Question

triangle rst is rotated 90° clockwise with the center of rotation at the origin, followed by a shift 1 unit right to form triangle lmn. which statement proves △rst ≅ △lmn? 〇 since \\(\overline{rs} \cong \overline{lm}\\), \\(\overline{st} \cong \overline{mn}\\), and \\(\overline{rt} \cong \overline{ln}\\), the two triangles are congruent by the sss postulate. 〇 since \\(\overline{rs} \cong \overline{mn}\\), \\(\overline{st} \cong \overline{lm}\\), and \\(\overline{rt} \cong \overline{ln}\\), the two triangles are congruent by the sss postulate. 〇 since \\(\angle rst \cong \angle lmn\\), \\(\angle trs \cong \angle mnl\\), and \\(\overline{rs} \cong \overline{mn}\\), the two triangles are congruent by the asa postulate. 〇 since \\(\angle rts \cong \angle mln\\), \\(\angle trs \cong \angle mnl\\), and \\(\overline{rt} \cong \overline{ln}\\), the two triangles are congruent by the asa postulate.

Explanation:

Step1: Recall Congruence Postulates

SSS (Side - Side - Side) postulate states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. ASA (Angle - Side - Angle) postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

Step2: Analyze Transformations

Rotation and translation are rigid transformations. Rigid transformations preserve the shape and size of the figure, so the corresponding sides of \(\triangle RST\) and \(\triangle LMN\) should be congruent.

Step3: Check Each Option

- Option 1: Check side congruence. Let's assume the coordinates (from the grid). For \(\triangle RST\) and \(\triangle LMN\), after rotation (90° clockwise) and translation (1 unit right), the corresponding sides: \(RS\) and \(MN\) (wait, no, let's re - evaluate). Wait, when we rotate 90° clockwise around the origin, the transformation of a point \((x,y)\) is \((y, - x)\), then translate 1 unit right (add 1 to x - coordinate). But maybe easier to look at the sides. The first option says \(RS\cong LM\), \(ST\cong MN\), \(RT\cong LN\). Wait, no, let's check the second option: \(RS\cong MN\), \(ST\cong LM\), \(RT\cong LN\). Wait, actually, from the grid, let's see the lengths. \(RS\): horizontal segment, length let's say 3 units (from x=-6 to x=-3, y = 1, for example). \(MN\): horizontal segment, length 3 units. \(ST\): vertical segment, length 4 units (from y = 1 to y = 5). \(LM\): vertical segment, length 4 units. \(RT\): the hypotenuse, length \(\sqrt{3^{2}+4^{2}} = 5\). \(LN\): hypotenuse, length 5. So \(RS\cong MN\), \(ST\cong LM\), \(RT\cong LN\). So by SSS, the triangles are congruent. Wait, no, the first option: \(RS\cong LM\), \(ST\cong MN\), \(RT\cong LN\). Wait, \(LM\) is vertical, \(RS\) is horizontal, so they can't be congruent. The second option: \(RS\cong MN\) (both horizontal, same length), \(ST\cong LM\) (both vertical, same length), \(RT\cong LN\) (both hypotenuse, same length). So the second option is correct? Wait, no, wait the options:

First option: "Since \(\overline{RS}\cong\overline{LM}\), \(\overline{ST}\cong\overline{MN}\), and \(\overline{RT}\cong\overline{LN}\), the two triangles are congruent by the SSS postulate."

Second option: "Since \(\overline{RS}\cong\overline{MN}\), \(\overline{ST}\cong\overline{LM}\), and \(\overline{RT}\cong\overline{LN}\), the two triangles are congruent by the SSS postulate."

Ah, here's the mistake. \(RS\) is a horizontal side, \(MN\) is a horizontal side (same length), \(ST\) is vertical, \(LM\) is vertical (same length), and \(RT\) and \(LN\) are hypotenuses (same length). So the second option has the correct corresponding sides. Wait, but wait the answer is the second option? Wait no, wait the first option: no, let's re - check the problem. Wait, the triangles: \(\triangle RST\) and \(\triangle LMN\). After rotation (90° clockwise) and translation, the corresponding sides: \(RS\) (bottom side of left triangle) corresponds to \(MN\) (bottom side of right triangle)? No, wait, when you rotate 90° clockwise, the horizontal side becomes vertical and vice - versa? Wait no, 90° clockwise rotation: a horizontal segment (along x - axis) with length \(a\) (from \((x,y)\) to \((x + a,y)\)) will become a vertical segment (along y - axis) with length \(a\) (from \((y, - x)\) to \((y, - x + a)\))? Wait, maybe I messed up. Let's take coordinates. Let's assume \(S\) is at \((-3,1)\), \(R\) is at \((-6,1)\), \(T\) is at \((-3,5)\). Then rotate 90° clockwise around origin: \((x,y)\to(y, - x)\). So \(S(-3,1)\to(1,3)\), \(R(-6,1)\to(1,6)\), \(T(-3,5)\to(5,3)\). Then translate 1 unit right: add 1 to x - coordinate. So \(S\) becomes \((2,3)\), \(R\) becomes \((2,6)\), \(T\) becomes \((6,3)\). Wait, but in the diagram, \(M\) is at \((2,1)\), \(L\) is at \((2,5)\), \(N\) is at \((5,1)\). Wait, maybe my coordinate assumption is wrong. Let's look at the diagram: \(R\), \(S\) are on the same horizontal line (y = 1), \(S\) is at x=-3, \(R\) at x=-6, \(T\) at x=-3, y = 5. Then \(M\) is at x = 2, y = 1; \(L\) at x = 2, y = 5; \(N\) at x = 5, y = 1. So after rotation 90° clockwise (around origin): \((x,y)\to(y, - x)\). So \(S(-3,1)\to(1,3)\), then translate 1 unit right: (2,3) – but \(M\) is at (2,1). Oh, maybe rotation is 90° clockwise around a different center? Wait, the problem says "center of rotation at the origin". Wait, maybe I made a mistake in rotation direction. 90° counter - clockwise is \((-y,x)\), 90° clockwise is \((y, - x)\). Wait, maybe the triangle \(RST\): \(R(-6,1)\), \(S(-3,1)\), \(T(-3,5)\). Rotate 90° clockwise around origin: \(R(-6,1)\to(1,6)\), \(S(-3,1)\to(1,3)\), \(T(-3,5)\to(5,3)\). Then translate 1 unit right: \(R(2,6)\), \(S(2,3)\), \(T(6,3)\). But in the diagram, \(L\) is at (2,5), \(M\) at (2,1), \(N\) at (5,1). So maybe the rotation is 90° clockwise around a different point? Wait, the problem says "center of rotation at the origin". Maybe the diagram is drawn with \(RST\) having \(S\) at (-2,1), \(R\) at (-5,1), \(T\) at (-2,5), and \(M\) at (2,1), \(L\) at (2,5), \(N\) at (5,1). Then rotation 90° clockwise around origin: \((x,y)\to(y, - x)\). So \(S(-2,1)\to(1,2)\), translate 1 unit right: (2,2) – no, \(M\) is at (2,1). I think I'm overcomplicating. The key is that rotation and translation are rigid motions, so corresponding sides are congruent. The SSS postulate requires three pairs of congruent sides. The second option: \(RS\cong MN\) (both horizontal, same length), \(ST\cong LM\) (both vertical, same length), \(RT\cong LN\) (both hypotenuse, same length). So by SSS, the triangles are congruent. Wait, but the first option says \(RS\cong LM\), \(ST\cong MN\), \(RT\cong LN\). \(LM\) is vertical, \(RS\) is horizontal, so they can't be congruent. The second option: \(RS\cong MN\) (horizontal, same length), \(ST\cong LM\) (vertical, same length), \(RT\cong LN\) (hypotenuse, same length). So the second option is correct? Wait, no, wait the options:

Wait the options are:

1. Since \(\overline{RS}\cong\overline{LM}\), \(\overline{ST}\cong\overline{MN}\), and \(\overline{RT}\cong\overline{LN}\), the two triangles are congruent by the SSS postulate.

2. Since \(\overline{RS}\cong\overline{MN}\), \(\overline{ST}\cong\overline{LM}\), and \(\overline{RT}\cong\overline{LN}\), the two triangles are congruent by the SSS postulate.

3. Since \(\angle RST\cong\angle LMN\), \(\angle TRS\cong\angle MNL\), and \(\overline{RS}\cong\overline{MN}\), the two triangles are congruent by the ASA postulate.

4. Since \(\angle RTS\cong\angle MLN\), \(\angle TRS\cong\angle MNL\), and \(\overline{RT}\cong\overline{LN}\), the two triangles are congruent by the ASA postulate.

Now, \(\angle RST\) is a right angle (since \(RS\) is horizontal and \(ST\) is vertical), \(\angle LMN\) is also a right angle. But in option 3, the angles: \(\angle TRS\) and \(\angle MNL\) – are they congruent? Probably not. Option 4: angles \(\angle RTS\) and \(\angle MLN\) – not sure. But for SSS, we need three sides. \(RS\) and \(MN\): both horizontal, length equal. \(ST\) and \(LM\): both vertical, length equal. \(RT\) and \(LN\): hypotenuse, length equal (by Pythagoras, since \(RS^{2}+ST^{2}=RT^{2}\) and \(MN^{2}+LM^{2}=LN^{2}\), and \(RS = MN\), \(ST = LM\), so \(RT = LN\)). So the second option has three pairs of congruent sides, so by SSS, the triangles are congruent. Wait, but earlier I thought the first option was wrong. Wait, maybe I mixed up the corresponding sides. Let's see the triangles: \(\triangle RST\) and \(\triangle LMN\). \(RS\) is the bottom side of \(\triangle RST\), \(MN\) is the bottom side of \(\triangle LMN\). \(ST\) is the left side of \(\triangle RST\), \(LM\) is the left side of \(\triangle LMN\). \(RT\) is the hypotenuse of \(\triangle RST\), \(LN\) is the hypotenuse of \(\triangle LMN\). So \(RS\cong MN\), \(ST\cong LM\), \(RT\cong LN\). So the second option is correct.

Wait, but the answer is the second option? Wait no, wait the first option: \(RS\cong LM\), \(ST\cong MN\), \(RT\cong LN\). \(LM\) is the left side of \(\triangle LMN\) (vertical), \(RS\) is the bottom side of \(\triangle RST\) (horizontal), so they are not congruent. The second option: \(RS\cong MN\) (both bottom sides, horizontal, same length), \(ST\cong LM\) (both left sides, vertical, same length), \(RT\cong LN\) (both hypotenuses, same length). So the second option is correct.

Answer:

B. Since \(\overline{RS}\cong\overline{MN}\), \(\overline{ST}\cong\overline{LM}\), and \(\overline{RT}\cong\overline{LN}\), the two triangles are congruent by the SSS postulate.