Question

in 10 - 13, \\( \triangle qrs\cong\triangle xyz \\). find the measures below.

1. \\( m\angle r\\)
2. \\( xy\\)
3. \\( m\angle x\\)
4. \\( m\angle s\\)

Explanation:

Step1: Use triangle angle - sum property

Since \(\triangle QRS\cong\triangle XYZ\), corresponding angles and sides are equal.
For \(\triangle QRS\), by the angle - sum property of a triangle (\(m\angle Q + m\angle R+m\angle S=180^{\circ}\)), and for \(\triangle XYZ\), \(m\angle X + m\angle Y+m\angle Z = 180^{\circ}\). Also, \(m\angle Q=m\angle X\), \(m\angle S=m\angle Z\), \(m\angle R=m\angle Y\), \(QR = XY\), \(QS=XZ\), \(RS = YZ\).

Step2: Solve for \(m\angle R\) (Question 10)

We know \(m\angle Q = 55^{\circ}\), \(m\angle S=m\angle Z = 50^{\circ}\).
Using \(m\angle Q + m\angle R+m\angle S=180^{\circ}\), substitute \(m\angle Q = 55^{\circ}\) and \(m\angle S = 50^{\circ}\).
\(m\angle R=180-(55 + 50)=75^{\circ}\)

Step3: Solve for \(XY\) (Question 11)

Since \(\triangle QRS\cong\triangle XYZ\), \(QR = XY\). Given \(QR = 6\), so \(XY = 6\)

Step4: Solve for \(m\angle X\) (Question 12)

Since \(\triangle QRS\cong\triangle XYZ\), \(m\angle X=m\angle Q\). Given \(m\angle Q = 55^{\circ}\), so \(m\angle X = 55^{\circ}\)

Step5: Solve for \(m\angle S\) (Question 13)

Since \(\triangle QRS\cong\triangle XYZ\), \(m\angle S=m\angle Z\). Given \(m\angle Z = 50^{\circ}\), so \(m\angle S = 50^{\circ}\)

Answer:

1. \(75\)
2. \(6\)
3. \(55\)
4. \(50\)