Question

what values of t and u make $\triangle efg \cong \triangle xzy$?
(image of two right triangles: $\triangle efg$ with right angle at e, leg ge = 23, hypotenuse fg = 3t + u + 7; $\triangle xzy$ with right angle at x, leg xy = u + 22, leg zy = 5u + t + 35)

Explanation:

Step1: Identify corresponding sides

Since \(\triangle EFG \cong \triangle XZY\), their corresponding sides are equal. The right - angled sides: \(EG = XY\) and \(EF = XZ\) (wait, no, let's re - identify. In \(\triangle EFG\), right angle at \(E\), so legs are \(EG = 23\) and \(EF\) (unknown), hypotenuse \(FG=3t + u+7\). In \(\triangle XZY\), right angle at \(X\), so legs are \(XY = u + 22\) and \(XZ\) (unknown), hypotenuse \(ZY=5u + t+35\). Since the triangles are congruent, the corresponding legs and hypotenuses are equal. So we have two equations:

1. \(EG=XY\): \(23=u + 22\)
2. \(FG = ZY\): \(3t+u + 7=5u + t+35\)

Step2: Solve for \(u\) from the first equation

From \(23=u + 22\), subtract 22 from both sides:
\(u=23 - 22=1\)

Step3: Substitute \(u = 1\) into the second equation and solve for \(t\)

Substitute \(u = 1\) into \(3t+u + 7=5u + t+35\):
\(3t+1 + 7=5\times1+t + 35\)
Simplify left side: \(3t+8\), right side: \(5 + t+35=t + 40\)
So the equation becomes \(3t+8=t + 40\)
Subtract \(t\) from both sides: \(3t - t+8=t - t + 40\), \(2t+8 = 40\)
Subtract 8 from both sides: \(2t=40 - 8=32\)
Divide both sides by 2: \(t=\frac{32}{2}=16\)

Answer:

\(t = 16\), \(u = 1\)