Question

1. given the similar triangles ( \triangle pqr-\triangle dfe ), find the following information entered as a reduced fraction, the scale factor ( mangle d= ) ( mangle r= ) ( mangle p= ) ( de= ) ( fe= )

Explanation:

Step1: Find the scale factor

Since \(\triangle PQR\sim\triangle DFE\), the scale factor is \(\frac{PQ}{DF}\). Given \(PQ = 20\) and \(DF=24\), the scale factor is \(\frac{20}{24}=\frac{5}{6}\).

Step2: Find \(m\angle D\)

Corresponding angles of similar triangles are equal. \(\angle D\) corresponds to \(\angle P\). First, find \(\angle R\) using the angle - sum property of a triangle (\(\angle Q = 89^{\circ}\), \(\angle E=35^{\circ}\)). In \(\triangle PQR\), \(\angle R+\angle Q+\angle P = 180^{\circ}\). In \(\triangle DFE\), \(\angle D+\angle F+\angle E=180^{\circ}\). Since \(\angle Q\) corresponds to \(\angle F\) (\(\angle Q = 89^{\circ}\)), \(\angle R\) corresponds to \(\angle E\) (\(\angle R = 35^{\circ}\)). Then \(m\angle D=180-(89 + 35)=56^{\circ}\).

Step3: Find \(m\angle R\)

Corresponding angles of similar triangles: \(\angle R\) and \(\angle E\) are corresponding. So \(m\angle R = 35^{\circ}\).

Step4: Find \(m\angle P\)

Using the angle - sum property of a triangle in \(\triangle PQR\): \(m\angle P=180-(89 + 35)=56^{\circ}\).

Step5: Find \(DE\)

Since \(\frac{PQ}{DF}=\frac{QR}{FE}=\frac{PR}{DE}\), \(\frac{40}{DE}=\frac{5}{6}\). Cross - multiply: \(5DE=40\times6\), \(DE=\frac{40\times6}{5}=48\).

Step6: Find \(FE\)

Using \(\frac{QR}{FE}=\frac{5}{6}\), \(QR = 36\). Cross - multiply: \(5FE=36\times6\), \(FE=\frac{36\times6}{5}=\frac{216}{5} = 43.2\).

Answer:

Scale factor: \(\frac{5}{6}\)
\(m\angle D = 56^{\circ}\)
\(m\angle R=35^{\circ}\)
\(m\angle P = 56^{\circ}\)
\(DE = 48\)
\(FE = 43.2\)