Answer :
The proof is as follows :
Step-by-step explanation:
Let NC = x
⇒ AB = 3x and AN = 2x
In Δ ABN, By using Pythagoras theorem,
AB² = BN² + AN²
⇒ BN² = AB² - AN²
⇒ BN² = (3x)² - (2x)²
⇒ BN² = 5x²
⇒ BN = x√5 .......................(1)
Now in ΔANC , Using Pythagoras theorem We have,
AC² = NC² + AN²
⇒ AC² = x² + (2x)²
⇒ AC² = 5x²
⇒ AC = x√5 ....................(2)
From equations (1) and (2) We get,
AC = BN , which is our required result
Answer:
A. △P'Q'R' does not equal △P''Q''R''.
B. Reflecting across UT would change the orientation of the figure.
C. The sequence does not include a reflection that exchanges U and S.
D. Rotating about point U is not a rigid motion because it changes the orientation of the figure.
E. Translating point R' to Q' is a non-invertible transformation because it changes the location of P'.
(D) Rotating about U is not a rigid motion because it changes the orientation of the figure. [I think D is an incorrect answer choice.]
Step-by-step explanation:
Proof No.1
Jordan wants to prove △PQR≅△STU using a sequence of rigid motions. This is Jordan's proof. Translate △PQR to get △P'Q'R' with R'=U. Then rotate △P'Q'R' about point U to get △P''Q''R''. Since translation and rotation preserve distance, R''Q''=RQ=UT, and Q''=T. Reflect △P''Q''R'' across UT to get △P'''Q'''R''. Since reflection preserves distance, P'''R'''=PR=US, and P'''=S. A sequence of rigid motions maps △PQR onto △STU, so △PQR≅△STU.
Proof No.2
Jordan wants to prove △PQR≅△STU using a sequence of rigid motions. This is Jordan's proof. Translate △PQR to get △P'Q'R' with R'=U. Then rotate △P'Q'R about point U to get △P''Q''R'' so that R''Q'' and UT coincide. Since translation and rotation preserve distance, R''Q''=RQ=UT, and Q''=T. Reflect △P''Q''R'' across UT to get △P'''Q'''R''. Since reflection preserves distance, P'''R'''=PR=US, and P'''=S. A sequence of rigid motions maps △PQR onto △STU, so △PQR≅△STU.
The answer should be C... :) your welcome!!!
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I’m sooo bored
