Question

are the polygons below similar?

(image of two polygons: left is a quadrilateral f g h i with sides fg=4, fi=6, hi=4, gh=2; right is a quadrilateral x y z w with sides xy=12, xw=16, wz=28, yz=16. corresponding angles are marked with red arcs. below the image are two options: yes, no)

Explanation:

Step1: Check Corresponding Angles

The marked angles (red arcs) indicate that corresponding angles are equal (since same number of arcs mean equal angles). So angle - angle similarity condition's angle part is satisfied.

Step2: Check Corresponding Sides Ratios

For the first quadrilateral (FGHI) and the second (WXYZ, need to match sides correctly). Let's match the sides:

• FG = 4, WX = 16? Wait, no, let's list sides:
• FG = 4, FI = 6, HI = 4, GH = 2.
• For the larger quadrilateral: XY = 12, YZ = 16, ZW = 28, WX = 16? Wait, no, let's pair corresponding sides. Let's assume the order of vertices: F - W, G - Y, H - Z, I - X? Wait, no, better to take the sides with same angle markings.
• Let's take the sides adjacent to equal angles. For the small quadrilateral: GH = 2, FG = 4, HI = 4, FI = 6.
• For the large quadrilateral: XY = 12, WX = 16, ZW = 28, YZ = 16. Wait, let's find the ratios of corresponding sides.
• Let's check the ratio of GH (2) to XY (12): $\frac{2}{12}=\frac{1}{6}$
• FG (4) to WX (16): $\frac{4}{16}=\frac{1}{4}$
• HI (4) to YZ (16): $\frac{4}{16}=\frac{1}{4}$
• FI (6) to ZW (28): $\frac{6}{28}=\frac{3}{14}$

Wait, that can't be. Wait, maybe I paired the sides wrong. Let's re - examine the angle markings. The small quadrilateral has angles at G, H, F, I. The large one at Y, Z, W, X. Let's match the sides with the same number of arcs. The angle at F (two arcs) and angle at W (two arcs) should be corresponding. Angle at G (one arc) and angle at Y (one arc) corresponding. Angle at H (one arc) and angle at Z (one arc) corresponding. Angle at I (two arcs) and angle at X (two arcs) corresponding? Wait, no, the small quadrilateral: G (1 arc), H (1 arc), F (2 arcs), I (2 arcs). The large quadrilateral: Y (1 arc), Z (1 arc), W (2 arcs), X (2 arcs). So corresponding vertices: G - Y, H - Z, F - W, I - X.
So sides:

• GH (2) corresponds to YZ (16): $\frac{2}{16}=\frac{1}{8}$? No, wait YZ is 16? Wait, the large quadrilateral: XY = 12, YZ = 16, ZW = 28, WX = 16. Wait, GH is 2, YZ is 16: $\frac{2}{16}=\frac{1}{8}$

FG is 4, WX is 16: $\frac{4}{16}=\frac{1}{4}$
HI is 4, ZW is 28: $\frac{4}{28}=\frac{1}{7}$
FI is 6, XY is 12: $\frac{6}{12}=\frac{1}{2}$
Wait, this is inconsistent. Wait, maybe I made a mistake in side pairing. Let's look again. The small quadrilateral: sides are 2, 4, 4, 6. The large one: 12, 16, 16, 28. Wait, let's check if the sides are in proportion. Let's group the sides: 2 and 12, 4 and 16, 4 and 16, 6 and 28.

• $\frac{2}{12}=\frac{1}{6}$, $\frac{4}{16}=\frac{1}{4}$, $\frac{4}{16}=\frac{1}{4}$, $\frac{6}{28}=\frac{3}{14}$. These ratios are not equal. Wait, but the angle markings: if the angles are equal (since same number of arcs), but the side ratios are not equal, then the polygons are not similar. Wait, but maybe I paired the sides wrong. Wait, maybe the small quadrilateral is a quadrilateral with sides 2, 4, 4, 6 and the large one with sides 12, 16, 16, 28. Let's check the ratios of consecutive sides. For the small one: $\frac{4}{2} = 2$, $\frac{4}{4}=1$, $\frac{6}{4}=1.5$. For the large one: $\frac{16}{12}=\frac{4}{3}\approx1.33$, $\frac{16}{16} = 1$, $\frac{28}{16}=1.75$. The ratios of consecutive sides are not equal, and the ratios of corresponding sides (when paired by angle) are not equal. So the polygons are not similar.

Answer:

no