Question

hw 4: congruent triangles
given $\triangle xps \cong \triangle dnf$, find the values of $x$ and $y$.
(image of two triangles: $\triangle xps$ with sides $4y - 3$ (xp), $17x + 3$ (xs), and $\triangle dnf$ with sides 51 (fn), 57 (nd), 54 (fd))
$x = \square$
$y = \square$

Explanation:

Step1: Find x using congruent sides

Since \( \triangle XPS \cong \triangle DNF \), corresponding sides are equal. So \( XS = ND \). Thus, \( 17x + 3 = 57 \).
Subtract 3: \( 17x = 57 - 3 = 54 \).
Divide by 17: Wait, 54? Wait, 57 - 3 is 54? Wait, no, 57 - 3 is 54? Wait, 17x + 3 = 57. Subtract 3: 17x = 54? Wait, no, 57 - 3 is 54? Wait, 57 - 3 is 54? Wait, 17x = 54? Wait, no, maybe I made a mistake. Wait, \( XS \) corresponds to \( ND \), which is 57. So \( 17x + 3 = 57 \). Subtract 3: \( 17x = 54 \)? Wait, no, 57 - 3 is 54? Wait, 57 - 3 is 54? Wait, 17x = 54? Then x = 54/17? That can't be. Wait, maybe I mixed up the correspondence. Wait, \( \triangle XPS \cong \triangle DNF \), so the order is X-P-S and D-N-F. So X corresponds to D, P to N, S to F? Wait, no, congruence notation: \( \triangle XPS \cong \triangle DNF \) means X ↔ D, P ↔ N, S ↔ F. So side XS (between X and S) corresponds to side DF? Wait, no, maybe the sides: X to P is 4y - 3, P to S, X to S is 17x + 3. In \( \triangle DNF \), D to N is 57? Wait, no, the triangle DNF has sides: N to F is 51, F to D is 54, N to D is 57. So \( \triangle XPS \cong \triangle DNF \), so X corresponds to D, P to N, S to F. So side XP (X to P) is 4y - 3, which should correspond to side DN? No, XP (X to P) corresponds to DN? Wait, no, X to P is a side, P to S, X to S. In \( \triangle DNF \), D to N is 57? Wait, no, N to D is 57, N to F is 51, F to D is 54. So \( \triangle XPS \): X-P: 4y - 3, P-S:?, X-S: 17x + 3. \( \triangle DNF \): D-N: 57, N-F: 51, F-D: 54. So if \( \triangle XPS \cong \triangle DNF \), then XS (X to S) corresponds to DF (D to F), which is 54? Wait, no, maybe I got the correspondence wrong. Wait, maybe X corresponds to D, P to F, S to N? Wait, the congruence is \( \triangle XPS \cong \triangle DNF \), so the vertices are X, P, S and D, N, F. So the order is X-P-S ≅ D-N-F. So X ↔ D, P ↔ N, S ↔ F. Therefore, side XP (X to P) ≅ side DN (D to N), side PS ≅ side NF, side XS ≅ side DF. Wait, XP is 4y - 3, DN is 57? No, DN is 57? Wait, in \( \triangle DNF \), DN is 57 (N to D), NF is 51 (N to F), FD is 54 (F to D). So XP (X to P) is 4y - 3, which should be equal to DN (D to N) = 57? Wait, no, 4y - 3 = 51? Because NF is 51, which is P to S? Wait, maybe XP corresponds to NF (51). So 4y - 3 = 51. Then 4y = 54, y = 54/4 = 13.5? No, 51 + 3 = 54, 54/4 = 13.5. And XS (17x + 3) corresponds to FD (54). So 17x + 3 = 54. Then 17x = 51, so x = 3. Ah, that makes sense. So I had the correspondence wrong. So \( \triangle XPS \cong \triangle DNF \), so X ↔ D, P ↔ N, S ↔ F. So side XP (X to P) is 4y - 3, which corresponds to side NF (N to F) = 51. So 4y - 3 = 51. Side XS (X to S) is 17x + 3, which corresponds to side FD (F to D) = 54. So 17x + 3 = 54. Let's redo:

Step1: Find x (XS = FD)

\( 17x + 3 = 54 \)
Subtract 3: \( 17x = 54 - 3 = 51 \)
Divide by 17: \( x = \frac{51}{17} = 3 \)

Step2: Find y (XP = NF)

\( 4y - 3 = 51 \)
Add 3: \( 4y = 51 + 3 = 54 \)
Divide by 4: Wait, 54/4? No, 51 + 3 is 54? Wait, 51 + 3 is 54? Then 4y = 54? y = 54/4 = 13.5? Wait, no, NF is 51, so XP should be 51. So 4y - 3 = 51. Then 4y = 54? That's 13.5. But maybe the correspondence is XP to DN? Wait, DN is 57? No, DN is 57? Wait, the triangle DNF: N to D is 57, N to F is 51, F to D is 54. So if \( \triangle XPS \cong \triangle DNF \), then X to P is 4y - 3, P to S, X to S is 17x + 3. So X to S (17x + 3) should equal D to F (54), so 17x + 3 = 54 → x = 3. X to P (4y - 3) should equal D to N (57)? No, D to N is 57? Wait, D to N is 57, N to F is 51, F to D is 54. So if X corresponds to D, P to N, S to F, then X to P (XP) is D to N (DN) = 57? So 4y - 3 = 57 → 4y = 60 → y = 15. Wait, now I'm confused. Wait, let's check the congruence notation again. \( \triangle XPS \cong \triangle DNF \): the order of the letters is X, P, S and D, N, F. So the sides:

- X to P: first two letters (X, P)
- P to S: middle and last (P, S)
- X to S: first and last (X, S)

In \( \triangle DNF \):

- D to N: first two letters (D, N)
- N to F: middle and last (N, F)
- D to F: first and last (D, F)

Therefore, by congruence, corresponding sides are equal:

- XP (X-P) = DN (D-N)
- PS (P-S) = NF (N-F)
- XS (X-S) = DF (D-F)

Ah! That's the key. So XP = DN, PS = NF, XS = DF.

So DN is 57 (D to N), NF is 51 (N to F), DF is 54 (D to F).

So:

XP = DN → 4y - 3 = 57

XS = DF → 17x + 3 = 54

Now solve for x and y:

Step1: Solve for x (XS = DF)

\( 17x + 3 = 54 \)
Subtract 3: \( 17x = 51 \)
Divide by 17: \( x = 3 \)

Step2: Solve for y (XP = DN)

\( 4y - 3 = 57 \)
Add 3: \( 4y = 60 \)
Divide by 4: \( y = 15 \)

Answer:

x = 3, y = 15