Rotating ΔSTU by 180° clockwise about the origin, we get its image ΔS'T'U' as shown below.Note that ΔSTU has vertices S (-5, 7), T(-6, 3), and U (-8, 1).Also, note that ΔS'T'U' has vertices S' (5, -7), I' (6, -3), and U' (8, -1).Complete the following questions a, b, and c.

Rotating ΔSTU by 180 clockwise about the origin we get its image ΔSTU as shown belowNote that ΔSTU has vertices S 5 7 T6 3 and U 8 1Also note that ΔSTU has vert class=

Respuesta :

To find the length of each side, we use the distance formula.

[tex]d=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2_{}}[/tex]

Let's find side CS first. The points are C(0,0) and S(-5,7). So, the coordinates are

[tex]x_1=0,x_2=-5,y_1=0,y_2=7[/tex]

Use the coordinates to find the length of CS.

[tex]\begin{gathered} CS=\sqrt[]{(7-0)^2+(-5-0)^2}=\sqrt[]{7^2+(-5)^2} \\ CS=\sqrt[]{49+25}=\sqrt[]{74} \end{gathered}[/tex]

Therefore, the length of CS is

[tex]CS=\sqrt[]{74}[/tex]

Let's repeat the same process for CT and CU. Given that C is just (0,0), we can use just T and U as coordinates.

[tex]CT=\sqrt[]{3^2+(-6)^2}=\sqrt[]{9+36}=\sqrt[]{45}=\sqrt[]{9\cdot5}=3\sqrt[]{5}[/tex]

Therefore, the length of CT is

[tex]CT=3\sqrt[]{5}[/tex]

Let's do the process one more time for CU.

[tex]CU=\sqrt[]{1^2+(-8)^2}=\sqrt[]{1+64}=\sqrt[]{65}[/tex]

Therefore, the length of CU is

[tex]CU=\sqrt[]{65}[/tex]