Question

test sss, sas, asa, aas, and hl

1. what values of y and z make \\( \triangle ijk \cong \triangle hfg \\)?

\\( y = \\)
\\( z = \\)

Explanation:

Step1: Set up equations based on congruent triangles

Since \(\triangle IJK\cong\triangle HFG\), corresponding sides are equal.
We have \(IJ = HF\) and \(IK = HG\).
Given \(IJ = 16\), \(HF=y + 2\), so \(y+2=16\).
Also, \(IK=y-2z + 50\) and \(HG=2y+z\).

Step2: Solve for \(y\) from the first - equation

From \(y + 2=16\), we subtract 2 from both sides:
\(y=16 - 2=14\).

Step3: Substitute \(y = 14\) into the second - equation

Substitute \(y = 14\) into \(y-2z + 50=2y+z\).
We get \(14-2z + 50=2\times14+z\).
Simplify the left - hand side: \(64-2z\).
Simplify the right - hand side: \(28+z\).
So, \(64-2z=28+z\).

Step4: Solve for \(z\)

Add \(2z\) to both sides: \(64=28 + 3z\).
Subtract 28 from both sides: \(3z=64 - 28=36\).
Divide both sides by 3: \(z = 12\).

Answer:

\(y = 14\), \(z = 12\)