Question

1. what values of y and z make △hij ≅ △pqr? y = z =

Explanation:

Step1: Use the property of congruent triangles

Since \(\triangle HIJ\cong\triangle PQR\), their corresponding sides are equal. So, \(20y - 16=16y\) (corresponding sides \(IJ\) and \(QR\)) and \(11z - 49 = 4z\) (corresponding sides \(HJ\) and \(PR\)).

Step2: Solve for \(y\)

For \(20y - 16=16y\), subtract \(16y\) from both sides: \(20y-16y-16 = 16y - 16y\), which gives \(4y-16=0\). Then add \(16\) to both sides: \(4y=16\). Divide both sides by \(4\): \(y=\frac{16}{4}=4\).

Step3: Solve for \(z\)

For \(11z - 49 = 4z\), subtract \(4z\) from both sides: \(11z-4z-49=4z - 4z\), which gives \(7z-49 = 0\). Then add \(49\) to both sides: \(7z=49\). Divide both sides by \(7\): \(z=\frac{49}{7}=7\).

Answer:

\(y = 4\), \(z = 7\)