Question

what values of y and z make \\( \triangle cde \cong \triangle kij \\)?
(there are two right triangles, one labeled cde with right angle at e, sides: ce = 2z, cd = y + 44; the other labeled kij with right angle at j, sides: ki = 12y, ji = z + 10. there are input boxes for y = \\( \square \\) and z = \\( \square \\))

Explanation:

Step1: Identify Corresponding Sides

Since \(\triangle CDE \cong \triangle KIJ\), their corresponding sides are equal. So, \(CE = KJ\) and \(CD = KI\). This gives the equations \(2z = z + 10\) and \(y + 44 = 12y\).

Step2: Solve for \(z\)

From \(2z = z + 10\), subtract \(z\) from both sides:
\(2z - z = z + 10 - z\)
\(z = 10\)

Step3: Solve for \(y\)

From \(y + 44 = 12y\), subtract \(y\) from both sides:
\(y + 44 - y = 12y - y\)
\(44 = 11y\)
Divide both sides by 11:
\(y = \frac{44}{11} = 4\)

Answer:

\(y = 4\), \(z = 10\)