Question

find the perimeter of \\( \triangle qrs \\). round your answer to the nearest tenth if necessary. figures are not necessarily drawn to scale.

Explanation:

Step1: Determine similarity of triangles

Since \(\angle P = 41^{\circ}\), \(\angle O=\angle R = 63^{\circ}\), in \(\triangle NOP\), \(\angle N=180^{\circ}-41^{\circ}-63^{\circ}=76^{\circ}\). In \(\triangle QRS\), \(\angle Q = 76^{\circ}\). So, \(\triangle NOP\sim\triangle QRS\) (by \(AAA\) similarity criterion).

Step2: Find the scale factor

The sides of \(\triangle NOP\) are \(NO = 16\), \(OP = 24\), \(PN=22\). The sides of \(\triangle QRS\) are \(QR = 60\), \(RS = 40\), \(SQ=x\). The scale factor \(k\) from \(\triangle NOP\) to \(\triangle QRS\):
For the sides \(OP\) and \(QR\), \(k=\frac{QR}{OP}=\frac{60}{24}=\frac{5}{2}\)

Step3: Find the lengths of the other sides

For side \(NO\) and \(SQ\): \(x=\frac{5}{2}\times16 = 40\)
For side \(PN\) and \(RS\): Let's check with another pair. For \(NO\) and \(RS\), if we use the scale - factor. The perimeter of \(\triangle NOP\) is \(P_{NOP}=16 + 22+24=62\)
Since the scale factor \(k = \frac{5}{2}\), the perimeter of \(\triangle QRS\) is \(P_{QRS}=k\times P_{NOP}\)
\(P_{QRS}=\frac{5}{2}\times62\)
\(P_{QRS}=155\)

Answer:

The perimeter of \(\triangle QRS\) is \(155\)