is 1
Step-by-step explanation:
