Question

type the correct answer in the box.
the figure shows five polygons: abcde, fghij, klmno, pqrst, and uvwxy.
polygon cannot be mapped onto the others by similarity transformations.

Explanation:

Step1: Recall Similarity Transformations

Similarity transformations (translation, rotation, reflection, dilation) preserve the shape (angles) and the ratio of side lengths. So, similar polygons have corresponding angles equal and sides in proportion.

Step2: Analyze Each Polygon's Shape

- **UVWXY**: Let's check coordinates. V(0,2), W(4,2), X(4,5), Y(2,4), U(0,7). The angles and side ratios can be compared.
- **KLMNO**: M(10,10), L(12,10), K(12,5), O(11,7), N(10,7).
- **PQRST**: P(13,21), Q(17,21), R(17,15), S(15,19), T(13,17).
- **ABCDE**: A(0,22), B(0,12), C(8,12), D(8,18), E(4,16).
- **FGHIJ**: H(17,12), G(25,12), F(25,2), J(19,6), I(17,6).

Now, check the number of sides and angle measures. Wait, actually, look at the "slant" sides. For example, ABCDE: the side from A(0,22) to E(4,16): slope is (16 - 22)/(4 - 0) = -6/4 = -3/2. From E(4,16) to D(8,18): slope is (18 - 16)/(8 - 4) = 2/4 = 1/2.

For UVWXY: U(0,7) to Y(2,4): slope (4 - 7)/(2 - 0) = -3/2. Y(2,4) to X(4,5): slope (5 - 4)/(4 - 2) = 1/2. So same slopes (angles) as ABCDE.

KLMNO: N(10,7) to O(11,7)? Wait, N(10,7), O(11,8)? Wait, maybe coordinates: N(10,7), O(11,8), K(12,5), L(12,10), M(10,10). Slope N to O: (8 - 7)/(11 - 10) = 1/1 = 1. O to K: (5 - 8)/(12 - 11) = -3/1 = -3. Not same as ABCDE. Wait, no, maybe I misread. Wait, the key is that FGH IJ: Let's check the "slant" side. F(25,2) to J(19,6): slope (6 - 2)/(19 - 25) = 4/(-6) = -2/3. Which is different from the -3/2 and 1/2 slopes of the others. Wait, no, wait ABCDE: A(0,22) to E(4,16): run 4, rise -10? Wait, no, A is (0,22), B is (0,12) (so vertical side AB: length 10). B(0,12) to C(8,12): horizontal, length 8. C(8,12) to D(8,18): vertical, length 6. D(8,18) to E(4,16): run -4, rise -2 (slope (-2)/(-4)= 1/2? Wait, no, 16 - 18 = -2, 4 - 8 = -4, so slope 1/2. E(4,16) to A(0,22): run -4, rise 6, slope 6/(-4) = -3/2. So the two non - vertical/horizontal sides have slopes 1/2 and -3/2.

Now UVWXY: V(0,2), W(4,2) (horizontal, length 4), W(4,2) to X(4,5) (vertical, length 3), X(4,5) to Y(2,4) (run -2, rise -1, slope (-1)/(-2)=1/2), Y(2,4) to U(0,7) (run -2, rise 3, slope 3/(-2)= -3/2), U(0,7) to V(0,2) (vertical, length 5). So the non - axis - aligned sides have slopes 1/2 and -3/2, same as ABCDE. The ratio of vertical/horizontal sides: ABCDE has AB (length 10), BC (length 8), CD (length 6). UVWXY has UV (length 5), VW (length 4), WX (length 3). So scale factor 0.5 (since 5 = 10*0.5, 4 = 8*0.5, 3 = 6*0.5). So similar to ABCDE.

KLMNO: M(10,10), L(12,10) (horizontal, length 2), L(12,10) to K(12,5) (vertical, length 5), K(12,5) to O(11,7) (run -1, rise 2, slope 2/(-1)= -2), O(11,7) to N(10,7) (horizontal, length 1), N(10,7) to M(10,10) (vertical, length 3). The non - axis - aligned side has slope -2, different from 1/2 and -3/2.

PQRST: P(13,21), Q(17,21) (horizontal, length 4), Q(17,21) to R(17,15) (vertical, length 6), R(17,15) to S(15,19) (run -2, rise 4, slope 4/(-2)= -2? Wait, no, 19 - 15 = 4, 15 - 17 = -2? Wait, S is (15,19), R is (17,15). So run -2, rise 4, slope -2. S(15,19) to T(13,17) (run -2, rise -2, slope (-2)/(-2)=1), T(13,17) to P(13,21) (vertical, length 4). Wait, maybe I messed up coordinates. But let's check FGH IJ: H(17,12), G(25,12) (horizontal, length 8), G(25,12) to F(25,2) (vertical, length 10), F(25,2) to J(19,6) (run -6, rise 4, slope 4/(-6)= -2/3), J(19,6) to I(17,6) (horizontal, length 2), I(17,6) to H(17,12) (vertical, length 6). The non - axis - aligned side has slope -2/3, which is different from the slopes of ABCDE's non - axis - aligned sides (1/2 and -3/2). Wait, no, earlier mistake: ABCDE's E to A: A(0,22), E(4,16): Δx = 4, Δy = -6, slope -6/4 = -3/2. E to D: D(8,18), E(4,16): Δx = -4, Δy = -2, slope (-2)/(-4)=1/2. FGH IJ: F(25,2), J(19,6): Δx = -6, Δy = 4, slope 4/(-6)= -2/3. Which is not equal to -3/2 or 1/2. Also, the other non - axis - aligned side? Wait, FGH IJ: J(19,6) to I(17,6) is horizontal, I(17,6) to H(17,12) is vertical, H(17,12) to G(25,12) is horizontal, G(25,12) to F(25,2) is vertical, F(25,2) to J(19,6) is the only non - axis - aligned side. Wait, no, ABCDE has two non - axis - aligned sides (E to D and E to A). KLMNO: let's re - check. M(10,10), N(10,7) (vertical, length 3), N(10,7) to O(11,8) (Δx = 1, Δy = 1, slope 1), O(11,8) to K(12,5) (Δx = 1, Δy = -3, slope -3), K(12,5) to L(12,10) (vertical, length 5), L(12,10) to M(10,10) (horizontal, length 2). So two non - axis - aligned sides with slopes 1 and -3. PQRST: P(13,21), T(13,17) (vertical, length 4), T(13,17) to S(15,19) (Δx = 2, Δy = 2, slope 1), S(15,19) to R(17,15) (Δx = 2, Δy = -4, slope -2), R(17,15) to Q(17,21) (vertical, length 6), Q(17,21) to P(13,21) (horizontal, length 4). So two non - axis - aligned sides with slopes 1 and -2.

But FGH IJ: Let's count the number of sides. ABCDE: 5 sides (pentagon). UVWXY: 5 sides. KLMNO: 5 sides. PQRST: 5 sides. FGH IJ: F, G, H, I, J: 5 sides? Wait, F(25,2), G(25,12)? No, G is (25,12)? Wait, the graph: H is (17,12), G is (25,12) (horizontal), G(25,12) to F(25,2) (vertical), F(25,2) to J(19,6) (slant), J(19,6) to I(17,6) (horizontal), I(17,6) to H(17,12) (vertical). So that's a pentagon. But the key is the slopes of the non - axis - aligned sides. In ABCDE and UVWXY, the two non - axis - aligned sides have slopes 1/2 and -3/2. In KLMNO, PQRST, the slopes are different. Wait, but the question is which one cannot be mapped. Wait, maybe FGH IJ: let's check the vertical/horizontal side lengths. ABCDE: vertical sides AB (length 10), CD (length 6); horizontal side BC (length 8). FGH IJ: vertical sides IH (length 6), FG (length 10); horizontal side HG (length 8). Wait, that's the same as ABCDE (AB = 10, CD = 6; IH = 10, FG = 6? Wait, no, I(17,6) to H(17,12): length 6. G(25,12) to F(25,2): length 10. H(17,12) to G(25,12): length 8. So same as ABCDE (AB = 10, CD = 6, BC = 8). But the slant side: F(25,2) to J(19,6): Δx = -6, Δy = 4, slope -2/3. ABCDE's E(4,16) to A(0,22): Δx = -4, Δy = 6, slope -3/2. These slopes are negative reciprocals? No, -2/3 and -3/2 multiply to 1, so they are negative reciprocals, meaning the angles are different (since slope is tan(theta), so if theta1 and theta2 have tan(theta1)= -3/2 and tan(theta2)= -2/3, then theta1 + theta2 = 90 degrees, so the angles are different).

Wait, maybe a better approach: similar polygons must have the same number of sides and corresponding angles equal. Let's check the "shape" of the pentagons. ABCDE, UVWXY, KLMNO, PQRST: let's see the number of "bends" and the direction of the slant. Wait, FGH IJ: when we look at the graph, FGH IJ has a slant side that goes from F (bottom right) to J (middle left - ish) with a positive slope (since y increases from F to J), while in ABCDE, the slant side from E to A has a negative slope (y decreases from E to A? No, E is (4,16), A is (0,22), so y increases from E to A. Wait, E(4,16) to A(0,22): x decreases, y increases, so slope negative. F(25,2) to J(19,6): x decreases, y increases, slope negative (4/(-6)= -2/3). Wait, maybe I made a mistake. Let's count the number of sides and the type of angles. Alternatively, notice that FGH IJ has a different "aspect ratio" or the slopes of the non - axis - aligned sides are different. After analyzing, the polygon that cannot be mapped by similarity transformations is FGH IJ (FGHIJ).

Answer:

FGHIJ