Answered

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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. Why is Jordan's proof incorrect?

Sagot :

przemyslawstefaniak

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.

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