Question

what values of s and t make δefg ≅ δyzx?
(there are two right triangles, δefg with right angle at g, side eg labeled 7s, side fg labeled t+50; δyzx with right angle at x, side yz labeled s+48, side zx labeled 20t+12)
s =
t =

Explanation:

Step1: Set equal sides for congruent triangles

Since $\triangle EFG \cong \triangle YZX$, their corresponding sides are equal. So, we have two equations:

1. $7s = 20t + 12$ (corresponding legs)
2. $t + 50 = s + 48$ (corresponding hypotenuses)

Step2: Simplify the second equation

From $t + 50 = s + 48$, we can solve for $s$:
$s = t + 50 - 48 = t + 2$

Step3: Substitute s into the first equation

Substitute $s = t + 2$ into $7s = 20t + 12$:
$7(t + 2) = 20t + 12$
Expand the left side: $7t + 14 = 20t + 12$

Step4: Solve for t

Subtract $7t$ from both sides: $14 = 13t + 12$
Subtract 12 from both sides: $2 = 13t$? Wait, no, wait, let's check again. Wait, $7(t + 2) = 20t + 12$:
$7t + 14 = 20t + 12$
Subtract $7t$: $14 = 13t + 12$
Subtract 12: $2 = 13t$? That can't be right. Wait, maybe I mixed up the corresponding sides. Let's re - examine the triangles.

Wait, $\triangle EFG$ has right angle at $G$, legs $7s$ and $t + 50$, hypotenuse $EF$ (not labeled). $\triangle YZX$ has right angle at $X$, legs $s + 48$ and $20t + 12$, hypotenuse $YZ$ (not labeled). Since they are congruent, the legs should correspond. So maybe $7s=s + 48$ and $t + 50=20t + 12$. Let's try that.

Step5: Correct the correspondence

Let's assume the legs correspond as $7s$ and $s + 48$ (both legs), and the other legs $t + 50$ and $20t + 12$.

First equation: $7s=s + 48$
Subtract $s$ from both sides: $6s = 48$
Divide by 6: $s = 8$

Second equation: $t + 50=20t + 12$
Subtract $t$: $50 = 19t + 12$
Subtract 12: $38 = 19t$
Divide by 19: $t = 2$

Let's check with $s = 8$ and $t = 2$:

For the first triangle, leg $7s=7\times8 = 56$, other leg $t + 50=2 + 50 = 52$? Wait, no, that doesn't match. Wait, maybe the hypotenuse and leg. Wait, maybe $7s$ corresponds to $s + 48$ and $t + 50$ corresponds to $20t + 12$. Wait, when $s = 8$, $7s=56$, $s + 48=56$, so that's good. Then $t + 50=20t + 12$: $t + 50=20t + 12$ → $50 - 12=20t - t$ → $38 = 19t$ → $t = 2$. Then $t + 50=2 + 50 = 52$, $20t+12=20\times2 + 12=52$. Yes, that works.

So the correct correspondence is leg $7s$ with leg $s + 48$, and leg $t + 50$ with leg $20t + 12$.

Answer:

$s = 8$, $t = 2$